Algebraic Expressions
Prepared by: Sa’diyya Hendrickson
Name:
Date:
** It’s all about the Distributive Property (DP)! **
Exercise E1: Expand the following expression: (x − 1)(x + 4)
a
c
b
! "# $ !"#$
!"#$
Solution: For the given expression: (x − 1)( x + 4 ), we can let a = x + 1,
b = x and c = 4. This results in the following expansion:
Notice that each term in the first pair of brackets was simply multiplied with
each term in the 2nd pair of brackets! In general, we have:
Many people are familiar with the acronym FOIL (First, Outer, Inner, Last).
But this is just a special case of the Distributive Property, when we have exactly
two elements in each pair of brackets!
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Expanding Expressions
Level: X
The Distributive Property tells us to simply take each term in
the 1st pair of brackets (one at a time) and multiply them with each
term in the 2nd pair of brackets (i.e. distribute them). Unlike FOIL,
this works for any number of elements in each pair of brackets!
For the above exercise, our strategy gives:
DP
(x − 1)(x + 4) = (x)(x) + (4)(x) − (1)(x) − (1)(4)
as before, but in one step!
− −4
= x2 + 4x
# $! x"
like terms
collect
= x2 + 3x − 4
Exercise E2: Expand (a − b)3
Solution: First recall that (a − b)3 = (a − b)(a − b)(a − b), by def ’n of exponentiation for a positive integer exponent. So, our approach is to use the associative
property of multiplication to group and expand two at a time!
(a − b)3 = (a − b)(a − b)(a − b)
#
$!
"
DP
= (a2 − ab − ba + b2 )(a − b)
Notice that ab = ba
When expanding, it’s a good idea to write the letters in alphabetical
order because it makes it easier to identify like terms!!
collect
= (a2 − 2ab + b2 )(a − b)
DP
= (a2 )(a) + (a2 )(−b) − (2ab)(a) − (2ab)(−b) + (b2 )(a) + (b2 )(−b)
= a3 − a2 b − 2a2 b + 2ab2 + ab2 − b3
collect
= a3 − 3a2 b + 3ab2 − b3
Try the following exercises on your own, using the same strategy:
1) Prove that (a + b)3 = a3 + 3a2 b + 3ab2 + b3 .
2) Prove that (a − b)(a2 + ab + b2 ) = a3 − b3 .
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Expanding Expressions
Level: X
Exercise E3: Expand the expression −2ab2 c (5a3 − b + 3b2 c)
As you distribute each term, it may be helpful to: first record the product
of the constants, while carefully determining the sign (i.e. is the product positive or
negative?). Then, determine and record the product of each variable, one by one.
(The chart below describes how to organize your mental math, as you expand.)
−2ab2 c (5a3 − b + 3b2 c)
Distribution
Constants
a
b
c
Summary
First
Second
Second
Solution:
−2ab2 c (5a3 − b + 3b2 c)
DP
=
P.O.E
− 10a4 b2 c + 2ab3 c − 6ab4 c2
Exercise E4: Expand the expression 4(x2 − 3x + 5) − 3(x2 − 2x + 1)
Solution:
DP
4(x2 − 3x + 5) − 3(x2 − 2x + 1) = 4x2 − 12x + 20 − 3x2 + 6x − 3
collect
= 4x2 − 3x2 −12x + 6x+20 − 3
collect
= x2 − 6x + 17
Exercise E5: Expand and evaluate
Solution:
(2 −
√
3)(2 +
√
(2 −
√
3)(2 +
√
3)
√
√
√ √
DP
3) = 2(2) + 2 3 − 2 3 − ( 3)( 3)
√
= 4 − ( 3)2
√
= 4−3
since ( n a)n = a, for a ≥ 0
=1
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Factoring Expressions
Level: X
** It’s still all about the Distributive Property (DP)! **
On the previous pages, we were going from factored form to expanded form.
Now, we will begin with expanded form and end in factored form.
We’ll know that we’ve achieved factored form once the original expression is a
product of terms and/or smaller expressions.
Exercise F1: Factor the expression −12x4 y 2 + 30xy 3 − 18xy 4 z.
S1 Decide if you’d like to factor out “−1” from all terms. This is
recommended if your leading term is negative.
Because our leading term is −12x4 y 2 , which is negative, we will pull
−1 as a common factor. Note that we can always do this!
e.g. −(12x4 y 2 −30xy 3 +18xy 4 z)
S2 Determine the GCF (Greatest Common Factor) among the
constants and each variable, one at a time. Note that for each variable,
we must factor the lowest power among all terms. Otherwise, we
cannot guarantee that we have a common factor.
Expression:
− 12x4 y 2 + 30xy 3 − 18xy 4 z
Elements
GCF
Factor −1?
N/A
Yes/No: −1
Constants
12, 30, 18
2·3=6
x
x4 , x1 , x1
x1
y
y2, y3, y4
z
z0, z0, z1
Summary
N/A
−6xy 2
This chart gives us a visual representation of how we can organize our
approach if we were to complete this strategy using mental math.
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Factoring Expressions
Level: X
Exercise F1 (Continued)
S3 Factor out each element in the GCF, starting with the constants.
Then complete the expression in the brackets by writing in the missing
part of the original term.
−12x4 y 2 + 30x3 − 18xy 4 z = − (12x4 y 2 −30xy 3 +18xy 4 z)
x4 y 2 −
−6(
=
xy 3 +
P.O.E
− 6x1 (2
P.O.E
=
− 6xy 2 (2x3
=
− 6xy 2 (2x3 − 5y + 3y 2 z)
=
y2 − 5
xy 4 z)
y3 + 3
−5
+3
y 4 z)
z)
With practice, you will be able to do this in one line!
Let’s begin by recalling the following definition:
Perfect square: a perfect square is an expression of form a2 , where
a is any integer. Note that there is nothing specific about our choice
of the letter a. Any letter will do! e.g. b2 , c2 , s2 , t2 , x2 , y 2 , etc.
Below are some square numbers that appear quite often. Therefore, it
may be worthwhile to commit them to memory.
02 = 0
62 = 36
12 = 1
72 = 49
22 = 4
82 = 64
I. Difference of Squares:
proof :
32 = 9
92 = 81
42 = 16
102 = 100
52 = 25
a2 − b2 = (a + b)(a − b)
DP
− −b2
(a − b)(a + b) = a2 + ab
# $! ab"
=0
collect
= a2 − b2
□
Exercise F2: Factor the expression x2 − 1.
S1 Determine if there are exactly two perfect squares, seperated by a “−” sign. Then, express the terms as powers of 2.
x2 ✓
1 = 12 ✓
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Factoring Expressions
Level: X
Exercise F2 (Continued)
S2
Identify “a” and “b” based on the given identity.
x2 − 1 = x2 − 1 2
S3
⇒
a = x and b = 1
Substitute into the identity to obtain the factored form.
x2 − 1 = x2 − 1 2
= (x
a = x and b = 1
1)(x
1)
= (x + 1)(x − 1)
Exercise F3: Factor the following expressions.
a)
b)
c)
a) Solution :
9 − z2
4k2 − 36
16u2 − 25v 2
9 − z 2 = 32 − z 2
= (3
a = 3 and b = z
z)(3
z)
= (3 + z)(3 − z)
b) Solution :
4k 2 − 36 = 22 k 2 − 62
P.O.E
= (2k)2 − 62
= (2k
6)(2k
a = 2k and b = 6
6)
= (2k + 6)(2k − 6)
c) Solution :
16u2 − 25v 2 = 42 u2 − 52 v 2
P.O.E
= (4u)2 − (5v)2
= (4u
5v)(4u
a = 4u and b = 5v
5v)
= (4u + 5v)(4u − 5v)
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Factoring Expressions
Level: X
II. Perfect Square:
proof :
(a + b)2 = a2 + 2ab + b2
(a + b)2 = (a + b)(a + b)
DP
= a2 + ab + ab + b2
collect
□
= a2 + 2ab + b2
Exercise F4: Factor the expression 4x2 − 12x + 9.
S1 Determine if the expression is a trinomial with perfect
squares as its first and last terms. Then, express the first and
last terms as powers of 2.
P.O.E
4x2 = 22 x2
=
9 = (±3)2
(2x)2
✓
✓
S2 Identify what “a” and “b” may be based on the identity
and info from S1. Note: we still have to verify the middle term!
4x2 − 12x + 9 = (2x)2 − 12x + (±3)2
S3
⇒
a = 2x and b = ±3
Verify that the middle term equals 2ab.
Take b = −3 ⇒ 2ab = 2(2x)(−3) = −12x
S4
✓
Substitute into the identity to obtain the factored form.
4x2 − 12x + 9 = (2x)2 + 2(−3)(2x) + (−3)2
= (2x + (−3))2
a = 2x and b = −3
by identity
= (2x − 3)2
Exercise F5: Factor 9s2 t2 − 6st + 1.
Solution :
P.O.E
9s2 t2 − 6st + 1 = (3st)2 − 2(3)st + (−1)2
= (3st + (−1))2
a = 3st and b = −1
by identity
= (3st − 1)2
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