CONTENTS
4 Polynomials
4.1 Adding & Subtracting Polynomials
4.2 Multiply Monomials & Binomials .
4.3 Multiplying Famous Polynomials .
4.4 Dividing by a Monomial . . . . . .
4.5 Dividing by Polynomials ii . . . . .
4.6 Factor Polynomials by DL . . . . .
4.7 Factor By Grouping . . . . . . . .
4.8 Factor by Splitting . . . . . . . . .
4.9 Factor Famous Polynomials . . . .
4.10 Chapter Review . . . . . . . . . . .
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1
2
9
14
21
29
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49
57
Bibliography
84
Index
85
1
2
CONTENTS
CHAPTER 4
POLYNOMIALS
We concluded chapter 3 with a modest discussion of solving linear equations, a very important milestone. In this chapter, we expand on that or
more accurately we begin sharpening our tools so that we can expand on
that, and solve not just linear equations, but many other types of equations
as well. Roughly speaking, this involves being comfortable when working
with ’polynomials’. Recall, chapter 2 was essentially a chapter on integers,
that is positive and negative numbers, and how to add, subtract, and multiply them. In chapter 3 we did the similar type of work but for fractions.
A good way to think about this chapter is that polynomials are to this
chapter, what rationals were to chapter 3, or what integers were to chapter
2. Once we get comfortable with polynomials, we will begin to solve much
more interesting equations, than the [linear] ones we solved at the end of
chapter 3.
1
2
CHAPTER 4. POLYNOMIALS
4.1
Adding & Subtracting Polynomials
Let us begin by defining a ’monomial’.
Definition A monomial in x with real coefficient is defined as an expression
that can be written as
Axn
Where A represents some constant real number, x is a variable, and n is a
whole number. While x is called the variable, n is called the degree of the
term, and A is called the coefficient. Another word for a monomial is a
term.
We could very easily make variations of this definition. For example:
Definition A monomial in x with integer coefficient is defined as an expression that can be written as
Axn
Where A represents some integer number, x is a variable, and n is a whole
number.
The sum or difference of two monomials is called a binomial , for example:
3x2 + 5x
The sum of difference of three monomials is called a trinomial. Generally,
we define,
Definition A polynomial is defined as the sum or difference of a collection
of one or more monomials.
For example,
x5 + 2x3 − 5
is a polynomial. Often, it is convenient to name the polynomials. In the
name we may often include some information about the polynomial, such
as what is the variable. For example we could name the x5 + 2x3 − 5 as
simply p(x), meaning its name is p and the variable is identified as x. Thus
we could write
p(x) = x5 + 2x3 − 5
Subsequently, we could just refer to x5 + 2x3 − 5 as simply p(x), under
the agreement that these are interchangeable. Meanwhile we could have a
different polynomial with a different name, such as
h(x) = x5 + x + 2
4.1. ADDING & SUBTRACTING POLYNOMIALS
3
Then when we write p(x) everyone should concur we are referring to x5 +
2x3 − 5 while if we mention h(x) everyone should think of x5 + x + 2
If we restrict the possible coefficients to only integers, and the possible
variables to only x’s, then we can call then entire set of such polynomials
Z[x]. Analogously, the set Q[t] represents all polynomials with only rational
coefficients and only t variables, while R[x] represents the set of all polynomials in x with
√ real coefficients. Moreover, we often abbreviate, instead of
”3x2 + 5x + 3 is a polynomial with real coefficients” we write:
3x2 + 5x +
√
3 ∈ R[x]
where the symbol ∈ is read as ’is an element of..’. Thus.
3x2 + 5x +
can be read as ”3x2 + 5x +
√
√
3 ∈ R[x]
3 is an element of the set of real polynomials”
Definition If q(x) is a polynomial, we define the degree of polynomial q(x)
as the highest of the degrees of any of its terms. If q(x) is a constant we say
that its degree is zero.
Examples
4x5 + 82x + 9
(degree for this polynomial is 5)
4x9 + 27x + 9
(degree for this polynomial is 9)
5x3 + 9 + 27
(degree for this polynomial is 3)
12
(degree for this polynomial is 0)
12 + x2 + 5x
(degree for this polynomial is 2)
Now that we have some of the basic vocabulary tackled, we are ready to
start adding/subtracting polynomials.
Example Add
(3x2 + 4x + 1) + (x2 − 4x + 5)
4
CHAPTER 4. POLYNOMIALS
solution:
(3x2 + 4x + 1) + (x2 − 4x + 5)
= (3x2 + 1x2 ) + (4x + −4x) + (1 + 5)
2
= (3 + 1)x + (4 + −4)x + (1 + 5)
(ALA, CoLA)
(DL)
2
(BI)
2
(0MT)
2
(AId)
= 4x + 0x + 6
= 4x + 0 + 6
= 4x + 6
It should be noted that based on the above example, it should be clear
that the underlying principle in adding polynomials is to use ALA and CoLA
to group associate like terms together, and like terms for us will mean same
degree. That is, we collect degree two terms together, degree one terms
together and so on. Then we can use DL to factor out the variable in effect
just adding the coefficients. This should be reminiscent of adding natural
numbers as you did in grade-school where you collect units, collect tens, and
collect 100’s more or less. Such as
+
4825
5307
10132
Based on this idea and the the above comments, we can alternatively add
polynomials the ’Kindergarden’ [KG] way such as:
3x2 + 4x + 1
+
x2 − 4x + 5
− − − − − − − − −−
4x2 + 6
Example Subtract
(3x2 + 4x + 1) − (x2 − 4x + 5)
(KG)
5
4.1. ADDING & SUBTRACTING POLYNOMIALS
solution:
(3x2 + 4x + 1) − (x2 − 4x + 5)
= (3x2 + 4x + 1) + −1 · (x2 + −4x + 5)
2
(def a-b)
2
= (3x + 4x + 1) + −1x + −1 · −4x + −1 · 5
2
(DL)
2
= 3x + 4x + 1 + −1x + 4x + −5
2
2
= (3x + −1x ) + (4x + 4x) + (1 + −5)
(DL)
(ALA, CoLA)
2
= (3 + −1)x + (4 + 4)x + (1 + −5)
2
= 2x + 8x + −4
(DL)
(BI)
OR we can try the KG method. .
3x2 + 4x + 1
−
x2 − 4x + 5
− − − − − − − − −−
2x2 + 8x − 4
4.1.1
Exercises
1. Compute and simplify:
1 − r4 + 7r6 + 2r5 + 6r4 + 5r3 + 3r2 + 2r + 4
2. Compute and simplify:
2q 3 + 4q + 2
+ 2q 4 − 4q 3 + q 2 + 7q
3. Compute and simplify:
4w6 + 6w5 + 7w4 + w2 − 5w + 3
−
−w6 + 5w5 + 5w3 + 5w − 4
(KG)
6
CHAPTER 4. POLYNOMIALS
4. Compute and simplify:
−
−3
(−5y − 5)
5. Compute and simplify:
−3w2 − 2w − 1 + 3w4 + 3w3 − 2w2 + 6w + 3
6. Compute and simplify:
4r3 + 2r2 + 4 + 3r3 − 4r2 − 2
7. Compute and simplify:
−
− 3r6 + 7r3 − 5r2 + 7r
−4r5 − 5r4 + r
8. Compute and simplify:
2t3 + 7t + 4 + 6t3 − 2t2 + 3t
9. Compute and simplify:
−
− y 4 + y 3 + 2y 2 − 4y + 5
3y 5 − y 4 + 7y 3 + 7
10. Compute and simplify:
7r6 − 5r5 + 5r4 − r2 + 5r − 4
−
r6 + 6r3 − 5r
11. Compute and simplify:
3r5 + 7r4 + 5r2 + 2
+
− 3r5 + 3r4 + r3 + r2 + 7r + 5
4.1. ADDING & SUBTRACTING POLYNOMIALS
12. Compute and simplify:
(3q − 5) + (6q − 2)
13. Compute and simplify:
−
− 4v 4 − 3v 3 − 3v 2 − 5v
−3v 4 − v 3 + 5v
14. Compute and simplify:
−
4q 6 − 5q 3 + 4q − 5
−2q 6 − q 5 + 7q 3 + 2
15. Compute and simplify:
(0) + (3t)
16. Compute and simplify:
−4z 2 − 5z + −2z 5 − z 4 − 4z 3 + 5z 2 − z + 7
17. Compute and simplify:
5v 2 − 3
+ 4v 4 + 6v 3 − 2v 2 − 3v + 7
18. Compute and simplify:
2y 2 + 3y + 2
+ 6y 3 + 6y 2 − y
19. Compute and simplify:
1
−
(w + 2)
7
8
CHAPTER 4. POLYNOMIALS
20. Compute and simplify:
7v 2 + 7v
+
6v 2
9
4.2. MULTIPLY MONOMIALS & BINOMIALS
4.2
Multiply Monomials & Binomials
At the heart of multiplication of polynomials is the distributive law. We
begin with the easiest case, namely a multiplication of a monomial times a
polynomial.
Example Multiply
(3x3 )(5x2 + 4x − 1)
solution:
(3x3 )(5x2 + 4x − 1)
= (3x3 )(5x2 + 4x + −1)
3
2
3
(def a-b)
3
= (3x )5x + (3x )4x + (3x )(−1)
3 2
3
(DL)
3
= (3 · 5)x x + (3 · 4)x x + (3 · −1)x
5
4
3
= 15x + 12x + −3x
(ALM, CoLM)
(BI)
Example Multiply
(−2x)(5x2 + 4x − 1)
solution:
(−2x)(5x2 + 4x − 1)
= (−2x)(5x2 + 4x + −1)
2
= (−2x)5x + (−2x)4x + (−2x)(−1)
2
= (−2 · 5)x · x + (−2 · 4)x · x + (−2 · −1)x
3
2
= −10x + −8x + 2x
Example Multiply
(−2x + 1)(5x2 + 4x − 1)
(def a-b)
(DL)
(ALM, CoLM)
(BI)
10
CHAPTER 4. POLYNOMIALS
solution:
(−2x + 1)(5x2 + 4x − 1)
= (−2x + 1)(5x2 + 4x + −1)
(def a-b)
2
= (−2x + 1)5x + (−2x + 1)4x + (−2x + 1)(−1)
2
(DL)
2
= (−2x)5x + (1)5x + (−2x)4x + (1)4x + (−2x)(−1) + (1)(−1)
3
2
= −10x + 5x + −8x + 4x + 2x + −1
3
(DL)
2
2
(DL)
2
= −10x + (5x + −8x ) + (4x + 2x) + −1
3
(ALA)
2
= −10x + (5 + −8)x + (4 + 2)x + −1
3
(DL)
2
= −10x + −3x + 6x + −1
(BI)
Alternative solution: Multiplying using the KinderGarden Method:
(−2x + 1)(5x2 + 4x − 1) =
+
5x2 + 4x + −1
×(−2x + 1)
5x2 + 4x + −1
− 10x3 − 8x2 + 2x
−10x3 + −3x2 + 6x + −1
4.2.1
Exercises
1. Compute and simplify using KG:
u4 + 6u2 + 3u − 5
×
7u2 − 5u + 5
2. Compute and simplify using KG:
−5w5 − 4w3 + 2w
×
6w6
(KG M)
11
4.2. MULTIPLY MONOMIALS & BINOMIALS
3. Compute and simplify use DL:
−2z 7 + 3z 6 + 7z
−2z 4
4. Compute and simplify using KG:
y 4 + 2y 3 + 7y 2 + 5y
5y 2 + 4y − 2
×
5. Compute and simplify using KG:
w6 − 5w4 + 7w3
×
7w6
6. Compute and simplify using KG:
4z 3 − 4z 2 − 5z + 7
×
3z 2 − 2z − 4
7. Compute and simplify using KG:
−w3 + 5w2 − 3w − 4
×
4 − 2w
8. Compute and simplify using KG:
−4v 5 + 6v 3 + 3v 2
×
3v 5
12
CHAPTER 4. POLYNOMIALS
9. Compute and simplify using KG:
3u4 + 4u2 − 2u − 1
u2 − 5u − 2
×
10. Compute and simplify use DL:
t2 + 5t − 5
t5 − 2t2
11. Compute and simplify using KG:
−4t3 − 5t2 − t − 2
×
6t2 + 3t
12. Compute and simplify using KG:
−3r3 + 7r2 + 1
7r3 − 5r2
×
13. Compute and simplify using KG:
−2z 2 + z + 4
×
2z 3 − 2z
14. Compute and simplify using KG:
4r5 + r3 − 5r
×
−r
13
4.2. MULTIPLY MONOMIALS & BINOMIALS
15. Compute and simplify using KG:
5w4 + 7w3 + 6w + 3
×
− 4w2 + 4w + 3
16. Compute and simplify use DL:
−y 5 − 5y 3 − 3y 2 (5y)
17. Compute and simplify use DL:
6z 5 + 4z 3 − 1
5z 7
18. Compute and simplify using KG:
6z 3 − 4z 2 − 5z
6z 3 − 2z
×
19. Compute and simplify use DL:
3y 2 + y − 4
−3y 3 − 2y
20. Compute and simplify using KG:
6y 4 + y + 4
×
− 4y 2
14
4.3
CHAPTER 4. POLYNOMIALS
Multiplying Famous Polynomials
In the previous section we introduced the kindergarten method for multiplying polynomials. In this section, we introduce a list of very famous
polynomials. In particular, we want to prove, understand, and own every
one of these famous products of polynomials. We will see these again, and it
will be essential that we recognize them when seen forwards and backwards.
Moreover, we introduce a very special kind of multiplication, namely the
type when we multiply a binomial times another binomial. For this type
of multiplication we introduce, as an option, the FOIL method. As we
shall soon see, the FOIL method is simply an abbreviated version of a the
distributive law applied a couple times.
Formally, and without further ado:
Theorem 4.3.1. FOIL
Suppose a, b, c, d are numbers or variables for which our axioms apply,
then:
(a + b)(c + d) = ac + ad + bc + bd
[FOIL]
The name is intended to help one remember the four terms on the right.
First we have the first terms from each of the binomials on the left, the a
and c are each the First terms thus the F , the the Outter, the Inner, and
the Last terms, thus the FOIL acronym.
Outer
First
(a + b)(c +
d)
Inner
Last
F
O
I
L
z}|{ z}|{ z}|{ z}|{
= ac + ad + bc + bd
Proof.
(a + b)(c + d) = a(c + d) + b(c + d)
= a·c+a·d+b·c+b·d
(DL)
(DL)
15
4.3. MULTIPLYING FAMOUS POLYNOMIALS
Example
(3x + 2)(4x + 1) = (3x)(4x) + (3x)(1) + (2)(4x) + (2)(1)
(FOIL)
2
(BI)
2
(BI)
= 12x + 3x + 8x + 2
= 12x + 11x + 2
One more time...
Example
(3x + 2)(5x2 + 7x) = (3x)(5x2 ) + (3x)(7x) + (2)(5x2 ) + (2)(7x) (FOIL)
= 15x3 + 21x2 + 10x2 + 14x
3
(BI)
2
= 15x + 31x + 14x
(BI)
It should be noted that the FOIL method is an option for the case when
we are multiplying binomial times a binomial. It becomes less relevant when
we are multiplying trinomials or monomials. Meanwhile, the methods from
the last section, DL and KG are good for all multiplication of all polynomials
from out class.
Next we turn our attention to some very famous polynomial products.
We will include here some of the proofs, while some of these and leave the
others as important exercises. For all of these, we assume our variables are
ones for which our axioms apply.
Theorem 4.3.2. Difference of Squares [DS]
(a − b)(a + b) = a2 − b2
Theorem 4.3.3. Difference of two Cubes [DC]
(a − b)(a2 + ab + b2 ) = a3 − b3
Proof.
(a − b)(a2 + ab + b2 ) =
+
a2 + ab + b2
(KG M)
×(a + −b)
−a2 b + −ab2 + −b3
a3 + a2 b
a3 +
= a3 − b 3
0+
+ ab2
0+
−b3
(BI)
16
CHAPTER 4. POLYNOMIALS
Theorem 4.3.4. Difference of two Cubes [SC]
(a + b)(a2 − ab + b2 ) = a3 + b3
To prove SC and each of the other polynomials here one may proceed by
using DL or KG. While we do a couple of these here, it is most important
and beneficial for these modest challenges to be taken on by the reader.
Most of these appear as important exercises.
Theorem 4.3.5. Pascal Polynomials
• (a + b)2 = a2 + 2ab + b2
3
3
2
[PP#2]
2
• (a + b) = a + 3a b + 3ab + b
3
[PP#3]
• (a + b)4 = a4 + 4a3 b + 6a2 b3 + 4ab4 + b4
[PP#4]
• (a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
[PP#5]
Theorem 4.3.6. Geometric Series Polynomials
• (x − 1)(x + 1) = x2 − 1
[GS#2]
• (x − 1)(x2 + x + 1) = x3 − 1
3
2
[GS#3]
4
• (x − 1)(x + x + x + 1) = x − 1
[GS#4]
• (x − 1)(x4 + x3 + x2 + x + 1) = x5 − 1
[GS#5]
As an example we can prove GS#5, and we can prove it at least a couple
ways, by KG method or by using DL a couple times. Here is the proof using
KG.
Proof.
x4 + x3 + x2 + x + 1
×
x−1
−x4 − x3 − x2 − x − 1
x5 + x4 + x3 + x2 + x
x5 − 1
4.3. MULTIPLYING FAMOUS POLYNOMIALS
17
Theorem 4.3.7. General Geometric Series Polynomials
• (x − y)(x + y) = x2 − y 2
[GGS#2]
• (x − y)(x2 + xy + y 2 ) = x3 − y 2
[GGS#3]
• (x − y)(x3 + x2 y + xy 2 + y 3 ) = x4 − y 4
[GGS#4]
• (x − y)(x4 + x3 y + x2 y 2 + xy 3 + y 4 ) = x5 − y 5
[GGS#5]
4.3.1
Exercises
1. State and prove the famous polynomial product, [GGS2]
2. Compute and simplify the following product, using FOIL:
(7w + 5) (5 − 4w)
3. Compute and simplify the following product, using FOIL:
(3t + 2) (4t − 5)
4. Use [PP2] to computing and simplify the following:
2
(v + 1)
5. prove [PP3] by computing and simplifying the following:
3
(w + y)
6. Compute and simplify the following [famous] product:
(β + 1) (β − 1)
7. Compute and simplify the following product, using FOIL:
(3q − 1) (3 − q)
8. Use [PP2] to computing and simplify the following:
2r3 + 3
2
18
CHAPTER 4. POLYNOMIALS
9. Compute and simplify the following product, using FOIL:
(5z + 1) (7z + 4)
10. Compute and simplify the following [famous] product:
5u2 + u
5u2 − u
11. prove [PP3] by computing and simplifying the following:
(r + z)3
12. Use [PP2] to computing and simplify the following:
4r3 + 7r
2
13. Prove [PP2] by computing and simplify the following:
(θ + φ)
2
14. Prove [PP2] by computing and simplify the following:
(β + φ)
2
15. Compute and simplify the following product, using FOIL:
6q 4 − q 3
7q 6 − 5q 7
16. Use [PP3] to computing and simplify the following:
3
(4 − y)
17. Use [PP3] to computing and simplify the following:
3
(4 − z)
18. Use [PP2] to computing and simplify the following:
7v 3 + 2
2
4.3. MULTIPLYING FAMOUS POLYNOMIALS
19. Use [PP2] to computing and simplify the following:
2
3 − 4t3
20. State and prove the famous polynomial product, [GS2]
21. Use [PP2] to computing and simplify the following:
2
(3y + 1)
22. Compute and simplify the following product, using FOIL:
(2v + 5) (4 − 5v)
23. Compute and simplify the following [famous] product:
2q 6 + 4q 2
2q 6 − 4q 2
24. Compute and simplify the following [famous] product:
7z 5 − z 4
7z 5 + z 4
25. Compute and simplify the following [famous] product:
11z 5 + 4z 2
11z 5 − 4z 2
26. Compute and simplify the following product, using FOIL:
(−4u − 3) (−5u − 3)
27. State and prove the famous polynomial product, [GS2]
28. Compute and simplify the following [famous] product:
(11v + 5) (11v − 5)
29. Prove [PP2] by computing and simplify the following:
2
(β + ω)
30. Compute and simplify the following product, using FOIL:
5y 5 − y 7
7y 6 − y 7
19
20
CHAPTER 4. POLYNOMIALS
31. Compute and simplify the following [famous] product:
(µ + 3) (µ − 3)
32. Use [PP2] to computing and simplify the following:
6q − 2q 3
2
33. Compute and simplify the following [famous] product:
(µ + 1) (µ − 1)
34. Compute and simplify the following product, using FOIL:
(3 − 4r) (1 − r)
35. Prove [DC] by computing and simplify the following:
(µ + φ) (φ − µ)
4.4. DIVIDING BY A MONOMIAL
4.4
21
Dividing by a Monomial
We will now learn a basic methods used to divide polynomials. The task can
be divided into two categories, dividing by a monomial, and dividing my a
non-monomial. We will practice dividing by a monomial in this section and
leave the dividing by a general polynomial for the next section. We do not
need to learn any new axioms or definition, as all the necessary axioms and
definition which apply to the real numbers apply to polynomials.
We have harnessed all the necessary ideas already to the point that all
left to do is to take a look a couple examples. Once you become good at
looking at the example you will be ready to practice doing some of these
yourself in the provided exercises.
Example divide 6x4 ÷ 2x
6x4
2x
3 · 2x3 x1
=
2x
3x3 · 2x
=
1 · 2x
3x3 2x
=
1 2x
3x3
=
1
= 3x3
6x4 ÷ 2x =
(def ÷)
(JAE,TT)
(CoLM, N-Expo, Mid)
(MAT)
(JOT,Mid)
(OUT)
Notice, this process is nearly identical to the work we did in simplifying
rational numbers, where we prime factorized each the numerator & denominator and we eliminated the gcd, ie the common prime pieces. Let’s try it
one more time.
Example divide 16x4 ÷ 2x3
22
CHAPTER 4. POLYNOMIALS
16x4
2x3
8 · 2x1 x3
=
1 · 2x3
8x · 2x3
=
1 · 2x3
8x 2x3
=
1 2x3
8x
·1
=
1
= 8x
16x4 ÷ 2x3 =
(Def ÷)
(JAE,TT)
(CoLM, +Expo)
(MAT)
(JOT)
(OUT,Mid)
Next, we turn our attention to diving a general polynomial by a monomial.
Example divide (4x3 + 10x2 + 8x + 1) ÷ 2x2
(4x3 + 10x2 + 8x + 1) ÷ 2x2
4x3 + 10x2 + 8x + 1
2x2
3
4x
10x2
8x
1
= 2+
+ 2+ 2
2x
2x2
2x
2x
4
1
= 2x + 5 + + 2
x 2x
=
Example divide (6x3 + 4x2 + 1) ÷ 2x
(def ÷)
(ATT)
(BI)
23
4.4. DIVIDING BY A MONOMIAL
6x3 + 4x2 + 1
2x
4x2
1
6x3
+
+
=
2x
2x
2x
2x · 3x2
2x · 2x
1
=
+
+
2x · 1
1 · 2x
2x
2x 2x
1
2x 3x2
+
+
=
2x 1
1 2x 2x
2x
1
3x2
+
·1+
=1·
1
1
2x
1
= 3x2 + 2x +
2x
(6x3 + 4x2 + 1) ÷ 2x =
(Def a/b, ÷)
(ATT)
(JAE,CoLM,BI)
(MAT)
(JOT)
(Mid, OUT)
1
. Alternatively, we say
We can conclude (6x3 +4x2 +1)÷2x = 3x2 +2x+ 2x
3
2
2
(6x + 4x + 1) ÷ 2x = 3x + 2x with reminder 1. This method used the idea
that every division problem can be expressed as a fraction, while making
use of our fraction skills. While it has its virtues, it also has drawbacks.
This method may not be very useful when we try to divide polynomial by
polynomial. Ultimately, we will have to resort to long division, which is
king in the world of Q[x] under division. Here is a second look at the same
problem above, solved using long division.
Example divide (6x3 + 4x2 + 1) ÷ 2x
First we set it up as, we do when we divide integers using long division,
2x
6x3 + 4x2 + 1
then the idea is to see howmany times 2x goes into the leading term 6x3 .
3
In other words we calculate 6x
2x . We can do this by inspection [BI] or as
3
2
example 1, above. In either case, we conclude 6x
2x = 3x . This becomes the
first part of the quotient.
3x2
2x
6x3 + 4x2 + 1
It is customary to try to keep columns ordered by degree. Observe the
3x2 was placed on the degree 2 column. The next step, as with integers, is
to multiply the 3x2 by the divisor 2x and subtract it from the leading term
6x3 . The result is shown,
24
CHAPTER 4. POLYNOMIALS
3x2
2x
6x3 + 4x2 + 1
− 6x3
We subtract and bring down the next term 4x2 to obtain:
3x2
2x
6x3 + 4x2 + 1
− 6x3
4x2
We now ask how many times will 2x go into 4x2 ? That is, we find
2x. This becomes the second part of the quotient. We obtain
4x2
2x
=
3x2 + 2x
2x
6x3 + 4x2 + 1
− 6x3
4x2
The next step is to multiply the 2x in the quotient by the divisor, 2x to
obtain 4x2 which we write down below and subtract from 4x2 to obtain
3x2 + 2x
2x
6x3 + 4x2 + 1
− 6x3
4x2
− 4x2
1
Finally, we bring down the 1. Since 2x does not divide 1 evenly ( 2x
does not reduce) it is left as a remainder. We conclude with the expected,
1
(6x3 + 4x2 + 1) ÷ 2x = 3x2 + 2x + 2x
OR (6x3 + 4x2 + 1) ÷ 2x = 3x2 + 2x
with reminder 1.
A couple more examples are in order
Example divide (3x3 + 15x2 − 6x − 12) ÷ (3x)
25
4.4. DIVIDING BY A MONOMIAL
3x
x2 + 5x − 2
3x3 + 15x2 − 6x − 12
− 3x3
15x2
− 15x2
− 6x
6x
Finally we bring down the -12 and find the remainder is -12, thus (3x3 +
−
12
15x2 − 6x − 12) ÷ (3x) = x2 + 5x − 2 + 3x
Example (3x3 + 5x2 − 16x − 2) ÷ (3x)
3x
x2 + 35 x −
16
3
3x3 + 5x2 − 16x − 2
− 3x3
5x2
− 5x2
− 16x
16x
Finally we bring down the (remainder) -2 and conclude, by Long Division
−
−
[LD] (3x3 + 5x2 − 16x − 2) ÷ (3x) = x2 + 35 x + 316 + 3x2
4.4.1
Exercises
1. Divide using Long Division:
36y 4 − 48y 3 − 30y 2 − 18y + 27 ÷ (3y)
2. Divide using Long Division:
−24t4 + 42t3 + 30t2 − 36t + 9 ÷ (−t)
3. Divide using [ATT] and [def ÷]:
−2q 9 + 8q 8 − 20q 2 ÷ 3q 2
26
CHAPTER 4. POLYNOMIALS
4. Divide using Long Division:
6u4 + 51u3 + 36u2 + 24u + 54 ÷ (2u)
5. Divide using Long Division:
51y 4 + 18y 3 + 36y 2 − 48y − 12 ÷ (−y)
6. Divide using [ATT] and [def ÷]:
23w8 − 2w6 + 11w5 ÷ −2w3
7. Divide using Long Division:
3u4 − 39u3 + 33u2 − 30u − 9 ÷ (2u)
8. Divide using [ATT] and [def ÷]:
−17y 6 − 22y 4 − 12y 3 ÷ −2y 3
9. Divide using Long Division:
−6y 4 + 48y 3 + 60y 2 + 39y − 33 ÷ (y)
10. Divide using [ATT] and [def ÷]:
−26v 5 + 11v 4 + 14v 2 ÷ −2v 2
11. Divide using Long Division:
45y 4 − 12y 3 − 24y 2 + 51y − 36 ÷ (−3y)
12. Divide using [ATT] and [def ÷]:
−20v 7 + 15v 5 − 18v 3 ÷ −3v 2
13. Divide using Long Division:
12x4 − 57x3 + 21x2 − 30x + 6 ÷ (−2x)
4.4. DIVIDING BY A MONOMIAL
14. Divide using [ATT] and [def ÷]:
2z 8 − 6z 7 + 26z 6 ÷ (3z)
15. Divide using Long Division:
−33x4 + 24x3 − 30x2 − 57x + 42 ÷ (3x)
16. Divide using Long Division:
−3t4 + 48t3 + 27t2 − 18t + 54 ÷ (−3t)
17. Divide using Long Division:
−48y 4 − 57y 3 + 12y 2 − 15y − 27 ÷ (3y)
18. Divide using [ATT] and [def ÷]:
−22r10 + 15r9 − 12r6 ÷ −2r3
19. Divide using [ATT] and [def ÷]:
26t7 + 13t5 + 23t3 ÷ (−3t)
20. Divide using Long Division:
−15u4 − 33u3 + 42u2 + 36u + 60 ÷ (−u)
21. Divide using [ATT] and [def ÷]:
−v 9 + 19v 7 − 14v 2 ÷ −v 3
22. Divide using Long Division:
33u4 − 27u3 − 30u2 − 60u − 54 ÷ (−u)
23. Divide using Long Division:
−27y 4 − 42y 3 + 18y 2 + 30y + 21 ÷ (2y)
27
28
CHAPTER 4. POLYNOMIALS
24. Divide using [ATT] and [def ÷]:
−11y 8 + 3y 5 − 22y 2 ÷ 3y 3
25. Divide using Long Division:
15x4 + 21x3 − 39x2 − 9x − 3 ÷ (−2x)
4.5. DIVIDING BY POLYNOMIALS II
4.5
29
Dividing by Polynomials ii
Last section we divided polynomials by monomials exclusively. We now
tackle the general problem of dividing by a binomial or any polynomial.
Consider dividing using long division 2 ÷ 5. If you think about it for a
second you will realize the quotient is 0 with a remainder of 2. This will
usually happen when trying to divide an small integer by a larger one. The
same holds for polynomials. You will not get very far if you are trying to
divide polynomial of small degree by a polynomial of large degree. The
quotient will be 0 with remainder equal to the remainder being the original
polynomial. For example, (5x + 1) ÷ x3 = 0 with remainder 5x + 1 OR
5x+1
x3 . Therefore, we will practice our diving skill mostly with cases where
we divide a polynomial of large degree by one of smaller degree. The Long
Division method will be the primary tool.
Example Divide
(3x3 + 5x2 − 6x − 2) ÷ (x − 1)
Most steps are identical. One small difference is that we always concentrate on the leading terms, momentarily ignoring the other term. For
example, we just concentrate on the leading term 3x3 inside and x, the lead3
ing term inside. We calculate by inspection 3xx = 3x2 to obtain our first
term of the quotient.
3x2
x−1
3x3 + 5x2 − 6x − 2
Now, we multiply the 3x2 by the entire divisor (x − 1), thus we have to
use DL, to get 3x2 (x − 1) = 3x3 − 3x2 . This is the quantity we subtract (i.e.
change the sign). We obtain...
3x2
x−1
3
3x + 5x2 − 6x − 2
− 3x3 + 3x2
Since we have changed the signs already (to subtract), we can simply add
the terms and bring down the next term to obtain,
3x2
x−1
3x3 + 5x2 − 6x − 2
− 3x3 + 3x2
8x2 − 6x
30
CHAPTER 4. POLYNOMIALS
We now calculate the quotient for leading terms
term of the quotient...
8x2
x
= 8x to get the next
3x2 + 8x
x−1
3x3 + 5x2 − 6x − 2
− 3x3 + 3x2
8x2 − 6x
We now multiply 8x(x − 1) = 8x2 − 8x and subtract, OR change sign and
add to obtain....
3x2 + 8x
x−1
Finally, we calculate
2x
x
3
3x + 5x2 − 6x − 2
− 3x3 + 3x2
8x2 − 6x
− 8x2 + 8x
2x − 2
= 2 for the last part of the quotient...
3x2 + 8x + 2
x−1
3x3 + 5x2 − 6x − 2
− 3x3 + 3x2
8x2 − 6x
− 8x2 + 8x
2x − 2
Then we have
3x2 + 8x + 2
x−1
3
3x + 5x2 − 6x − 2
− 3x3 + 3x2
8x2 − 6x
− 8x2 + 8x
2x − 2
− 2x + 2
0
Therefore, by [LD] we have (3x3 + 5x2 − 6x − 2) ÷ (x − 1) = 3x2 + 8x + 2
Example Divide (10x4 + 13x3 + 8x2 + 8x + 3) ÷ (2x + 1)
31
4.5. DIVIDING BY POLYNOMIALS II
5x3 + 4x2 + 2x + 3
2x + 1
10x4 + 13x3 + 8x2 + 8x + 3
− 10x4 − 5x3
8x3 + 8x2
− 8x3 − 4x2
4x2 + 8x
− 4x2 − 2x
6x + 3
− 6x − 3
0
Example
Divide (4x3 +− 4x2 +− 23x + 22) ÷ (2x − 5)
2x − 5
2x2 + 3x − 4
4x3 − 4x2 − 23x + 22
− 4x3 + 10x2
6x2 − 23x
− 6x2 + 15x
− 8x + 22
8x − 20
2
thus by [LD] (4x3 +− 4x2 +− 23x + 22) ÷ (2x − 5) = 2x2 + 3x − 4 +
Example Divide (4x4 + 8x3 + −3x2 − x − 4) ÷ (2x2 + 3x − 4)
2x2 + x + 1
2x2 + 3x − 4
4x4 + 8x3 − 3x2 − x − 4
− 4x4 − 2x3 − 2x2
6x3 − 5x2 − x
− 6x3 − 3x2 − 3x
− 8x2 − 4x − 4
8x2 + 4x + 4
0
thus by [LD] (4x3 +− 4x2 +− 23x + 22) ÷ (2x − 5) = 2x2 + 3x − 4
Example Divide (4x4 + 8x3 + −3x2 − x − 4) ÷ (3x + 2)
2
2x−5
32
CHAPTER 4. POLYNOMIALS
3x + 2
4 3
3x
3
16 2
9 x
2
4x4 + 8x − 3x
− 4x4 − 38 x3
−
4.5.1
+
16 3
3 x
16 3
3 x
−
− 3x2
2
− 32
9 x
−
59 2
9 x
59 2
9 x
59
27 x
91
81
−4
−x
−x
118
27 x
91
27 x
91
− 27
x
+
+
−
−
−4
182
81
506
81
Exercises
1. Divide using Long Division:
−96t2 + 40t − 1 ÷ (8t − 2)
2. Divide using Long Division:
−84x6 + 68x5 + 92x4 − 59x3 + 26x2 − x − 50 ÷ −7x2 + x + 6
3. Divide using Long Division:
30y 6 + 76y 5 + 81y 4 + 40y 3 − 20y 2 + 19y − 9 ÷ −6y 2 − 8y + 3
4. Divide using Long Division:
90y 3 + 48y 2 + 33y + 12 ÷ (−9y − 3)
5. Divide using Long Division:
−12y 4 + 70y 3 − 88y 2 − 42y + 71 ÷ (6y − 11)
6. Divide using Long Division:
−20x5 − 105x3 − 95x2 + 60x + 31 ÷ (10x + 5)
4.5. DIVIDING BY POLYNOMIALS II
7. Divide using Long Division:
132u5 + 88u4 + 12u3 − 16u2 + 128u + 98 ÷ (−12u − 8)
8. Divide using Long Division:
40y 4 − 72y 3 − 78y 2 + 36y − 34 ÷ (4y − 10)
9. Divide using Long Division:
−9u2 + 6u + 23 ÷ (−3u − 4)
10. Divide using Long Division:
5y 5 − 10y 4 + 6y 3 − 24y 2 + 15y + 25 ÷ (2 − y)
11. Divide using Long Division:
99y 5 + 76y 4 + 84y 3 − 11y 2 − 87y − 25 ÷ (−9y − 2)
12. Divide using Long Division:
−28y 5 + 63y 4 − 20y 3 + 13y 2 + 56y + 28 ÷ (9 − 4y)
13. Divide using Long Division:
−24u4 − 36u3 − 48u2 − 96u + 88 ÷ (−6u − 12)
14. Divide using Long Division:
99y 3 − 63y 2 − 33y + 17 ÷ 3 − 9y 2
15. Divide using Long Division:
−55u5 + 96u4 + 90u3 − 62u2 − 53u + 45 ÷ (5u − 11)
33
34
4.6
CHAPTER 4. POLYNOMIALS
Factor Polynomials by DL
Recall ’To factor’ means to break up into multiples. The main tool here will
be the distributive law. To factor completely means to factor, factor until
you can factor no more. Another name for factoring completely is prime
factorization. Here we may not learn to all methods to factor polynomials
completely. However, we will learn one method of factoring polynomials.
This method called the DL METHOD involves looking at each of the terms,
finding the gcd for the collection of terms, and factoring this gcd using DL
among other axioms and theorems. If the gcd is 1, then this method does
not provide any non-trivial factorization of the polynomial.
Example factor 6x + 4
note the gcd(6x, 4) = 2, thus using the DL method we can factor 2 from
each of the terms.
6x + 4 = 2 · 3x + 2 · 2
= 2(3x + 2)
(T.T)
(D.L.)
Example factor 16x2 + 40x + 24.
note the gcd(16x2 , 40x, 24) = 8, thus using the DL method we can factor
8 from each of the terms.
16x2 + 40x + 24 = 8 · 2x2 + 8 · 5x + 8 · 3
2
= 8(2x + 5x + 3)
(T.T)
(D.L.)
Example factor 16x5 + 40x4 + 24x3
note the gcd(16x5 , 40x4 , 24x3 ) = 8x3 , thus using the DL method we can
factor 8x3 from each of the terms.
16x5 + 40x4 + 24x3 = 8x3 · 2x2 + 8x3 · 5x + 8x3 · 3
3
2
= 8x (2x + 5x + 3)
(B.I.)
(D.L.)
Example factor (blah)x3 + (blah)2x
(blah)x3 + (blah)2x = (blah)(x3 + 2x)
2
= (blah)(x + 2)x
(D.L.)
(D.L., BI)
4.6. FACTOR POLYNOMIALS BY DL
35
Example factor (x + 5)x3 + (x + 5)2x
(x + 5)x3 + (x + 5)2x
= (x + 5)(x3 + 2x)
(given)
(D.L.)
= (x + 5)(x2 · x2 + 2x)
(J.A.E.)
= (x + 5)(x2 · x + 2x)
(N.Expo)
= (x + 5)(x2 + 2)x
(D.L., BI)
It should be noted that this method for factoring is not always successful.
Specially if there is nothing common amongst the terms.
Example factor (x + 4)x3 + (x + 5)2x
(x + 4)x3 + (x + 5)2
3
=(x + 4)x + (x + 5)2
(given)
(nothing we can factor)
This expression is as factored as it will get, at least using the DL method
from this section.
As always, the homework is an essential part of the learning process. At
this point the only thing left to do is to get to it. The exercises should help
you practice spotting gcd’s and factoring it from each of the terms using
DL.
4.6.1
Exercises
1. Factor using DL:
8t11 − 11t6 − 7t3
2. Factor using DL:
3u5 − 5u4
36
CHAPTER 4. POLYNOMIALS
3. Factor using DL:
9t12 + 7t7 + 8t4
4. Factor using DL:
−8t6 + 11t3 + 10t
5. Factor using DL:
12y 7 − 7y 4 − 8y 2
6. Factor using DL:
−5 t3 − 9 t3 u
7. Factor using DL:
11u5 u3 − 12u2 u3 + 7 u3
8. Factor using DL:
9t8 + 5t5 − 4t3
9. Factor using DL:
−6y 12 + 4y 7 + 7y 4
10. Factor using DL:
12t5 − 2t2 + 7t
11. Factor using DL:
12 x3 + 8 + 8 x3 + 8 u5 − x3 + 8 u2
12. Factor using DL:
t12 + 12t7 + 8t4
13. Factor using DL:
3 (5x + 12) u − 8 (5x + 12)
4.6. FACTOR POLYNOMIALS BY DL
14. Factor using DL:
7t8 + 3t5 − 8t3
15. Factor using DL:
5u2 ω 2 + uω 2 − 8ω 2
16. Factor using DL:
y 4 − 12y 5
17. Factor using DL:
7t2 − 9t
18. Factor using DL:
10µ4 + 8µ4 x3 − 12µ4 x2 + 3µ4 x
19. Factor using DL:
11t8 (11u + 2) − 7t3 (11u + 2) − (11u + 2)
20. Factor using DL:
7t5 x3 − 2t2 x3 − 11 x3
21. Factor using DL:
2 x3 − 4 x3 t4 + x3 t
22. Factor using DL:
−7λ4 − 8λ4 x
23. Factor using DL:
−11θ2 − 2θ2 x
24. Factor using DL:
−7 (7y + 11) + 4 (7y + 11) t8 + 6 (7y + 11) t3
37
38
CHAPTER 4. POLYNOMIALS
25. Factor using DL:
−4µ2 − 6µ2 t2 − 3µ2 t
26. Factor using DL:
11x2 + 4x
27. Factor using DL:
−10u7 + 9u4 + 11u2
28. Factor using DL:
11u4 + 6u3
29. Factor using DL:
6u4 x6 − 9u x6 − 10 x6
30. Factor using DL:
12 y 2 + 11 y 2 x5 + 9 y 2 x2
39
4.7. FACTOR BY GROUPING
4.7
Factor By Grouping
In the our section, we learned to factor using DL, once we identified the
gcd for the terms of a polynomial. It turns out this method often fails.
Particularly, if there is nothing in common within the terms, that is, it fails
if the gcd is 1. Such as:
3 + 5y + 6y 3 + 10y 4
In this section, we discuss one possible remedy for this, It is called factor
by grouping. While it works with many polynomials, we will pay special
attention to the case when we have a quad-nominal, that is a polynomial
with four terms such as the one state above. The basic idea is to group the
terms, and see where that leads.
Example
3 + 5y + 6y 3 + 10y 4
(given)
3
= (3 + 5y) + 6y + 10y
4
(ALA)
We have grouped but nothing has happened... but wait..
3 + 5y + 6y 3 + 10y 4
3
= (3 + 5y) + 6y + 10y
(given)
4
3
=1 · (3 + 5y) + 2y (3 + 5y)
(ALA)
(BI, DL)
But now we see, both of these groups have something in common, indeed,
we proceed,
3 + 5y + 6y 3 + 10y 4
= (3 + 5y) + 6y 3 + 10y
3
(given)
4
=1 · (3 + 5y) + 2y (3 + 5y)
= 1 + 2y 3 (3 + 5y)
(ALA)
(BI, DL)
(DL)
... and the factor by grouping method has succeeded! It does not always
work, and it is not always clear when it will work. You must try it and
practice. If you can’t factor a polynomial using this method, relax. Later
we will learn a few more ideas to factor.
40
CHAPTER 4. POLYNOMIALS
Example factor 1 + 3x + x2 + 3x3
1 + 3x + x2 + 3x3
= (1 + 3x) + (1 · x2 + 3x · x2 )
2
= (1 + 3x) · 1 + (1 + 3x)x
2
= (1 + 3x)(1 + x )
(ALA and M.Id)
(MId, D.L.)
(D.L.)
Example factor 1 + 3x − x2 − 3x3
1 + 3x − x2 − 3x3
= (1 + 3x) + (−x2 + −3x3 )
(ALA)
2
2
= (1 + 3x) + (1 · −x + 3x · −x )
2
= (1 + 3x) + (1 + 3x)(−x )
2
= (1 + 3x) · 1 + (1 + 3x)(−x )
2
= (1 + 3x)(1 + (−x ))
2
= (1 + 3x)(1 + −x )
(MId, N-Expo, CLM)
(D.L.)
(M.Id)
(D.L.)
(D.L.(to be continued...))
At last, the most important thing to do is for you to practice the factor
by grouping method.
4.7.1
Exercises
1. Factor by grouping:
2 + −4x2 + −4x3 + 8x5
2. Factor by grouping:
12 + 4ρ2 + 3ρ3 + ρ5
3. Factor by grouping:
1 + −t3 + −t2 + t5
4.7. FACTOR BY GROUPING
4. Factor by grouping:
4 + 3Q2 + 4Q3 + 3Q5
5. Factor by grouping:
20 + 8w + 5w2 + 2w3
6. Factor by grouping:
2 + 10x + 2x2 + 10x3
7. Factor by grouping:
6 + 10v 2 + 6v 3 + 10v 5
8. Factor by grouping:
6 + 3v + 10v 2 + 5v 3
9. Factor by grouping:
16 + 12θ2 + 16θ3 + 12θ5
10. Factor by grouping:
−2 + 5T 2 + −8T 3 + 20T 5
11. Factor by grouping:
10 + 20z + 10z 2 + 20z 3
12. Factor by grouping:
15 + 3T 2 + 15T 3 + 3T 5
13. Factor by grouping:
3 + 12x2 + 4x3 + 16x5
14. Factor by grouping:
4 + −10β 3 + −6β 2 + 15β 5
41
42
CHAPTER 4. POLYNOMIALS
15. Factor by grouping:
3 + 15λ3 + 5λ2 + 25λ5
16. Factor by grouping:
12 + 12ξ 3 + 6ξ 2 + 6ξ 5
17. Factor by grouping:
20 + 15q + 16q 2 + 12q 3
18. Factor by grouping:
12 + 4φ2 + 12φ3 + 4φ5
19. Factor by grouping:
−1 + 2λ2 + −4λ3 + 8λ5
20. Factor by grouping:
−1 + −2U 3 + 4U 2 + 8U 5
4.8. FACTOR BY SPLITTING
4.8
43
Factor by Splitting
In this section, we consider trinomials such as
x2 + 5x + 4
which can not be factored by DL, nor by grouping. For these we introduce
some brand new and very creative medicine. We call this medicine split the
middle term. Here is roughly what the process looks like in action.
x2 + 5x + 4
=x2 + (1 + 4)x + 4
2
=x + 1x + 4x + 4
(given)
(AT)
(DL)
We have indeed split the middle term. You may wonder, while all steps
look legal, what madness would drive one to do such a silly thing as splitting
the middle term? As it turns out this becomes the key, turning the trinomial
into a quadnomial on which we can factor by grouping, much like we did on
the last section. Observe:
x2 + 5x + 4
(given)
2
(AT)
2
(DL)
=x + (1 + 4)x + 4
=x + 1x + 4x + 4
= x2 + 1x + (4x + 4)
(ALA)
= (x + 1) x + (x + 1) 4
= (x + 1) (x + 4)
(BI, DL)
(BI, DL)
viola! We factored by splitting the middle term. It may help to re-read
this example, and try to pinpoint the key to its success. The key is arguably
the splitting of 5x into 1x + 4x. This may lead you to wonder, how did we
know that would work? would other splitting also work? where did this
come from? For starters, not all splitting works to end up with a nicely
factored polynomial, for example had we tried to split 5x into 3x + 2x,
though legal, we would have gotten stuck on this step:
44
CHAPTER 4. POLYNOMIALS
x2 + 5x + 4
(given)
2
(AT)
2
(DL)
=x + (1 + 4)x + 4
=x + 1x + 4x + 4
= x2 + 3x + (2x + 4)
(ALA)
= (x + 3) x + (x + 2) 2
(ALA)
Because the terms do not match, we can not complete the factorization by
grouping strategy. This shows that not all splitting works, which begs the
question, which do?
Here is one way to see which splitting work. Consider working backwards.
That is suppose we had already factored, using some numbers we yet don’t
know but imagine, a, b, c, d it might look like this:
x2 + 5x + 4 = (ax + b)(cx + d)
Then we try to gather hints about these possible numbers, a, b, c, d.
x2 + 5x + 4 = (ax + b)(cx + d)
2
= acx + bcx + dax + bd
(Assume)
(FOIL)
Now comes the hint, if you multiply bc · da you get the same thing as if you
multiply ac · bd Said differently, the split coefficients, when multiplied give
you the same as the outer coefficients when multiplied, when multiplied.
Said differently the split 5 pieces must be the same as 4, the product of
outer coefficients. So to look for the possible split pieces of 5 we look to the
possible factors of 4. Namely,
4= 2·2
4 = −2 · −2
4= 1·4
(or)
(or)
(or)
4 = −1 · −4
We see the product 1 · 4 = 4 and the sum 1 + 4 = 5, thus we have winner
winner chicken dinner.
4.8. FACTOR BY SPLITTING
45
That was lots of explaining, and maybe good to come back and re-read
after a couple example. Lets try one more,
Example Factor 3x2 + 2x − 16
First consider the outer product 3 · −16 = −48 then consider the ways of
factoring it
−48 = 1 · −48
−48 = 2 · −24
−48 = 3 · −16
......
(or)
(or)
(or)
we’re looking for a pair,w/ product ”-48” & w/sum ”2”, so we can split the
2x
......
−48 = 4 · −12
−48 = −6 · 8
(or)
(winner winner!)
That was the harder part, now, we do the splitting and all will be good..
3x2 + 2x − 16
(AT)
2
(DL)
=3x + (−6 + 8)x + −16
=3x + −6x + 8x + −16
= 3x2 + −6x + (8x + −16)
= (x + −2) 3x + (x + −2) 8
= (x + −2) (3x + 8)
4.8.1
(given)
2
Exercises
1. Factor by Splitting The Middle Term:
4λ2 + 20λ + 24
(ALA)
(DL, BI)
(DL)
46
CHAPTER 4. POLYNOMIALS
2. Factor by Splitting The Middle Term:
6x4 + 42x2 + 60
3. Factor by Splitting The Middle Term:
z 2 + −4z + 4
4. Factor by Splitting The Middle Term:
6β 2 + 26β + 8
5. Factor by Splitting The Middle Term:
6µ6 + 30µ3 z 2 + 36z 4
6. Factor by Splitting The Middle Term:
2ρ6 + 16ρ3 r2 + 30r4
7. Factor by Splitting The Middle Term:
2q 2 + 27q + 55
8. Factor by Splitting The Middle Term:
2x4 + 23x2 + 45
9. Factor by Splitting The Middle Term:
3γ 6 + 21γ 3 q 2 + 30q 4
10. Factor by Splitting The Middle Term:
t2 + 19t + 84
11. Factor by Splitting The Middle Term:
3µ6 + 10µ3 t2 + 3t4
12. Factor by Splitting The Middle Term:
6x2 + −6x + −120
4.8. FACTOR BY SPLITTING
13. Factor by Splitting The Middle Term:
2λ2 + 17λ + 21
14. Factor by Splitting The Middle Term:
2x2 + 33x + 121
15. Factor by Splitting The Middle Term:
−w2 + 11w + 12
16. Factor by Splitting The Middle Term:
−2v 2 + 5v + 3
17. Factor by Splitting The Middle Term:
4q 2 + 12q + 8
18. Factor by Splitting The Middle Term:
2w2 + 22w + 60
19. Factor by Splitting The Middle Term:
6r4 + 31r2 + 35
20. Factor by Splitting The Middle Term:
3ω 6 + 23t2 ω 3 + 40t4
21. Factor by Splitting The Middle Term:
2β 2 + 19β + 42
22. Factor by Splitting The Middle Term:
3λ2 + 19λ + 6
23. Factor by Splitting The Middle Term:
2t4 + 25t2 + 72
47
48
CHAPTER 4. POLYNOMIALS
24. Factor by Splitting The Middle Term:
2z 2 + −2z + −12
25. Factor by Splitting The Middle Term:
3γ 2 + 4γ + 1
26. Factor by Splitting The Middle Term:
6θ6 + 33θ3 q 2 + 45q 4
27. Factor by Splitting The Middle Term:
4x2 + −30x + 44
28. Factor by Splitting The Middle Term:
−r2 + −9r + −18
29. Factor by Splitting The Middle Term:
3µ2 + 5µ + 2
30. Factor by Splitting The Middle Term:
−2v 2 + 15v + −27
49
4.9. FACTOR FAMOUS POLYNOMIALS
4.9
Factor Famous Polynomials
In this section, we simply pause to reflect on some very famous polynomials. Recall earlier in this chapter we studying the multiplication of famous
polynomials. Such as the Pascal Polynomial #2 [PP2] which summarized
that
(x + y)2 = x2 + 2yx + y 2
Along with that one we studied many other famous polynomials and we verified these by doing KG multiplication on them, or by using DL to distribute
and multiply. In this section, we review the exact same polynomials, but
rather than seeing them as famous multiplications, we look at them backwards as famous factorizations. In other words, instead of seeing PP2 as
(x + y)(x + y) = x2 + 2yx + y 2
We see it the other way around as a famous factorization:
x2 + 2yx + y 2 = (x + y)(x + y)
Since it is really the same statement with a little symmetric property ??
added, we do not need to prove it. We simply want to practice spotting
these and become very good and fluent at factoring them. Not only do we
want to review and spot all PP2 polynomial forms, but we also want to do
the same for the other dozen or so famous polynomials. Here is the list of
famous polynomials from earlier this chapter.
Example Suppose we wanted to factor x2 + 6x + 9
While we could split the middle as done in the previous section, we want
to become good at spotting patterns. The first term is a square ’x2 ’, the last
terms seems to be a square too, 9 = 32 and the middle has the right form,
2 times each of these, thus it has the perfect PP2 pattern. We proceed as
follows:
( + △)2 = 2 + 2△ + △2
2
2
x + 6x + 9 = x + 2 · 3 · x + 3
2
= (x + 3)
(..recall the pp2 Pattern)
2
(BI (spotted pattern))
(PP2)
.. and we are done! We factored this famous polynomial. We want to
practice this sort of factoring with all of our famous polynomials. Here is a
list for review. Followed by a couple more examples then the very important
exercises.
50
CHAPTER 4. POLYNOMIALS
Very Famous Polynomials
name
Short
Example
Difference of Squares
[DS]
x2 − y 2 = (x − y)(x + y)
Difference of Cubes
[DS]
x3 − y 3 = (x − y)(x2 + xy + y 2 )
Sum of Cubes
[SC]
x3 + y 3 = (x + y)(x2 − xy + y 2 )
Sum of Cubes
[P P 2]
(x + y)2 = x2 + 2xy + y 2
Pascal Polynomials
name
Short
Example
Pascal Polynomial #2
[P P 2]
(x + y)2 = x2 + 2yx + y 2
Pascal Polynomial #3
[P P 3]
(x + y)3 = x3 + 3x2 y + 3xy 2 + y 3
Pascal Polynomial #4
[P P 4]
(x + y)4 = x4 + 4x3 y + 6x2 y 2 + 6xy 3 + y 4
Pascal Polynomial #5
[P P 5]
(x + y)5 = x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5
Geometric Series Polynomials
name
Short
Example
Geometric Series #2
[GS2]
x2 − 1 = (x − 1)(x + 1)
Geometric Series #3
[GS3]
x3 − 1 = (x − 1)(x2 + x + 1)
Geometric Series #4
[GS4]
x4 − 1 = (x − 1)(x3 + x2 + x + 1)
Geometric Series #5
[GS5]
x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1)
Generalized Geometric Series Polynomials
name
Short
Example
Geometric Series #2
[GGS2]
x2 − y 2 = (x − y)(x + y)
Geometric Series #3
[GGS3]
x3 − y 3 = (x − y)(x2 + xy + y 2 )
Geometric Series #4
[GGS4]
x4 − y 4 = (x − 1)(x3 + x2 y + xy 2 + y 3 )
Geometric Series #5
[GGS5]
x5 − y 4 = (x − 1)(x4 + x3 y + x2 y 2 + xy 3 + y 4 )
51
4.9. FACTOR FAMOUS POLYNOMIALS
Example Suppose we wanted to factor 4x2 − 10x + 25
We note the first term is a square, 4x2 = (2x)2 , the last terms seems to be a square too,
25 = (−5)2 and the middle has the right form, 2 times each of these, thus it has the
perfect PP2 pattern. We proceed as follows:
( + △)2 = 2 + 2△ + △2
2
2
(..recall the pp2 Pattern)
2
4x − 10x + 25 = (2x) + 2 · (−5) · x + (−5)
(BI (identified PP2 pattern))
= (2x + −5)2
(PP2)
.. and we are done! We factored this famous polynomial.
1
Example Suppose we wanted to factor 8x2 + 125
3
3
We note the first term is a cube 8x = (2x) , the last terms seems to be a cube too,
3
1
= 15 , thus it has the perfect Sum of Cubes pattern. We proceed as follows:
125
3 + △3 = ( + △) 2 − △ + △2
(..recall the SC Pattern)
3
1
1
(BI (identified SC pattern))
= (2x)3 +
8x3 +
125
5
"
2 #
1
1
1
(2x)2 − (2x)
+
(SC)
= 2x +
5
5
5
.. and we are done! We factored another famous polynomial.
Example Suppose we wanted to factor 32x4 − 243
We note the first term is a fifth power.. 32x5 = (2x)5 , the last terms seems to be a fifth
power too, 243 = 35 , thus it has the perfect General Geometric #5 pattern. We proceed
as follows:
5 − △5 = ( − △) 4 + 3 △ + 2 △2 + △3 + △4
(..recall the GGS5 Pattern)
32x5 − 243 = (2x)5 − 35
(BI (identified GG5 pattern))
4
3
2
2
= (2x − 3) (2x) + (2x) (3) + (2x) (3) + (2x)(3)3 + (3)4
(SC)
.. and we are done! We factored another famous polynomial.
4.9.1
Exercises
1. Factor by recognizing the famous polynomial:
27v 3 + −216
52
CHAPTER 4. POLYNOMIALS
2. Factor by recognizing the famous polynomial:
9w2 +
21w 49
+
2
16
3. Factor by recognizing the famous polynomial:
4t2 + 4t + 1
4. Factor by recognizing the famous polynomial:
y 3 + −30y 2 + 300y + −1000
5. Factor by recognizing the famous polynomial:
4z 2 + 16z + 16
6. Factor by recognizing the famous polynomial:
θ3 + 6θ2 + 12θ + 8
7. Factor by recognizing the famous polynomial:
9V 2 + 54V + 81
8. Factor by recognizing the famous polynomial:
4v 2 − 25
9. Factor by recognizing the famous polynomial:
x2 + 2x + 1
10. Factor by recognizing the famous polynomial:
4µ2 +
1
4µ
+
7
49
11. Factor by recognizing the famous polynomial:
27λ3 + −81λ2 + 81λ + −27
4.9. FACTOR FAMOUS POLYNOMIALS
12. Factor by recognizing the famous polynomial:
x2 − 81
13. Factor by recognizing the famous polynomial:
8q 3 + 72q 2 + 216q + 216
14. Factor by recognizing the famous polynomial:
v2 − 4
15. Factor by recognizing the famous polynomial:
9r2 + 42r + 49
16. Factor by recognizing the famous polynomial:
X 2 + 10X + 25
17. Factor by recognizing the famous polynomial:
27w3 + −135w2 + 225w + −125
18. Factor by recognizing the famous polynomial:
8ω 3 + 108ω 2 + 486ω + 729
19. Factor by recognizing the famous polynomial:
27µ3 + 216µ2 + 576µ + 512
20. Factor by recognizing the famous polynomial:
9r2 + 36r + 36
21. Factor by recognizing the famous polynomial:
X 3 + 1000
22. Factor by recognizing the famous polynomial:
V 2 + 4V + 4
53
54
CHAPTER 4. POLYNOMIALS
23. Factor by recognizing the famous polynomial:
9Y 2 + 60Y + 100
24. Factor by recognizing the famous polynomial:
ξ 3 + 30ξ 2 + 300ξ + 1000
25. Factor by recognizing the famous polynomial:
27ρ3 + −108ρ2 + 144ρ + −64
26. Factor by recognizing the famous polynomial:
27ρ3 + −512
27. Factor by recognizing the famous polynomial:
R3 + 12R2 + 48R + 64
28. Factor by recognizing the famous polynomial:
9u2 + 60u + 100
29. Factor by recognizing the famous polynomial:
27W 3 + −54W 2 + 36W + −8
30. Factor by recognizing the famous polynomial:
4W 2 +
16
16W
+
9
81
31. Factor by recognizing the famous polynomial:
9x2 − 16
32. Factor by recognizing the famous polynomial:
9q 2 +
14q 49
+
3
81
4.9. FACTOR FAMOUS POLYNOMIALS
33. Factor by recognizing the famous polynomial:
9t2 − 9
34. Factor by recognizing the famous polynomial:
w3 + −216
35. Factor by recognizing the famous polynomial:
4r2 − 64
55
56
CHAPTER 4. POLYNOMIALS
57
4.10. CHAPTER REVIEW
4.10
Chapter Review
Very Famous Polynomials
name
Short
Example
FOIL
[F OIL]
(A + B)(C + D) = AC + AD + BC + BD
Difference of Squares
[DS]
x2 − y 2 = (x − y)(x + y)
Difference of Cubes
[DS]
x3 − y 3 = (x − y)(x2 + xy + y 2 )
Sum of Cubes
[SC]
x3 + y 3 = (x + y)(x2 − xy + y 2 )
Sum of Cubes
[P P 2]
(x + y)2 = x2 + 2xy + y 2
Pascal Polynomials
name
Short
Example
Pascal Polynomial #2
[P P 2]
(x + y)2 = x2 + 2yx + y 2
Pascal Polynomial #3
[P P 3]
(x + y)3 = x3 + 3x2 y + 3xy 2 + y 3
Pascal Polynomial #4
[P P 4]
(x + y)4 = x4 + 4x3 y + 6x2 y 2 + 6xy 3 + y 4
Pascal Polynomial #5
[P P 5]
(x + y)5 = x5 + 5x4 y + 10x3 y 2 + 10x2 y 3 + 5xy 4 + y 5
Geometric Series Polynomials
name
Short
Example
Geometric Series #2
[GS2]
x2 − 1 = (x − 1)(x + 1)
Geometric Series #3
[GS3]
x3 − 1 = (x − 1)(x2 + x + 1)
Geometric Series #4
[GS4]
x4 − 1 = (x − 1)(x3 + x2 + x + 1)
Geometric Series #5
[GS5]
x5 − 1 = (x − 1)(x4 + x3 + x2 + x + 1)
Generalized Geometric Series Polynomials
name
Short
Example
Geometric Series #2
[GGS2]
x2 − y 2 = (x − y)(x + y)
Geometric Series #3
[GGS3]
x3 − y 3 = (x − y)(x2 + xy + y 2 )
Geometric Series #4
[GGS4]
x4 − y 4 = (x − 1)(x3 + x2 y + xy 2 + y 3 )
Geometric Series #5
[GGS5]
x5 − y 4 = (x − 1)(x4 + x3 y + x2 y 2 + xy 3 + y 4 )
58
CHAPTER 4. POLYNOMIALS
Some Answers
2.
try:
−5w
Section 4.1
5
− 4w
3
+ 2w
×
1. 7r6 + 2r5 + 5r4 + 5r3 + 3r2 + 2r + 5
2. 2q4 − 2q3 + q2 + 11q + 2
3. 5w6 + w5 + 7w4 − 5w3 + w2 − 10w + 7
4. 5y + 2
5. 3w4 + 3w3 − 5w2 + 4w + 2
6. 7r3 − 2r2 + 2
7. −3r6 + 4r5 + 5r4 + 7r3 − 5r2 + 6r
8. 8t3 − 2t2 + 10t + 4
9. −3y5 − 6y3 + 2y2 − 4y − 2
10. 6r6 − 5r5 + 5r4 − 6r3 − r2 + 10r − 4
11. 10r4 + r3 + 6r2 + 7r + 7
12. 9q − 7
13. −v4 − 2v3 − 3v2 − 10v
14. 6q6 + q5 − 12q3 + 4q − 7
15. 3t
16. −2z5 − z4 − 4z3 + z2 − 6z + 7
17. 4v4 + 6v3 + 3v2 − 3v + 4
18. 6y3 + 8y2 + 2y + 2
19. −w − 1
20. 13v2 + 7v
−30w
3.
11
− 24w
6w
9
+ 12w
6
7
try:
(−2z
7
+ 3z
7
6
4
+ 7z)(−2z )
6
4
4
=(−2z )(−2z ) + (3z )(−2z ) + (7z)(−2z )
(DL)
=4z
4.
11
4
− 6z
10
− 14z
5
(BI)
try:
y
4
+ 2y
×
−2y
4y
5y
5y
5.
6
6
5
+ 8y
+ 10y
+ 14y
5
4
5
+ 41y
4
3
4
2
3
3
2
+ 20y
+ 25y
+ 49y
2
+ 5y
+ 4y − 2
− 14y
+ 28y
+ 35y
+ 7y
5y
− 4y
4
3
− 10y
2
3
+ 6y
2
− 10y
try:
w
6
− 5w
4
+ 7w
×
7w
12
− 35w
10
7w
+ 49w
3
6
9
Section 4.2
1.
6.
try:
try:
4
2
u + 6u + 3u − 5
×
4z
2
7u − 5u + 5
×
4
2
5u + 30u + 15u − 25
−16z
−5u5 − 30u3 − 15u2 + 25u
6
4
3
2
7u + 42u + 21u − 35u
6
5
4
3
2
7u − 5u + 47u − 9u − 20u + 40u − 25
3
3
− 4z
3z
+ 16z
2
2
− 5z + 7
2
− 2z − 4
+ 20z − 28
−8z 4 + 8z 3 + 10z 2 − 14z
12z
12z
5
5
− 12z
− 20z
4
4
− 15z
− 23z
3
3
+ 21z
+ 47z
2
2
+ 6z − 28
59
Answers for Section 4.2
7.
12.
try:
−w
3
+ 5w
2
−4w
2w
8.
4
4
3
+ 20w
− 10w
3
− 14w
3
2
4 − 2w
2
2
−21r
+ 8w
−21r
− 4w − 16
13.
6
+ 64r
5
6
3
+ 7r
×
7r
5
− 35r
15r
− 12w − 16
+ 6w
+ 26w
−3r
− 3w − 4
×
2w
try:
+ 49r
− 35r
4
5
2
4
− 5r
2
3
3
− 5r
2
try:
try:
−2z
−4v
5
+ 6v
3
×
+ 3v
3v
2
10
+ 18v
8
+ 9v
2
×
+z+4
2z
3
− 2z
2
− 8z
5
4z
−12v
9.
+1
− 5r
+ 7r
+ 7r
2
3
7
−4z
−4z
5
+ 2z
4
5
3
+ 2z
+ 12z
− 2z
4
3
+ 8z
− 2z
3
2
− 8z
try:
4
2
3u + 4u − 2u − 1
14.
try:
2
u − 5u − 2
×
4r
5
+r
3
− 5r
×
−r
4
2
−6u − 8u + 4u + 2
5
3
2
−15u − 20u + 10u + 5u
−4r
6
−r
4
+ 5r
2
6
4
3
2
3u + 4u − 2u − u
6
5
4
3
2
3u − 15u − 2u − 22u + u + 9u + 2
15.
try:
5w4 + 7w3 + 6w + 3
×
10.
− 4w
2
+ 4w + 3
try:
15w
2
5
2
(t + 5t − 5)(t − 2t )
20w
=(t2 )(t5 − 2t2 ) + (5t)(t5 − 2t2 ) + (−5)(t5 − 2t2 )
(DL)
5
+ 28w
4
4
+ 21w
+ 24w
3
2
+ 18w + 9
+ 12w
−20w6 − 28w5 − 24w3 − 12w2
2
5
2
2
5
2
5
2
=(t )(t ) + (t )(−2t ) + (5t)(t ) + (5t)(−2t ) + (−5)(t ) + (−5)(−2t )
6
5
4
3
2
−20w − 8w + 43w − 3w + 12w + 30w + 9
(DL)
7
6
5
4
3
2
=t + 5t − 5t − 2t − 10t + 10t
(BI)
16.
11.
try:
(−y
try:
5
− 5y
3
2
− 3y )(5y)
5
3
2
=(−y )(5y) + (−5y )(5y) + (−3y )(5y)
3
2
−4t − 5t − t − 2
×
= − 5y
6
− 25y
4
− 15y
3
(DL)
(BI)
2
6t + 3t
−12t4 − 15t3 − 3t2 − 6t
5
4
3
2
−24t − 30t − 6t − 12t
17.
try:
(6z
5
+ 4z
5
3
7
− 1)(5z )
3
7
7
=(6z )(5z ) + (4z )(5z ) + (−1)(5z )
5
4
3
2
−24t − 42t − 21t − 15t − 6t
=30z
12
7
+ 20z
10
− 5z
7
(DL)
(BI)
60
Answers for Section 4.2
18.
4.
try:
6z
3
− 4z
×
−12z
36z
36z
6
− 24z
5
6
4
− 24z
− 42z
4
6z
+ 8z
5
3
2
− 5z
3
− 2z
+ 10z
− 30z
+ 8z
3
using [PP2] ...
2
(v + 1)
= (v)
=v
2
2
2
(given)
2
+ 2 (1) (v) + (1)
(PP2)
+ 2v + 1
(BI)
4
+ 10z
2
5.
using [PP3] ...
(w + y)3
19.
(3y
2
+ y − 4)(−3y
3
(given)
2
1
(w + y) (w + y)
2
2
w + 2wy + y
(w + y)
try:
− 2y)
2
3
3
3
=(3y )(−3y − 2y) + (y)(−3y − 2y) + (−4)(−3y − 2y)
(DL)
now we use KG
2
3
2
3
3
=(3y )(−3y ) + (3y )(−2y) + (y)(−3y ) + (y)(−2y) + (−4)(−3y ) + (−4)(−2y)
(DL)
= − 9y
5
20.
try:
− 3y
4
+ 6y
3
− 2y
2
w
+ 8y
(BI)
2
(JAE)
(PP2,BI)
+ 2wy + y
×
w
3
2
2
2
+ 2w y + wy
w y + 2wy
6y
4
×
−24y
6
− 4y
3
2
+y
3
+y+4
− 4y
2
− 16y
2
w
6.
3
2
2
3
+ 3w y + 3wy + y
this is famous [PP2] goes like this...
(β + 1) (β − 1)
(given)
= (β) (β) + (β) (−1) + (1) (β) + (1) (−1)
Section 4.3
= β
= β
1.
2.
2
w+y
(FOIL)
2
+ −β + β + −1
(BI)
2
−1
(BI)
vary, use DL or KG for example..
7.
(7w + 5) (5 − 4w)
(given)
(3q − 1) (3 − q)
= (7w) (−4w) + (7w) (5) + (5) (−4w) + (5) (5)
(FOIL)
= −28w
2
= −28w
2
+ 35w + −20w + 25
(BI)
+ 15w + 25
(BI)
3.
= (3q) (−q) + (3q) (3) + (−1) (−q) + (−1) (3)
(FOIL)
= −3q2 + 9q + q + −3
= −3q
8.
(3t + 2) (4t − 5)
(given)
2
(BI)
+ 10q − 3
(BI)
using [PP2] ...
(given)
= (3t) (4t) + (3t) (−5) + (2) (4t) + (2) (−5) (FOIL)
2r
3
2
= 12t + −15t + 8t + −10
(BI)
=
2
= 12t − 7t − 10
(BI)
= 4r
2
+3
2r
6
3 2
(given)
+ 2 (3) 2r
+ 12r
3
+9
3
2
+ (3)
(PP2)
(BI)
61
Answers for Section 4.3
14.
9.
(5z + 1) (7z + 4)
(given)
= (5z) (7z) + (5z) (4) + (1) (7z) + (1) (4)
= 35z
2
(FOIL)
+ 20z + 7z + 4
10.
this is famous [DC] goes like this...
2
= 5u
2
5u
+
2
5u
(N.Expo)
= (β) (β) + (β) (φ) + (φ) (β) + (φ) (φ)
=β
2
+ βφ + βφ + φ
(FOIL)
2
(BI)
= β 2 + 2βφ + φ2
(BI)
15.
(given)
4
3
6
7
6q − q
7q − 5q
2
(−u) + (u) 5u
(given)
+ (u) (−u)
(FOIL)
4
3
3
2
= 25u + −5u + 5u + −u
4
7
4
6
3
7
3
6
= 6q
−5q
+ 6q
7q
+ −q
−5q
+ −q
7q
(FOIL)
(BI)
4
2
= 25u − u
11.
(given)
(β + φ) (β + φ)
(BI)
2
2
5u + u
5u − u
(β + φ)2
(BI)
= 35z 2 + 27z + 4
this is famous [DC] goes like this...
(BI)
= −30q
11
= −30q
11
+ 42q
+ 47q
10
10
+ 5q
10
− 7q
+ −7q
9
(BI)
9
(BI)
using [PP3] ...
16.
3
(r + z)
using [PP3] ...
(given)
2
1
(r + z) (r + z)
2
2
r + 2rz + z
(r + z)
(JAE)
(PP2,BI)
(4 − y)3
= (−y)
= −y
3
3
(given)
+ 3 (4)
+ 12y
2
2
(−y) + 3 (4) (−y)
2
+ (4)
3
− 48y + 64
(PP2)
(BI)
now we use KG
r
2
+ 2rz + z
×
2
17.
using [PP3] ...
r+z
3
(4 − z)
r3 + 2r2 z + rz 2
2
2
3
r z + 2rz + z
r
3
= −z
4r
=
3
+ 7r
2
(given)
3 2
3
2
4r
+ 2 (7r) 4r
+ (7r)
= 16r
6
+ 56r
4
+ 12z
2
− 48z + 64
(PP2)
(BI)
using [PP2] ...
using [PP2] ...
13.
3
2
2
3
+ 3r z + 3rz + z
18.
12.
(given)
3
2
2
3
= (−z) + 3 (4) (−z) + 3 (4) (−z) + (4)
+ 49r
2
7v
=
(PP2)
3
2
+2
(given)
3 2
3
2
7v
+ 2 (2) 7v
+ (2)
= 49v
6
+ 28v
3
+4
(PP2)
(BI)
(BI)
19.
using [PP2] ...
this is famous [DC] goes like this...
2
(θ + φ)
(θ + φ) (θ + φ)
=θ
2
2
+ θφ + θφ + θ
+ 2θφ + φ
2
2
3 2
3 − 4t
3 2
3
2
= −4t
+ 2 (3) −4t
+ (3)
(N.Expo)
= (φ) (φ) + (φ) (θ) + (θ) (φ) + (θ) (θ)
=φ
(given)
(FOIL)
6
3
= 16t − 24t + 9
(BI)
(BI)
20.
vary, use DL or KG for example..
(given)
(PP2)
(BI)
62
Answers for Section 4.3
21.
using [PP2] ...
(3y + 1)
= (3y)
= 9y
2
2
2
(given)
+ 2 (1) (3y) + (1)
2
+ 6y + 1
27.
28.
(PP2)
vary, use DL or KG for example..
this is famous [DC] goes like this...
(11v + 5) (11v − 5)
(given)
(BI)
= (11v) (11v) + (11v) (−5) + (5) (11v) + (5) (−5)
(FOIL)
22.
= 121v
(2v + 5) (4 − 5v)
(given)
= (2v) (−5v) + (2v) (4) + (5) (−5v) + (5) (4)
(FOIL)
= −10v
2
+ 8v + −25v + 20
(BI)
= −10v
2
− 17v + 20
(BI)
= 121v
29.
2
+ −55v + 55v + −25
(BI)
2
− 25
(BI)
this is famous [DC] goes like this...
(β + ω)
2
(given)
(β + ω) (β + ω)
(N.Expo)
= (ω) (ω) + (ω) (β) + (β) (ω) + (β) (β)
23.
2q
this is famous [DC] goes like this...
6
+ 4q
2
2q
6
− 4q
2
= ω
= β
(given)
2
2
+ βω + βω + β
+ 2βω + ω
2
(FOIL)
(BI)
2
(BI)
6
6
6
2
2
6
2
2
= 2q
2q
+ 2q
−4q
+ 4q
2q
+ 4q
−4q
30.
(FOIL)
= 4q
= 4q
12
+ −8q
12
− 16q
8
+ 8q
8
+ −16q
4
(BI)
4
(BI)
5y
5
−y
7
6
7
7y − y
(given)
7
7
7
6
5
7
5
6
= −y
−y
+ −y
7y
+ 5y
−y
+ 5y
7y
(FOIL)
24.
7z
this is famous [DC] goes like this...
5
−z
4
5
4
7z + z
= y
(given)
= y
5
5
5
4
4
5
4
4
= 7z
7z
+ 7z
z
+ −z
7z
+ −z
z
(FOIL)
= 49z
= 49z
25.
11z
10
10
+ 7z
−z
9
+ −7z
9
+ −z
8
(BI)
8
(BI)
31.
+ 4z
2
5
2
11z − 4z
+ −7y
14
− 7y
13
13
+ −5y
− 5y
12
12
+ 35y
+ 35y
11
11
(BI)
(BI)
this is famous [PP2] goes like this...
(µ + 3) (µ − 3)
(given)
= (µ) (µ) + (µ) (−3) + (3) (µ) + (3) (−3)
this is famous [DC] goes like this...
5
14
(FOIL)
2
= µ + −3µ + 3µ + −9
(BI)
= µ2 − 9
(BI)
(given)
32.
using [PP2] ...
2
2
5
2
2
5
5
5
−4z
+ 4z
11z
+ 4z
−4z
+ 11z
11z
= 11z
3 2
6q − 2q
(FOIL)
= 121z
= 121z
10
+ −44z
10
− 16z
7
+ 44z
4
7
+ −16z
4
(BI)
(BI)
26.
(−4u − 3) (−5u − 3)
=
−2q
= 4q
33.
6
3 2
(given)
3
2
+ 2 (6q) −2q
+ (6q)
− 24q
4
+ 36q
2
(PP2)
(BI)
this is famous [PP2] goes like this...
(given)
(µ + 1) (µ − 1)
= (−4u) (−5u) + (−4u) (−3) + (−3) (−5u) + (−3) (−3)
(FOIL)
= (µ) (µ) + (µ) (−1) + (1) (µ) + (1) (−1)
(given)
(FOIL)
2
= 20u + 12u + 15u + 9
(BI)
2
= µ + −µ + µ + −1
(BI)
2
= 20u + 27u + 9
(BI)
2
=µ −1
(BI)
63
Answers for Section 4.4
3.
34.
(3 − 4r) (1 − r)
(given)
− 2q
= (−4r) (−r) + (−4r) (1) + (3) (−r) + (3) (1)
(FOIL)
= 4r
= 4r
2
+ −4r + −3r + 3
(BI)
2
− 7r + 3
(BI)
9
+ 8q
+
3q2
this is famous [DC] goes like this...
(given)
=−
2q7
8q8
= φ
= φ
÷ 3q
2
(given)
(def ÷)
−20q2
+
3q2
+
3
= (φ) (φ) + (φ) (−µ) + (µ) (φ) + (µ) (−µ)
2
3q2
−2q9
(µ + φ) (φ − µ)
− 20q
−2q9 + 8q8 − 20q2
=
=
35.
8
8q6
(ATT)
3q2
+−
3
20
(BI)
3
(FOIL)
2
2
+ −µφ + µφ + −µ
(BI)
2
2
−µ
(BI)
4.
2u
Section 4.4
3u3 + 51 u2 + 18u + 12
2
6u4 + 51u3 + 36u2 + 24u + 54
4
− 6u
51u3
− 51u3
1.
36u2
− 36u2
12y3 − 16y2 − 10y
3y
−6
24u
− 24u
36y4 − 48y3 − 30y2 − 18y + 27
− 36y4
− 48y3
48y3
thus final asnwers is
− 30y2
30y2
− 18y
18y
27
51u2
3
+ 18u +
+ 12
3u +
2
u
thus final asnwers is
12y
3
− 16y
2
− 10y +
9
5.
−6
y
− 51y3 − 18y2 − 36y + 48
2.
−y
24t3 − 42t2 − 30t + 36
− t − 24t4 + 42t3 + 30t2 − 36t + 9
24t4
51y4 + 18y3 + 36y2 − 48y − 12
− 51y4
18y3
− 18y3
36y2
− 36y2
42t3
− 42t3
− 48y
48y
30t2
− 30t2
− 36t
36t
thus final asnwers is
thus final asnwers is
9
3
2
24t − 42t − 30t −
+ 36
t
−51y
3
− 18y
2
− 36y +
12
y
+ 48
64
Answers for Section 4.4
6.
thus final asnwers is
23w
8
− 2w
6
+ 11w
5
÷ −2w
3
(given)
−6y
=
=
=−
23w5
3
11w5
+
−2w3
+w
(ATT)
−2w3
+−
11w2
2
− 26v
=
(BI)
5
+ 39
+ 11v
3 u3 − 39 u2 + 33 u − 15
2
2
2
3u4 − 39u3 + 33u2 − 30u − 9
4
− 3u
+ 14v
2
÷ −2v
2
11v4
+
−2v2
=13v
3
+−
−2v2
11v2
+
(given)
(def ÷)
−2v2
−26v5
7.
4
−26v5 + 11v4 + 14v2
2
=
2u
33
+ 60y −
10.
−2w6
+
2
(def ÷)
−2w3
−2w3
+ 48y
y
23w8 − 2w6 + 11w5
23w8
3
14v2
(ATT)
−2v2
+ −7
(BI)
2
− 39u3
39u3
33u2
− 33u2
11.
− 30u
30u
− 15y3
− 3y
thus final asnwers is
3u3
−
2
39u2
+
33u
2
9
−
2
− 24y2
24y2
2u
8.
=
−17y6 − 22y4 − 12y3
−17y6
=
=
−2y3
17y3
−22y4
+
−2y3
51y
− 51y
(given)
(def ÷)
−2y3
thus final asnwers is
−15y
−12y3
+
−2y3
3
+ 4y
2
+ 8y +
12
− 17
y
(ATT)
12.
+ 11y + 6
(BI)
2
9.
+ 8y − 17
− 12y3
12y3
− 15
− 17y6 − 22y4 − 12y3 ÷ −2y3
+ 4y2
45y4 − 12y3 − 24y2 + 51y − 36
− 45y4
− 20v
=
7
+ 15v
5
− 18v
3
÷ −3v
−20v7 + 15v5 − 18v3
−3v2
2
(given)
(def ÷)
− 6y3 + 48y2 + 60y + 39
y − 6y4 + 48y3 + 60y2 + 39y − 33
6y4
−20v7
=
48y3
− 48y3
60y2
− 60y2
=
−3v2
20v5
3
39y
− 39y
+
15v5
−3v2
+ −5v
3
−18v3
+
+ 6v
−3v2
(ATT)
(BI)
65
Answers for Section 4.4
13.
thus final asnwers is
− 6x3 + 57 x2 − 21 x + 15
2
2
− 2x
12x4 − 57x3 + 21x2 − 30x + 6
4
− 12x
18
3
2
+6
t − 16t − 9t −
t
− 57x3
57x3
17.
21x2
− 21x2
− 16y3 − 19y2 + 4y − 5
3y − 48y4 − 57y3 + 12y2 − 15y − 27
48y4
− 30x
30x
− 57y3
57y3
thus final asnwers is
−6x
3
57x2
+
21x
−
2
3
−
2
12y2
− 12y2
+ 15
x
− 15y
15y
14.
2z
=
8
− 6z
7
+ 26z
6
÷ 3z
(given)
thus final asnwers is
2z 8 − 6z 7 + 26z 6
(def ÷)
3z
−16y
=
2z 8
−6z 7
+
+
3z
=
3z
2z 7
+ −2z
6
+
− 19y
2
+ 4y −
9
−5
(ATT)
3z
18.
26z 5
3
3
y
26z 6
(BI)
− 22r
3
=
10
24x3
− 24x3
− 12r
6
÷ −2r
−2r3
=11r
7
+
+−
15r9
−2r3
15r6
3
(given)
(def ÷)
−2r3
−22r10
=
9
−22r10 + 15r9 − 12r6
15.
− 11x3 + 8x2 − 10x − 19
3x − 33x4 + 24x3 − 30x2 − 57x + 42
33x4
+ 15r
−12r6
+
(ATT)
−2r3
+ 6r
3
(BI)
2
− 30x2
30x2
− 57x
57x
19.
thus final asnwers is
−11x
3
+ 8x
2
− 10x +
14
7
5
3
26t + 13t + 23t ÷ −3t
− 19
(given)
x
16.
=
26t7 + 13t5 + 23t3
(def ÷)
−3t
t3 − 16t2
− 9t
+6
− 3t − 3t4 + 48t3 + 27t2 − 18t + 54
3t4
48t3
− 48t3
27t2
− 27t2
=
26t7
−3t
=−
+
26t6
3
− 18t
18t
13t5
−3t
+−
+
23t3
13t4
3
(ATT)
−3t
+−
23t2
3
(BI)
66
Answers for Section 4.4
20.
thus final asnwers is
15u3 + 33u2 − 42u − 36
−
− u − 15u4 − 33u3 + 42u2 + 36u + 60
15u4
27y3
− 21y
2
+ 9y +
2
− 33u3
33u3
21
+ 15
2y
24.
42u2
− 42u2
− 11y
36u
− 36u
=
8
+ 3y
5
− 22y
2
÷ 3y
3
(given)
−11y8 + 3y5 − 22y2
(def ÷)
3y3
thus final asnwers is
−11y8
60
3
2
15u + 33u − 42u −
− 36
u
=
3y3
21.
−v
=
9
+ 19v
7
− 14v
2
÷ −v
3
−v9 + 19v7 − 14v2
=−
(given)
11y5
+
3y5
3y3
+y
2
−22y2
+
+−
3
(ATT)
3y3
22
(BI)
3y
(def ÷)
−v3
25.
−v9
=
−v3
+
19v7
−v3
−14v2
+
(ATT)
−v3
− 15 x3 − 21 x2 + 39 x + 9
2
2
2
2
− 2x
=v
6
+ −19v
4
+
14
15x4 + 21x3 − 39x2
− 15x4
(BI)
− 9x − 3
21x3
− 21x3
v
− 39x2
39x2
22.
− 9x
9x
− 33u3 + 27u2 + 30u + 60
−u
33u4 − 27u3 − 30u2 − 60u − 54
− 33u4
thus final asnwers is
− 27u3
27u3
−
− 30u2
30u2
15x3
−
21x2
2
+
2
39x
+
2
3
2x
− 60u
60u
thus final asnwers is
54
3
2
+ 60
−33u + 27u + 30u +
u
Section 4.5
1.
23.
− 27 y3 − 21y2 + 9y + 15
2
2y − 27y4 − 42y3 + 18y2 + 30y + 21
4
27y
− 12t + 2
8t − 2 − 96t2 + 40t − 1
96t2 − 24t
16t − 1
− 16t + 4
− 42y3
42y3
3
18y2
− 18y2
thus final asnwers is
30y
− 30y
2 − 12t +
3
8t − 2
+
9
2
67
Answers for Section 4.5
2.
thus final asnwers is
12x4
− 8x3
− 4x2
+x −7
− 7x2 + x + 6 − 84x6 + 68x5 + 92x4 − 59x3 + 26x2
84x6 − 12x5 − 72x4
− x − 50
56x5 + 20x4 − 59x3
− 56x5 + 8x4 + 48x3
−2y
3
+ 8y
2
−7+−
6
6y − 11
6.
28x4 − 11x3 + 26x2
− 28x4 + 4x3 + 24x2
− 2x4
+ x3
10x + 5 − 20x5
− 105x3
− 7x3 + 50x2 − x
20x5 + 10x4
7x3
− x2 − 6x
10x4 − 105x3
49x2 − 7x − 50
− 10x4
− 5x3
− 49x2 + 7x + 42
− 110x3
−8
110x3
4
− 8x
3
− 4x
2
+x−7+−
− 4x
+8
− 95x2
+ 55x2
− 40x2 + 60x
40x2 + 20x
thus final asnwers is
12x
− 11x2
− 95x2 + 60x + 31
8
80x + 31
− 80x − 40
−7x2 + x + 6
−9
3.
− 5y4
− 6y2 − 8y + 3
− 6y3
− 8y2
thus final asnwers is
+y−2
30y6 + 76y5 + 81y4 + 40y3 − 20y2 + 19y − 9
4
3
2
−2x + x − 11x − 4x + 8 + −
− 30y6 − 40y5 + 15y4
36y5 + 96y4 + 40y3
− 36y5 − 48y4 + 18y3
48y4 + 58y3 − 20y2
− 48y4 − 64y3 + 24y2
− 6y3
6y3
7.
+ 4y2 + 19y
+ 8y2 −
− 12u
3y − 8
12y2 + 16y − 9
− 12y2 − 16y + 6
− 11u4
− u2
12u3 − 16u2
− 12u3 − 8u2
− 24u2 + 128u
24u2 + 16u
thus final asnwers is
4
− 6y
3
− 8y
4.
2
+y−2+−
− 10y2
− 9y − 3
144u + 98
− 144u − 96
3
−6y2 − 8y + 3
− 2y
−3
2
thus final asnwers is
90y3 + 48y2 + 33y + 12
− 90y3 − 30y2
18y2 + 33y
− 18y2 − 6y
27y + 12
− 27y − 9
−10y
2
−12u − 8
8.
10y3
4y − 10
− 2y − 3 +
2
4
2
−11u − u + 2u − 12 +
3
thus final asnwers is
+ 2u − 12
132u5 + 88u4 + 12u3 − 16u2 + 128u + 98
− 132u5 − 88u4
−3
−5y
9
10x + 5
+ 7y2
− 2y
+4
40y4 − 72y3 − 78y2 + 36y − 34
− 40y4 + 100y3
3
28y3 − 78y2
− 28y3 + 70y2
−9y − 3
− 8y2 + 36y
8y2 − 20y
5.
− 2y3 + 8y2
−7
6y − 11 − 12y4 + 70y3 − 88y2 − 42y + 71
4
3
12y − 22y
48y3 − 88y2
− 48y3 + 88y2
16y − 34
− 16y + 40
6
thus final asnwers is
− 42y + 71
42y − 77
−6
10y
3
+ 7y
2
− 2y + 4 +
6
4y − 10
68
Answers for Section 4.5
13.
9.
3u − 6
− 3u − 4 − 9u2 + 6u + 23
9u2 + 12u
4u3 − 2u2
− 6u − 12 − 24u4 − 36u3 − 48u2
24u4 + 48u3
18u + 23
− 18u − 24
−8
12u3 − 48u2
− 12u3 − 24u2
−1
thus final asnwers is
3u − 6 + −
+ 12u
− 96u + 88
− 72u2 − 96u
72u2 + 144u
1
−3u − 4
48u + 88
− 48u − 96
10.
− 5y4
−y+2
− 6y2 + 12y
−8
+9
5y5 − 10y4 + 6y3 − 24y2 + 15y + 25
− 5y5 + 10y4
thus final asnwers is
6y3 − 24y2
− 6y3 + 12y2
3
2
4u − 2u + 12u − 8 + −
− 12y2 + 15y
12y2 − 24y
− 9y + 25
9y − 18
8
−6u − 12
14.
7
− 11y
thus final asnwers is
−5y4 − 6y2 + 12y + 9 +
− 9y2 + 3
7
2−y
− 63y2
63y2
11.
− 11y4
− 9y − 2
− 6y3
− 8y2
+ 3y
+ 17
− 21
−4
+9
99y5 + 76y4 + 84y3 − 11y2 − 87y − 25
− 99y5 − 22y4
thus final asnwers is
54y4 + 84y3
− 54y4 − 12y3
7 − 11y + −
72y3 − 11y2
− 72y3 − 16y2
− 27y2 − 87y
27y2 + 6y
+7
99y3 − 63y2 − 33y + 17
− 99y3
+ 33y
4
3 − 9y2
15.
− 81y − 25
81y + 18
−7
thus final asnwers is
−11y
4
− 6y
3
− 8y
2
+ 3y + 9 + −
7
− 11u4 − 5u3 + 7u2 + 3u − 4
5u − 11 − 55u5 + 96u4 + 90u3 − 62u2 − 53u + 45
55u5 − 121u4
− 25u4 + 90u3
25u4 − 55u3
−9y − 2
35u3 − 62u2
− 35u3 + 77u2
12.
7y4
+ 5y2 + 8y + 4
− 4y + 9 − 28y5 + 63y4 − 20y3 + 13y2 + 56y + 28
28y5 − 63y4
15u2 − 53u
− 15u2 + 33u
− 20u + 45
20u − 44
− 20y3 + 13y2
20y3 − 45y2
1
− 32y2 + 56y
32y2 − 72y
thus final asnwers is
− 16y + 28
16y − 36
−8
4
3
2
−11u − 5u + 7u + 3u − 4 +
1
5u − 11
thus final asnwers is
7y
4
+ 5y
2
+ 8y + 4 + −
8
9 − 4y
69
Answers for Section 4.6
Section 4.6
7.
3
5
11u u
1.
11
6
3
8t
− 11t − 7t
3
3 8
3 3
+ 8 t t − 11 t t
=−7 t
3
2
− 12u u
3
+7 u
5
2
3
= 11u − 12u + 7
u
(given)
(given)
(DL, BI)
(Bi, factor gcd)
8.
8
3
3
= 8t − 11t − 7
t
(DL, BI)
9t8 + 5t5 − 4t3
2.
(given)
3
3 5
3 2
+9 t t +5 t t
=−4 t
5
4
3u − 5u
4
4
=3 u u − 5 u
4
= (3u − 5) u
(Bi, factor gcd)
(given)
5
2
3
= 9t + 5t − 4
t
(DL, BI)
(Bi, factor gcd)
(DL, BI)
9.
− 6y
12
+ 4y
7
+ 7y
4
(given)
3.
12
7
4
9t
+ 7t + 8t
4
3 4
8 4
+ 7t t
+8 t
=9t t
8
3
4
= 9t + 7t + 8
t
(given)
(Bi, factor gcd)
(DL, BI)
4
4
4
8
3
−6 y
y +4 y
y
=7 y
(Bi, factor gcd)
4
8
3
y
= −6y + 4y + 7
(DL, BI)
10.
5
2
12t − 2t + 7t
4.
6
3
− 8t + 11t + 10t
(given)
5
2
=10(t) − 8(t)t + 11(t)t
(Bi, factor gcd)
5
2
= −8t + 11t + 10 (t)
(DL, BI)
4
=7(t) + 12(t)t − 2(t)t
(given)
(Bi, factor gcd)
4
= 12t − 2t + 7 (t)
(DL, BI)
11.
5.
12y
7
− 7y
4
− 8y
2
2
2
2
5
2
=12 y
y −8 y
−7 y
y
5
2
2
= 12y − 7y − 8
y
(given)
(Bi, factor gcd)
(DL, BI)
3
3
3
5
2
12 x + 8 + 8 x + 8 u − x + 8 u
5
2
3
= 8u − u + 12
x +8
3
3
−9 t u
−5 t
= (−9u − 5) t3
(given)
(DL, BI)
(DL, BI)
12.
12
7
4
t
+ 12t + 8t
6.
(given)
4
4 3
4 8
+ t t + 12 t t
=8 t
t4
= t8 + 12t3 + 8
(given)
(Bi, factor gcd)
(DL, BI)
70
Answers for Section 4.6
20.
13.
3(5x + 12)u − 8(5x + 12)
(given)
= (3u − 8) (5x + 12)
3
2
5
3
3
− 2t x
− 11 x
7t x
(given)
(DL, BI)
5
2
3
= 7t − 2t − 11
x
(DL, BI)
14.
7t8 + 3t5 − 8t3
5 3
3
2 3
=7t t
+ 3t t
−8 t
(given)
21.
2 x3 − 4 x3 t4 + x3 t
(Bi, factor gcd)
5
2
3
= 7t + 3t − 8
t
x3
= −4t4 + t + 2
(DL, BI)
(given)
(DL, BI)
22.
15.
2 2
2
2
5u ω + uω − 8ω
(given)
2
2
= 5u + u − 8
ω
(DL, BI)
16.
− 7λ
4
4
− 8λ x
4
= (−8x − 7) λ
(given)
(DL, BI)
23.
y
4
− 12y
5
(given)
− 11θ
4
4
− 12 y
y
= y
(Bi, factor gcd)
4
= (1 − 12y) y
2
2
− 2θ x
2
= (−2x − 11) θ
(given)
(DL, BI)
(DL, BI)
24.
17.
8
3
− 7(7y + 11) + 4(7y + 11)t + 6(7y + 11)t
2
7t − 9t
(given)
(given)
=7(t)t − 9(t)
(Bi, factor gcd)
= (7t − 9) (t)
(DL, BI)
8
3
= 4t + 6t − 7 (7y + 11)
(DL, BI)
25.
18.
2
2 2
2
− 4µ − 6µ t − 3µ t
4
4 3
4 2
4
10µ + 8µ x − 12µ x + 3µ x
= 8x
3
− 12x
2
4
µ
+ 3x + 10
(given)
(given)
2
2
µ
= −6t − 3t − 4
(DL, BI)
(DL, BI)
26.
19.
11t8 (11u + 2) − 7t3 (11u + 2) − (11u + 2)
11x
2
+ 4x
(given)
(given)
= 11t8 − 7t3 − 1 (11u + 2)
(DL, BI)
=4(x) + 11(x)x
(Bi, factor gcd)
= (11x + 4) (x)
(DL, BI)
71
Answers for Section 4.7
2.
27.
7
4
2
− 10u + 9u + 11u
(given)
2
3
5
12 + 4ρ + 3ρ + ρ
2
2
2
5
2
− 10 u u + 9 u u
=11 u
(given)
(Bi, factor gcd)
5
2
2
= −10u + 9u + 11
u
2
3
5
= 12 + 4ρ
+ 3ρ + ρ
(DL, BI)
(ALA)
h
i
h i
2
3
2
3
= 3 (4) + ρ (4) + 3 ρ
+ρ
ρ
28.
4
3
11u + 6u
(given)
3
3
=6 u
+ 11 u u
(Bi, factor gcd)
3
= (11u + 6) u
(BI)
2
2
3
= 3+ρ
(4) + 3 + ρ
ρ
(DL)
2
3
= 3+ρ
4+ρ
(DL)
(DL, BI)
3.
29.
6
4
6u x
6
6
− 9u x
− 10 x
(given)
1 + −t3 + −t2 + t5
4
6
= 6u − 9u − 10
x
(DL, BI)
2
3
5
=1 + −t + −t + t
2
= 1 + −t
30.
2
2
2
5
2
+ 11 y
x +9 y
x
12 y
5
2
2
= 11x + 9x + 12
y
(given)
(DL, BI)
(given)
+
(ColA, ALA)
3
5
−t + t
(ALA)
h
i
h
i
2
3
2
3
= −1 (−1) + t (−1) + −1 t
+t
t
(BI)
= −1 + t2 (−1) + −1 + t2
t3
(DL)
= −1 + t2
−1 + t3
(DL)
Section 4.7
4.
1.
2 + −4x
2
+ −4x
3
+ 8x
5
(given)
4 + 3Q
2
+
= −2 (−1) + 4x
2
= 2 + −4x
h
−4x
3
+ 8x
i
h
5
(−1) + −2 2x
(ALA)
2
= 4 + 3Q
3
+ 4x
2
2x
3
i
(BI)
+ 4Q
2
+
3
+ 3Q
4Q
3
5
+ 3Q
(given)
5
h
i
h i
2
3
2
3
= 4 (1) + 3Q (1) + 4 Q
+ 3Q
Q
(ALA)
(BI)
2
2
3
= −2 + 4x
(−1) + −2 + 4x
2x
(DL)
2
2
3
= 4 + 3Q
(1) + 4 + 3Q
Q
(DL)
2
3
= −2 + 4x
−1 + 2x
(DL)
2
3
= 4 + 3Q
1+Q
(DL)
72
Answers for Section 4.7
8.
5.
20 + 8w + 5w
= (20 + 8w) +
2
+ 2w
3
6 + 3v + 10v
(given)
2
3
5w + 2w
= (6 + 3v) +
(ALA)
h i
2
2
= [5 (4) + 2w (4)] + 5 w
+ 2w w
2
+ 5v
3
(given)
2
3
10v + 5v
(ALA)
h i
2
2
= [2 (3) + v (3)] + 2 5v
+ v 5v
(BI)
(BI)
2
= (5 + 2w) (4) + (5 + 2w) w
(DL)
2
= (2 + v) (3) + (2 + v) 5v
(DL)
2
= (5 + 2w) 4 + w
(DL)
2
= (2 + v) 3 + 5v
(DL)
9.
6.
2 + 10x + 2x
= (2 + 10x) +
2
+ 10x
2x
2
3
+ 10x
(given)
3
(ALA)
i
h 2
2
+ 5x 2x
= [1 (2) + 5x (2)] + 1 2x
(BI)
16 + 12θ
2
+ 16θ
= 16 + 12θ2
+
3
+ 12θ
5
(given)
16θ3 + 12θ5
h
i
h i
= 4 (4) + 3θ2 (4) + 4 4θ3 + 3θ2 4θ3
(ALA)
(BI)
2
= (1 + 5x) (2) + (1 + 5x) 2x
(DL)
3
2
2
4θ
(4) + 4 + 3θ
= 4 + 3θ
(DL)
2
= (1 + 5x) 2 + 2x
(DL)
2
3
= 4 + 3θ
4 + 4θ
(DL)
10.
7.
6 + 10v
2
+ 6v
3
+ 10v
5
− 2 + 5T
2
+ −8T
3
+ 20T
5
(given)
(given)
2
3
5
= 6 + 10v
+ 6v + 10v
h
i
h i
2
3
2
3
= 3 (2) + 5v (2) + 3 2v
+ 5v
2v
(ALA)
(BI)
2
3
5
= −2 + 5T
+ −8T + 20T
(ALA)
h
i
h
i
2
3
2
3
= −2 (1) + 5T (1) + −2 4T
+ 5T
4T
(BI)
2
2
3
= 3 + 5v
(2) + 3 + 5v
2v
(DL)
2
2
3
= −2 + 5T
(1) + −2 + 5T
4T
(DL)
2
3
= 3 + 5v
2 + 2v
(DL)
2
3
= −2 + 5T
1 + 4T
(DL)
73
Answers for Section 4.7
14.
11.
4 + −10β
10 + 20z + 10z
2
+ 20z
3
(given)
=4 + −6β
= (10 + 20z) +
10z
2
+ 20z
3
(ALA)
h i
2
2
= [2 (5) + 4z (5)] + 2 5z
+ 4z 5z
(BI)
3
2
+ −6β
2
+ −10β
3
+ 15β
+ 15β
5
5
(given)
(ColA, ALA)
= 4 + −6β 2 + −10β 3 + 15β 5
(ALA)
h
i
h
i
2
3
2
3
= −2 (−2) + 3β (−2) + −2 5β
+ 3β
5β
(BI)
2
= (2 + 4z) (5) + (2 + 4z) 5z
= (2 + 4z) 5 + 5z
2
(DL)
(DL)
= −2 + 3β 2 (−2) + −2 + 3β 2
5β 3
(DL)
2
3
= −2 + 3β
−2 + 5β
(DL)
15.
12.
3 + 15λ
15 + 3T
2
+ 15T
3
+ 3T
5
(given)
=3 + 5λ
= 15 + 3T
2
3
5
+ 15T + 3T
(ALA)
i
i
h 3
2
2
3
3T
+T
= 5 (3) + T (3) + 5 3T
h
2
2
= 5+T
= 5+T
(3) +
5+T
3 + 3T
3
2
3T
3
(BI)
(DL)
(DL)
3
2
+ 15λ
3
+ 25λ
+ 25λ
5
5
(given)
(ColA, ALA)
2
3
5
= 3 + 5λ
+ 15λ + 25λ
(ALA)
h
i
h i
2
3
2
3
= 3 (1) + 5λ (1) + 3 5λ
+ 5λ
5λ
(BI)
2
2
3
= 3 + 5λ
(1) + 3 + 5λ
5λ
(DL)
2
3
= 3 + 5λ
1 + 5λ
(DL)
12 + 12ξ
2
16.
13.
3 + 12x
+ 5λ
2
= 3 + 12x
+ 4x
2
+
3
+ 16x
4x
3
5
+ 16x
(given)
5
h
i
h i
2
3
2
3
= 1 (3) + 4x (3) + 1 4x
+ 4x
4x
(ALA)
(BI)
=12 + 6ξ
3
2
+ 6ξ
+ 12ξ
2
3
+ 6ξ
+ 6ξ
5
5
(given)
(ColA, ALA)
2
3
5
= 12 + 6ξ
+ 12ξ + 6ξ
h
i
h i
2
3
2
3
= 4 (3) + 2ξ (3) + 4 3ξ
+ 2ξ
3ξ
(ALA)
(BI)
2
2
3
= 1 + 4x
(3) + 1 + 4x
4x
(DL)
2
2
3
= 4 + 2ξ
(3) + 4 + 2ξ
3ξ
(DL)
2
3
= 1 + 4x
3 + 4x
(DL)
2
3
= 4 + 2ξ
3 + 3ξ
(DL)
74
Answers for Section 4.7
20.
17.
20 + 15q + 16q
= (20 + 15q) +
2
+ 12q
− 1 + −2U 3 + 4U 2 + 8U 5
3
2
3
16q + 12q
h i
2
2
= [4 (5) + 3q (5)] + 4 4q
+ 3q 4q
= (4 + 3q) (5) + (4 + 3q) 4q
(given)
(given)
2
2
= (4 + 3q) 5 + 4q
= − 1 + 4U
2
+ −2U
3
+ 8U
5
(ColA, ALA)
(ALA)
(BI)
(DL)
(DL)
2
3
5
= −1 + 4U
+ −2U + 8U
(ALA)
h
i
h
i
2
3
2
3
= −1 (1) + 4U (1) + −1 2U
+ 4U
2U
(BI)
2
2
3
= −1 + 4U
(1) + −1 + 4U
2U
(DL)
2
3
= −1 + 4U
1 + 2U
(DL)
18.
12 + 4φ
2
+ 12φ
3
+ 4φ
5
(given)
= 12 + 4φ2 + 12φ3 + 4φ5
h
i
h i
= 3 (4) + φ2 (4) + 3 4φ3 + φ2 4φ3
(ALA)
(BI)
3
2
2
4φ
(4) + 3 + φ
= 3+φ
(DL)
2
3
= 3+φ
4 + 4φ
(DL)
Section 4.8
1.
First take the product of the outside terms,
4λ2 · 24 = 96λ2 Then we look at all the way we
can factor this coefficent, 96 7→ as: ±[1, 96], ±[2, 48],
±[3, 32], ±[4, 24], ±[6, 16], ±[8, 12] We scan for a
suitable pair to split 20λ into. Idenfiying 20λ =
8λ + 12λ as a suitable candidate. Then we stop talking start the doing!
4λ
19.
=4λ
− 1 + 2λ
2
+ −4λ
3
+ 8λ
5
2
3
5
= −1 + 2λ
+ −4λ + 8λ
2
2
= 4λ
(given)
+ 20λ + 24
+ (8λ + 12λ) + 24
2
+ 8λ
+ (12λ + 24)
= (2λ + 4) 2λ + (2λ + 4) 6
(ALA)
= (2λ + 4) (2λ + 6)
(given)
(Bi, The Split!)
(ALA)
(DL, BI)
(DL)
h
i
h
i
2
3
2
3
= −1 (1) + 2λ (1) + −1 4λ
+ 2λ
4λ
(BI)
2
2
3
= −1 + 2λ
(1) + −1 + 2λ
4λ
(DL)
2
3
= −1 + 2λ
1 + 4λ
(DL)
2.
First take the product of the outside terms,
6x4 · 60 = 360x4 Then we look at all the way
we can factor this coefficent, 360 7→ as: ±[1, 360],
±[2, 180], ±[3, 120], ±[4, 90], ±[5, 72], ±[6, 60],
±[8, 45], ±[9, 40], ±[10, 36], ±[12, 30], ±[15, 24],
±[18, 20] We scan for a suitable pair to split 42x2
into. Idenfiying 42x2 = 30x2 + 12x2 as a suitable
75
Answers for Section 4.8
candidate. Then we stop talking start the doing!
6x
=6x
4
4
+ 42x
+
2
30x
+ 60
2
+ 12x
(given)
2
+ 60
(ALA)
= 2x2 + 10
3x2 + 6
(DL, BI)
(DL)
3.
First take the product of the outside terms,
z 2 · 4 = 4z 2 Then we look at all the way we can
factor this coefficent, 4 7→ as: ±[1, 4], ±[2, 2] We
scan for a suitable pair to split −4z into. Idenfiying −4z = −2z + −2z as a suitable candidate. Then
we stop talking start the doing!
z
2
+ −4z + 4
=z 2 + (−2z + −2z) + 4
6
=6µ +
(given)
3 2
3 2
4
12µ z + 18µ z
+ 36z
(Bi, The Split!)
= 2x2 + 10 3x2 + 2x2 + 10 6
6
3 2
4
6µ + 30µ z + 36z
(Bi, The Split!)
4
2
2
= 6x + 30x
+ 12x + 60
as: ±[1, 216], ±[2, 108], ±[3, 72], ±[4, 54], ±[6, 36],
±[8, 27], ±[9, 24], ±[12, 18] We scan for a suitable
pair to split 30µ3 z 2 into.
Idenfiying 30µ3 z 2 =
12µ3 z 2 + 18µ3 z 2 as a suitable candidate. Then we
stop talking start the doing!
(given)
(Bi, The Split!)
6
3 2
3 2
4
= 6µ + 12µ z
+ 18µ z + 36z
(ALA)
3
2
3
3
2
2
= 2µ + 4z
3µ + 2µ + 4z
9z
(DL, BI)
3
2
3
2
= 2µ + 4z
3µ + 9z
(DL)
6.
First take the product of the outside terms,
2ρ6 · 30r4 = 60ρ6 r4 Then we look at all the way
we can factor this coefficent, 60 7→ as: ±[1, 60],
±[2, 30], ±[3, 20], ±[4, 15], ±[5, 12], ±[6, 10] We scan
for a suitable pair to split 16ρ3 r2 into. Idenfiying
16ρ3 r2 = 6ρ3 r2 + 10ρ3 r2 as a suitable candidate.
Then we stop talking start the doing!
6
3 2
4
2ρ + 16ρ r + 30r
2
= z + −2z + (−2z + 4)
= (z + −2) z + (z + −2) − 2
= (z + −2) (z + −2)
(given)
(ALA)
6
=2ρ +
(DL, BI)
(DL)
3 2
3 2
6ρ r + 10ρ r
6
3 2
= 2ρ + 6ρ r
+
+ 30r
4
3 2
4
10ρ r + 30r
(Bi, The Split!)
3
2
3
3
2
2
= 2ρ + 6r
ρ + 2ρ + 6r
5r
4.
First take the product of the outside terms,
6β 2 · 8 = 48β 2 Then we look at all the way we can
factor this coefficent, 48 7→ as: ±[1, 48], ±[2, 24],
±[3, 16], ±[4, 12], ±[6, 8] We scan for a suitable pair
to split 26β into. Idenfiying 26β = 2β + 24β as a
suitable candidate. Then we stop talking start the
doing!
6β
=6β
2
2
+ 26β + 8
+ (2β + 24β) + 8
(given)
(Bi, The Split!)
3
2
3
2
= 2ρ + 6r
ρ + 5r
(ALA)
= (3β + 1) 2β + (3β + 1) 8
(DL, BI)
= (3β + 1) (2β + 8)
2
2
+ 27q + 55
+ (22q + 5q) + 55
2
= 2q + 22q + (5q + 55)
= (q + 11) 2q + (q + 11) 5
5.
First take the product of the outside terms,
6µ6 · 36z 4
=
216µ6 z 4 Then we look at all
the way we can factor this coefficent, 216 7→
(DL)
First take the product of the outside terms,
2q2 · 55 = 110q2 Then we look at all the way we can
factor this coefficent, 110 7→ as: ±[1, 110], ±[2, 55],
±[5, 22], ±[10, 11] We scan for a suitable pair to split
27q into. Idenfiying 27q = 22q + 5q as a suitable
candidate. Then we stop talking start the doing!
=2q
(DL)
(DL, BI)
7.
2q
= 6β 2 + 2β + (24β + 8)
(ALA)
= (q + 11) (2q + 5)
(given)
(Bi, The Split!)
(ALA)
(DL, BI)
(DL)
76
Answers for Section 4.8
8.
First take the product of the outside terms,
2x4 · 45 = 90x4 Then we look at all the way we
can factor this coefficent, 90 7→ as: ±[1, 90], ±[2, 45],
±[3, 30], ±[5, 18], ±[6, 15], ±[9, 10] We scan for a
suitable pair to split 23x2 into. Idenfiying 23x2 =
18x2 + 5x2 as a suitable candidate. Then we stop
talking start the doing!
start the doing!
2
t + 19t + 84
(given)
=t2 + (7t + 12t) + 84
(Bi, The Split!)
2
= t + 7t + (12t + 84)
2x
=2x
4
4
+ 23x
+
2
18x
+ 45
2
+ 5x
= (t + 7) t + (t + 7) 12
(given)
2
2
2
2
= x + 9 2x + x + 9 5
2
2
= x +9
2x + 5
(ALA)
(DL, BI)
(DL)
11.
First take the product of the outside terms,
3µ6 · 3t4 = 9µ6 t4 Then we look at all the way we
can factor this coefficent, 9 7→ as: ±[1, 9], ±[3, 3] We
scan for a suitable pair to split 10µ3 t2 into. Idenfiying 10µ3 t2 = µ3 t2 + 9µ3 t2 as a suitable candidate.
Then we stop talking start the doing!
6
3 2
4
3µ + 10µ t + 3t
=3µ6 +
9.
First take the product of the outside terms,
3γ 6 · 30q4 = 90γ 6 q4 Then we look at all the way
we can factor this coefficent, 90 7→ as: ±[1, 90],
±[2, 45], ±[3, 30], ±[5, 18], ±[6, 15], ±[9, 10] We scan
for a suitable pair to split 21γ 3 q2 into. Idenfiying
21γ 3 q2 = 6γ 3 q2 + 15γ 3 q2 as a suitable candidate.
Then we stop talking start the doing!
=3γ
6
3 2
4
+ 21γ q + 30q
+
(given)
4
3 2
3 2
+ 30q
(Bi, The Split!)
6γ q + 15γ q
6
3 2
3 2
4
= 3γ + 6γ q
+ 15γ q + 30q
2
3
3
2
3
2
5q
γ + 3γ + 6q
= 3γ + 6q
(ALA)
µ3 t2 + 9µ3 t2
(given)
+ 3t4
(Bi, The Split!)
6
3 2
3 2
4
= 3µ + µ t
+ 9µ t + 3t
(ALA)
3
2
3
3
2
2
= 3µ + t
µ + 3µ + t
3t
3
2
3
2
µ + 3t
= 3µ + t
(DL, BI)
(DL)
12.
First take the product of the outside terms,
6x2 · −120 = −720x2 Then we look at all the way
we can factor this coefficent, −720 7→ as: ±[1, 720],
±[2, 360], ±[3, 240], ±[4, 180], ±[5, 144], ±[6, 120],
±[8, 90], ±[9, 80], ±[10, 72], ±[12, 60], ±[15, 48],
±[16, 45], ±[18, 40], ±[20, 36], ±[24, 30] We scan for
a suitable pair to split −6x into. Idenfiying −6x =
24x + −30x as a suitable candidate. Then we stop
talking start the doing!
6x
2
+ −6x + −120
2
+ (24x + −30x) + −120
(given)
(DL, BI)
=6x
3
2
3
2
= 3γ + 6q
γ + 5q
(DL)
(Bi, The Split!)
6
(DL, BI)
= (t + 7) (t + 12)
+ 45
4
2
2
= 2x + 18x
+ 5x + 45
3γ
(ALA)
(Bi, The Split!)
(DL)
2
= 6x + 24x + (−30x + −120)
= (3x + 12) 2x + (3x + 12) − 10
= (3x + 12) (2x + −10)
(ALA)
(DL, BI)
(DL)
10.
First take the product of the outside terms,
t2 · 84 = 84t2 Then we look at all the way we can
factor this coefficent, 84 7→ as: ±[1, 84], ±[2, 42],
±[3, 28], ±[4, 21], ±[6, 14], ±[7, 12] We scan for a
suitable pair to split 19t into.
Idenfiying 19t =
7t + 12t as a suitable candidate. Then we stop talking
13.
First take the product of the outside terms,
2λ2 · 21 = 42λ2 Then we look at all the way we
77
Answers for Section 4.8
can factor this coefficent, 42 7→ as: ±[1, 42], ±[2, 21],
±[3, 14], ±[6, 7] We scan for a suitable pair to split
17λ into. Idenfiying 17λ = 14λ + 3λ as a suitable
candidate. Then we stop talking start the doing!
2λ
=2λ
2
+ 17λ + 21
2
= 2λ
+ (14λ + 3λ) + 21
2
+ 14λ
scan for a suitable pair to split 5v into. Idenfiying
5v = 6v + −v as a suitable candidate. Then we stop
talking start the doing!
− 2v
2
= − 2v
2
+ 5v + 3
(given)
(given)
+ (6v + −v) + 3
(Bi, The Split!)
(Bi, The Split!)
+ (3λ + 21)
= −2v2 + 6v + (−v + 3)
(ALA)
(ALA)
= (v + −3) − 2v + (v + −3) − 1
= (λ + 7) 2λ + (λ + 7) 3
(DL, BI)
(DL, BI)
= (v + −3) (−2v + −1)
= (λ + 7) (2λ + 3)
(DL)
(DL)
17.
14.
First take the product of the outside terms,
2x2 · 121 = 242x2 Then we look at all the way we can
factor this coefficent, 242 7→ as: ±[1, 242], ±[2, 121],
±[11, 22] We scan for a suitable pair to split 33x into.
Idenfiying 33x = 22x + 11x as a suitable candidate.
Then we stop talking start the doing!
2x
2
+ 33x + 121
First take the product of the outside terms,
4q2 · 8 = 32q2 Then we look at all the way we can factor this coefficent, 32 7→ as: ±[1, 32], ±[2, 16], ±[4, 8]
We scan for a suitable pair to split 12q into. Idenfiying 12q = 4q + 8q as a suitable candidate. Then we
stop talking start the doing!
4q
(given)
=4q
=2x
2
+ (22x + 11x) + 121
= 2x
2
+ 22x
(Bi, The Split!)
+ (11x + 121)
2
2
+ 12q + 8
+ (4q + 8q) + 8
= 4q2 + 4q + (8q + 8)
(given)
(Bi, The Split!)
(ALA)
(ALA)
= (2q + 2) 2q + (2q + 2) 4
= (x + 11) 2x + (x + 11) 11
(DL, BI)
= (x + 11) (2x + 11)
= (2q + 2) (2q + 4)
(DL, BI)
(DL)
(DL)
18.
15.
First take the product of the outside terms,
−w2 · 12 = −12w2 Then we look at all the way we
can factor this coefficent, −12 7→ as: ±[1, 12], ±[2, 6],
±[3, 4] We scan for a suitable pair to split 11w into.
Idenfiying 11w = 12w + −w as a suitable candidate.
Then we stop talking start the doing!
First take the product of the outside
terms, 2w2 · 60 = 120w2 Then we look at all
the way we can factor this coefficent, 120
7→
as: ±[1, 120], ±[2, 60], ±[3, 40], ±[4, 30], ±[5, 24],
±[6, 20], ±[8, 15], ±[10, 12] We scan for a suitable
pair to split 22w into. Idenfiying 22w = 10w + 12w
as a suitable candidate. Then we stop talking start
the doing!
2w
−w
2
=−w
2
+ 11w + 12
=2w
+ (12w + −w) + 12
2
+ 22w + 60
2
+ (10w + 12w) + 60
2
= 2w + 10w + (12w + 60)
(ALA)
= (w + 5) 2w + (w + 5) 12
= (w + −12) − w + (w + −12) − 1
(ALA)
(DL, BI)
(DL, BI)
= (w + 5) (2w + 12)
(DL)
(DL)
19.
16.
(Bi, The Split!)
(Bi, The Split!)
2
= −w + 12w + (−w + 12)
= (w + −12) (−w + −1)
(given)
(given)
First take the product of the outside terms,
−2v2 · 3 = −6v2 Then we look at all the way we can
factor this coefficent, −6 7→ as: ±[1, 6], ±[2, 3] We
First take the product of the outside terms,
6r4 · 35
=
210r4 Then we look at all the
way we can factor this coefficent, 210
7→ as:
±[1, 210], ±[2, 105], ±[3, 70], ±[5, 42], ±[6, 35],
±[7, 30], ±[10, 21], ±[14, 15] We scan for a suitable
78
Answers for Section 4.8
pair to split 31r2 into. Idenfiying 31r2 = 21r2 +10r2
as a suitable candidate. Then we stop talking start
the doing!
6r
=6r
4
4
+ 31r
+
2
21r
+ 35
2
(given)
+ 10r
2
+ 35
22.
First take the product of the outside terms,
3λ2 ·6 = 18λ2 Then we look at all the way we can factor this coefficent, 18 7→ as: ±[1, 18], ±[2, 9], ±[3, 6]
We scan for a suitable pair to split 19λ into. Idenfiying 19λ = 18λ + λ as a suitable candidate. Then we
stop talking start the doing!
3λ
(Bi, The Split!)
4
2
2
= 6r + 21r
+ 10r + 35
=3λ
(ALA)
2
2
2
= 2r + 7 3r + 2r + 7 5
2
2
+ 19λ + 6
(given)
+ (18λ + λ) + 6
2
= 3λ + 18λ + (λ + 6)
(DL, BI)
= (λ + 6) 3λ + (λ + 6) 1
2
2
= 2r + 7
3r + 5
= (λ + 6) (3λ + 1)
First take the product of the outside terms,
3ω6 · 40t4
=
120t4 ω6 Then we look at all
the way we can factor this coefficent, 120
7→
as: ±[1, 120], ±[2, 60], ±[3, 40], ±[4, 30], ±[5, 24],
±[6, 20], ±[8, 15], ±[10, 12] We scan for a suitable
pair to split 23t2 ω3 into.
Idenfiying 23t2 ω3 =
8t2 ω3 + 15t2 ω3 as a suitable candidate. Then we
stop talking start the doing!
=3ω
6
2 3
4
+ 23t ω + 40t
6
4
2 3
2 3
+ 40t
+ 8t ω + 15t ω
6
3
3
= 3ω
= 3ω
= 3ω
2 3
+ 8t ω
2 3
4
15t ω + 40t
2
+ 8t
2
+ 8t
3
2
ω + 5t
+
3ω
3
2
+ 8t
2
5t
=2β
2
2
+ 19β + 42
+ (7β + 12β) + 42
2
= 2β + 7β + (12β + 42)
= (2β + 7) β + (2β + 7) 6
= (2β + 7) (β + 6)
as a suitable candidate. Then we stop talking start
the doing!
4
2
2t + 25t + 72
4
=2t +
(ALA)
+
+ 72
9t2 + 72
(Bi, The Split!)
(ALA)
(DL, BI)
(DL)
24.
First take the product of the outside terms,
2z 2 ·−12 = −24z 2 Then we look at all the way we can
factor this coefficent, −24 7→ as: ±[1, 24], ±[2, 12],
±[3, 8], ±[4, 6] We scan for a suitable pair to split
−2z into. Idenfiying −2z = 4z + −6z as a suitable
candidate. Then we stop talking start the doing!
2z
(Bi, The Split!)
=2z
(DL)
2
+ 9t
2
2
= t +8
2t + 9
(given)
(DL, BI)
(given)
= t2 + 8 2t2 + t2 + 8 9
(DL)
(ALA)
2
16t
= 2t4 + 16t2
(DL, BI)
First take the product of the outside terms,
2β 2 · 42 = 84β 2 Then we look at all the way we can
factor this coefficent, 84 7→ as: ±[1, 84], ±[2, 42],
±[3, 28], ±[4, 21], ±[6, 14], ±[7, 12] We scan for a
suitable pair to split 19β into. Idenfiying 19β =
7β + 12β as a suitable candidate. Then we stop talking start the doing!
(DL)
First take the product of the outside
terms, 2t4 · 72 = 144t4 Then we look at all
the way we can factor this coefficent, 144 7→
as: ±[1, 144], ±[2, 72], ±[3, 48], ±[4, 36], ±[6, 24],
±[8, 18], ±[9, 16], ±[12, 12] We scan for a suitable
pair to split 25t2 into. Idenfiying 25t2 = 16t2 + 9t2
(Bi, The Split!)
21.
2β
(DL, BI)
23.
(given)
ω
3
+
(ALA)
(DL)
20.
3ω
(Bi, The Split!)
2
+ −2z + −12
2
+ (4z + −6z) + −12
2
= 2z + 4z + (−6z + −12)
= (z + 2) 2z + (z + 2) − 6
= (z + 2) (2z + −6)
(given)
(Bi, The Split!)
(ALA)
(DL, BI)
(DL)
79
Answers for Section 4.8
25.
First take the product of the outside terms,
3γ 2 · 1 = 3γ 2 Then we look at all the way we can
factor this coefficent, 3 7→ as: ±[1, 3] We scan for a
suitable pair to split 4γ into. Idenfiying 4γ = γ + 3γ
as a suitable candidate. Then we stop talking start
the doing!
3γ
2
+ 4γ + 1
= 3γ
2
+γ
First take the product of the outside terms,
−r2 · −18 = 18r2 Then we look at all the way we
can factor this coefficent, 18 7→ as: ±[1, 18], ±[2, 9],
±[3, 6] We scan for a suitable pair to split −9r into.
Idenfiying −9r = −3r + −6r as a suitable candidate.
Then we stop talking start the doing!
(given)
=3γ 2 + (γ + 3γ) + 1
28.
−r
2
+ −9r + −18
=−r
2
+ (−3r + −6r) + −18
(Bi, The Split!)
+ (3γ + 1)
(ALA)
= (3γ + 1) γ + (3γ + 1) 1
(DL, BI)
= (3γ + 1) (γ + 1)
2
= −r + −3r + (−6r + −18)
= (−r + −3) r + (−r + −3) 6
(given)
(Bi, The Split!)
(ALA)
(DL, BI)
(DL)
= (−r + −3) (r + 6)
(DL)
26.
First take the product of the outside terms,
6θ6 · 45q4
=
270θ6 q4 Then we look at all
the way we can factor this coefficent, 270 7→
as: ±[1, 270], ±[2, 135], ±[3, 90], ±[5, 54], ±[6, 45],
±[9, 30], ±[10, 27], ±[15, 18] We scan for a suitable
pair to split 33θ3 q2 into.
Idenfiying 33θ3 q2 =
18θ3 q2 + 15θ3 q2 as a suitable candidate. Then we
stop talking start the doing!
6θ
=6θ
6
3 2
4
+ 33θ q + 45q
6
+
29.
First take the product of the outside terms,
3µ2 · 2 = 6µ2 Then we look at all the way we can
factor this coefficent, 6 7→ as: ±[1, 6], ±[2, 3] We
scan for a suitable pair to split 5µ into. Idenfiying
5µ = 3µ + 2µ as a suitable candidate. Then we stop
talking start the doing!
2
3µ + 5µ + 2
(given)
3 2
3 2
18θ q + 15θ q
+ 45q
4
2
=3µ + (3µ + 2µ) + 2
2
= 3µ + 3µ + (2µ + 2)
(ALA)
= (µ + 1) 3µ + (µ + 1) 2
3
2
3
3
2
2
= 3θ + 9q
2θ + 3θ + 9q
5q
= 3θ
3
+ 9q
2
2θ
3
+ 5q
2
(Bi, The Split!)
(Bi, The Split!)
6
3 2
3 2
4
= 6θ + 18θ q
+ 15θ q + 45q
(given)
(ALA)
(DL, BI)
(DL, BI)
= (µ + 1) (3µ + 2)
(DL)
(DL)
30.
27.
First take the product of the outside terms,
4x2 · 44 = 176x2 Then we look at all the way
we can factor this coefficent, 176 7→ as: ±[1, 176],
±[2, 88], ±[4, 44], ±[8, 22], ±[11, 16] We scan for a
suitable pair to split −30x into. Idenfiying −30x =
−8x + −22x as a suitable candidate. Then we stop
talking start the doing!
4x
=4x
2
2
= 4x
+ −30x + 44
+ (−8x + −22x) + 44
2
+ −8x
− 2v
2
+ 15v + −27
= − 2v
2
+ (9v + 6v) + −27
(given)
(Bi, The Split!)
+ (−22x + 44)
= (2x + −4) 2x + (2x + −4) − 11
= (2x + −4) (2x + −11)
First take the product of the outside terms,
−2v2 · −27 = 54v2 Then we look at all the way we
can factor this coefficent, 54 7→ as: ±[1, 54], ±[2, 27],
±[3, 18], ±[6, 9] We scan for a suitable pair to split
15v into. Idenfiying 15v = 9v + 6v as a suitable candidate. Then we stop talking start the doing!
2
= −2v + 9v + (6v + −27)
(given)
(Bi, The Split!)
(ALA)
(ALA)
= (−2v + 9) v + (−2v + 9) − 3
(DL, BI)
(DL)
= (−2v + 9) (v + −3)
(DL, BI)
(DL)
80
Answers for Section 4.9
Section 4.9
6.
3
( + △)
1.
3
=
3
2
2
+ △ = ( + △) − △ + △
3
+ −216
2
2
3
+ 3 △ + 3△ + △
..recall the pp3 Pattern)
(
..recall the SC Pattern)
(
27v
3
θ
(given)
3
+ 6θ
= (θ)
3
3
+ (−6)
(Bi, see SC pattern)
h
i
2
2
= (3v + −6) (3v) − (3v) (−6) + (−6)
= (3v)
3
2
+ +12θ + 8
+ 3 · (θ)
2
(given)
2
2
(2) + 3 · (θ) (2) + (2)
(Bi, seePP3 pattern)
3
= (θ + 2)
(PP3)
(SC)
h
i
2
= (3v + −6) 9v + 18v + 36
(BI)
7.
2
2
2
( + △) = + 2△ + △
2.
..recall the pp2 Pattern)
(
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
(
9w
2
21w
+
+
2
= (3w)
2
49
(given)
7
+
+ 54V + 81
= (3V )
16
+ 2 · (3w)
2
9V
2
(given)
2
+ 2 · (3V ) (9) + (9)
(Bi, identified PP2 pattern)
= (3V + 9)2
7 2
4
4
(Bi, identified PP2 pattern)
7 2
(PP2)
= 3w +
4
(PP2)
8.
2
2
− △ = ( − △)( + △)
..recall the DS Pattern)
(
3.
4v
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
= (2t)
2
2
− (5)
(given)
2
(Bi, see DS pattern)
= (2v + 5) (2v − 5)
(DS)
(given)
2
+ 2 · (2t) (1) + (1)
(Bi, identified PP2 pattern)
= (2t + 1)
+ −25
= (2v)
(
2
4t + 4t + 1
2
2
(PP2)
9.
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
4.
(
x
3
( + △)
=
3
2
+ 3 △ + 3△
2
+△
2
+ 2x + 1
..recall the pp3 Pattern)
(
y
3
+ −30y
= (y)
3
2
+ +300y + −1000
+ 3 · (y)
= (y + −10)
2
= (x + 1)
2
(PP2)
(given)
2
2
(−10) + 3 · (y) (−10) + (−10)
(Bi, seePP3 pattern)
3
(given)
2
2
= (x) + 2 · (x) (1) + (1)
(Bi, identified PP2 pattern)
3
(PP3)
10.
( + △)2 = 2 + 2△ + △2
5.
..recall the pp2 Pattern)
(
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
(
4z
2
+ 16z + 16
(given)
2
2
= (2z) + 2 · (2z) (4) + (4)
(Bi, identified PP2 pattern)
= (2z + 4)
2
(PP2)
4µ
1
4µ2 +
+
7
49
= (2µ)
2
+ 2 · (2µ)
(given)
1
1 2
+
7
7
(Bi, identified PP2 pattern)
2
1
(PP2)
= 2µ +
7
81
Answers for Section 4.9
11.
17.
3
( + △)
= 3
3
( + △)
2
2
3
+ 3 △ + 3△ + △
..recall the pp3 Pattern)
(
27λ
3
= (3λ)
+ −81λ
3
2
+ +81λ + −27
+ 3 · (3λ)
= (3λ + −3)
2
(given)
2
2
(−3) + 3 · (3λ) (−3) + (−3)
(Bi, seePP3 pattern)
3
(PP3)
3
2
2
3
+ 3 △ + 3△ + △
3
+ −135w
= ..recall the pp3 Pattern)
(
27w
2
+ +225w + −125
(given)
3
2
2
2
= (3w) + 3 · (3w) (−5) + 3 · (3w) (−5) + (−5)
(Bi, seePP3 pattern)
= (3w + −5)
3
(PP3)
12.
2
18.
2
− △ = ( − △)( + △)
..recall the DS Pattern)
(
x
2
+ −81
( + △)3
(given)
2
2
= (x) − (9)
=
(Bi, see DS pattern)
= (x + 9) (x − 9)
3
..recall the pp3 Pattern)
(
(DS)
8ω
3
+ 108ω
= (2ω)
13.
= 3
2
+ +486ω + 729
+ 3 · (2ω)
2
2
+ 3 △ + 3△
2
+△
(given)
2
2
(9) + 3 · (2ω) (9) + (9)
(Bi, seePP3 pattern)
3
= (2ω + 9)
3
( + △)
3
2
2
3
+ 3 △ + 3△ + △
(PP3)
3
..recall the pp3 Pattern)
(
8q
3
+ 72q
2
+ +216q + 216
(given)
19.
= (2q)3 + 3 · (2q) 2 (6) + 3 · (2q) (6)2 + (6)2
(Bi, seePP3 pattern)
3
( + △)
3
= (2q + 6)
= (PP3)
3
2
2
3
+ 3 △ + 3△ + △
..recall the pp3 Pattern)
(
3
2
27µ + 216µ + +576µ + 512
14.
= (3µ)
2
3
+ 3 · (3µ)
2
2
− △ = ( − △)( + △)
..recall the DS Pattern)
= (3µ + 8)3
(
v
2
+ −4
= (v)
2
(given)
2
2
(8) + 3 · (3µ) (8) + (8)
(Bi, seePP3 pattern)
(PP3)
(given)
2
− (2)
(Bi, see DS pattern)
= (v + 2) (v − 2)
(DS)
20.
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
15.
(
( + △)2 = 2 + 2△ + △2
9r
..recall the pp2 Pattern)
(
2
9r
+ 42r + 49
= (3r)
2
(given)
2
(PP2)
16.
= (X)
2
(given)
2
+ 2 · (X) (5) + (5)
(Bi, identified PP2 pattern)
= (X + 5)
2
(PP2)
3
3
2
2
+ △ = ( + △) − △ + △
(
..recall the pp2 Pattern)
(
+ 10X + 25
(given)
2
+ 2 · (3r) (6) + (6)
(Bi, identified PP2 pattern)
..recall the SC Pattern)
2
2
2
( + △) = + 2△ + △
2
2
21.
X
+ 36r + 36
2
= (3r + 6)
2
+ 2 · (3r) (7) + (7)
(Bi, identified PP2 pattern)
= (3r + 7)
2
= (3r)
(PP2)
X
3
+ 1000
3
(given)
3
+ (10)
(Bi, see SC pattern)
h
i
2
2
= (X + 10) (X) − (X) (10) + (10)
(SC)
h
i
= (X + 10) X 2 − 10X + 100
(BI)
= (X)
82
Answers for Section 4.9
27.
22.
2
2
2
( + △) = + 2△ + △
3
( + △)
..recall the pp2 Pattern)
(
2
V
+ 4V + 4
= 3 + 32 △ + 3△2 + △3
..recall the pp3 Pattern)
(given)
(
2
2
= (V ) + 2 · (V ) (2) + (2)
(Bi, identified PP2 pattern)
= (V + 2)
R
3
+ 12R
= (R)
2
3
(PP2)
2
+ +48R + 64
(given)
2
2
2
+ 3 · (R) (4) + 3 · (R) (4) + (4)
(Bi, seePP3 pattern)
= (R + 4)
3
(PP3)
23.
28.
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
(
9Y
2
+ 60Y + 100
= (3Y )
2
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
(given)
(
2
+ 2 · (3Y ) (10) + (10)
(Bi, identified PP2 pattern)
= (3Y + 10)
2
9u + 60u + 100
= (3u)
2
2
(PP2)
(given)
2
+ 2 · (3u) (10) + (10)
(Bi, identified PP2 pattern)
= (3u + 10)
2
(PP2)
24.
29.
3
( + △)
= 3
2
+ 3 △ + 3△
2
+△
3
..recall the pp3 Pattern)
(
ξ
3
+ 30ξ
2
+ +300ξ + 1000
= (ξ + 10)
3
2
2
3
+ 3 △ + 3△ + △
3
+ −54W
..recall the pp3 Pattern)
(given)
= (ξ)3 + 3 · (ξ)2 (10) + 3 · (ξ) (10)2 + (10)2
(Bi, seePP3 pattern)
3
3
( + △)
=
(PP3)
(
27W
2
+ +36W + −8
(given)
3
2
2
2
= (3W ) + 3 · (3W ) (−2) + 3 · (3W ) (−2) + (−2)
(Bi, seePP3 pattern)
3
= (3W + −2)
(PP3)
25.
30.
3
( + △)
= 3
2
2
3
+ 3 △ + 3△ + △
2
2
2
( + △) = + 2△ + △
..recall the pp3 Pattern)
(
3
2
27ρ + −108ρ + +144ρ + −64
= (3ρ)
3
+ 3 · (3ρ)
= (3ρ + −4)
2
..recall the pp2 Pattern)
(
(given)
4W
2
2
(−4) + 3 · (3ρ) (−4) + (−4)
(Bi, seePP3 pattern)
3
3
..recall the SC Pattern)
(
3
27ρ + −512
(given)
3
+ (−8)
(Bi, see SC pattern)
h
i
= (3ρ + −8) (3ρ)2 − (3ρ) (−8) + (−8)2
= (3ρ)
3
(SC)
i
= (3ρ + −8) 9ρ2 + 24ρ + 64
h
(BI)
16W
2
+
16
(given)
81
+ 2 · (2W )
4
4 2
+
9
9
(Bi, identified PP2 pattern)
4 2
= 2W +
(PP2)
9
= (2W )
(PP3)
3
2
2
+ △ = ( + △) − △ + △
+
9
26.
2
31.
2
2
− △ = ( − △)( + △)
..recall the DS Pattern)
(
9x
2
+ −16
= (3x)
2
2
− (4)
= (3x + 4) (3x − 4)
(given)
(Bi, see DS pattern)
(DS)
83
Answers for Section 4.9
32.
2
2
2
( + △) = + 2△ + △
..recall the pp2 Pattern)
(
9q
2
+
14q
+
49
(given)
81
7
7 2
2
= (3q) + 2 · (3q)
+
9
9
(Bi, identified PP2 pattern)
7 2
(PP2)
= 3q +
9
3
33.
2
2
− △ = ( − △)( + △)
..recall the DS Pattern)
(
2
9t + −9
= (3t)
2
(given)
2
− (3)
(Bi, see DS pattern)
= (3t + 3) (3t − 3)
(DS)
34.
3
3
2
2
+ △ = ( + △) − △ + △
..recall the SC Pattern)
(
w
3
+ −216
= (w)
3
= (w +
3
+ (−6)
h
2
−6) (w)
(given)
−
(Bi, see SC pattern)
i
2
(w) (−6) + (−6)
(SC)
h
i
2
= (w + −6) w + 6w + 36
(BI)
35.
2
2
− △ = ( − △)( + △)
..recall the DS Pattern)
(
4r
2
+ −64
= (2r)
2
2
− (8)
= (2r + 8) (2r − 8)
(given)
(Bi, see DS pattern)
(DS)
Bibliography
[1] Wikipedia Euclid’s Elements Retrieved August 10, 2011
[2] Wikipedia Georg Cantor Retrieved August 2011
[3] Wikipedia Cantor’s Diagonal Argument Retrieved August 2011
84
Index
∈, 3
Q[t], 3
R[x], 3
Z[x], 3
binomial, 2
degree of polynomial, 3
degree of the term, 2
Difference of Squares [DS], 15
Difference of two Cubes [DC], 15
Difference of two Cubes [SC], 16
factor by grouping, 39
FOIL, 14
General Geometric Series Polynomials, 17
Geometric Series Polynomials, 16
GGS#2, 17
GGS#3, 17
GGS#4, 17
GGS#5, 17
GS#2, 16
GS#3, 16
GS#4, 16
GS#5, 16
like terms, 4
monomial in x with integer coefficient, 2
monomial in x with real coefficient, 2
Pascal Polynomials, 16
polynomial, 2
PP#2, 16
PP#3, 16
PP#4, 16
PP#5, 16
split the middle, 43
variable, 2
85