N
O
SS
LE
4
Factoring and Solving Polynomials
The Factor Theorem
UNDERSTAND To factor an expression means to write it as the product of factors. In the past,
you have factored quadratic expressions. For example, x2 1 x 2 12 can be written as the
product of two factors: (x 1 4)(x 2 3). Larger-degree polynomials, such as cubic and quartic
expressions, can also be factored.
UNDERSTAND A zero of a function is an input that results in an output of 0. In other words,
when that value is substituted for the variable, the value of the function is 0.
The Factor Theorem: If (x 2 k) is a factor of a function f (x), then k is a zero of the function.
Writing a function in factored form helps identify the zeros. Write the quadratic expression
above as a function: f (x) 5 x 2 1 x 2 12. Now, write it in factored form: f (x) 5 (x 1 4)(x 2 3).
The zeros of the function are 24 and 3.
The zeros of a function are the values for x, such that f (x) 5 0. Zeros are also roots, which are
all values for the variable x that make the equation true. Both roots and zeros are considered
solutions.
Consider this cubic polynomial: 6x3 1 3x2 1 24x 1 12.
The first step in factoring a polynomial is to factor out the greatest common factor (GCF)
from all terms if one exists. This cubic polynomial has a GCF of 3.
6x3 1 3x2 1 24x 1 12 5 3(2x3 1 x2 1 8x 1 4)
2x3 1 x2 1 8x 1 4 5 (2x3 1 x2) 1 (8x 1 4) 2
5 x (2x 1 1) 1 4(2x 1 1) Add parentheses.
Factor.
The terms have a common factor of (2x 1 1), which can be factored out. Remember to include
the GCF when writing the complete factorization.
6x3 1 3x2 1 24x 1 12 5 3(2x 1 1)(x2 1 4)
Note that only certain polynomials can be factored by grouping.
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Now factor the expression in parentheses. One method for factoring polynomials is to factor
by grouping. Group the terms into two pairs so that the pairs have a common factor.
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Connect
Factor 5x4 2 35x3 2 45x2 1 315x by grouping.
1
Factor out the GCF, if possible.
The terms have a common factor of 5x.
5x4 2 35x3 2 45x2 1 315x
3
2
5 5x(x 2 7x 2 9x 1 63)
2
Factor the remaining polynomial by
grouping.
Rewrite the subtractions as additions to
avoid errors.
x3 1 (27x2) 1 (29x) 1 63
Group the first two terms and the last
two terms.
[x3 1 (27x2)] 1 [(29x) 1 63]
Factor the pairs.
x2[x 1 (27)] 1 (29)[x 1 (27)]
Factor out the common factor of [x 1 (27)].
[x2 1 (29)][x 1 (27)]
3
Rewrite the additions as subtractions.
Include the GCF in the final factorization.
▸ 5x
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4
3
2
2 35x 2 45x 1 315x
2
5 5x(x 2 9)(x 2 7)
TRY
Factor 2x5 1 4x4 1 5x 1 10 by grouping.
Try grouping the first and third terms and
the second and last terms.
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Polynomial Identities
UNDERSTAND Identities show equivalent ways of writing the same expression. They can be
useful in factoring polynomials.
Here are some important identities for quadratic as well as higher degree polynomials.
Polynomial Identities
Square of a Sum
(x 1 y)2 5 x2 1 2xy 1 y2
Square of a Difference
(x 2 y)2 5 x2 2 2xy 1 y2
Difference of Squares
x2 2 y2 5 (x 1 y)(x 2 y)
Sum of Cubes
x3 1 y3 5 (x 1 y)(x2 2 xy 1 y2)
Difference of Cubes
x3 2 y3 5 (x 2 y)(x2 1 xy 1 y2)
Difference of Quartics
x4 2 y4 5 (x2 1 y2)(x2 2 y2)
Sum of Quintics
x5 1 y5 5 (x 1 y)(x4 2 x3y 1 x2y2 2 xy3 1 y4)
Difference of Quintics
x5 2 y5 5 (x 2 y)(x4 1 x3y 1 x2y2 2 xy3 1 y4)
UNDERSTAND Not all polynomials can be factored using real numbers. Some polynomials
can be factored using complex numbers.
2
2
The quadratic x 1 25 is a sum of squares. Rewrite it as a difference of squares: x 2 (225).
The identity for the difference of squares is x2 2 y 2 5 (x 1 y)(x 1 y). There is no real number
that can be multiplied by itself to equal 225, but there is an imaginary number that can be
used to get this product. The imaginary number is 5i.
x2 1 25 5 x2 2 (225)
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5 (x 1 5i)(x 2 5i)
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Connect
Show that the expression (x 2 y)(x2 1 xy 1 y2) is equivalent to x3 2 y3.
1
Multiply the terms in the first expression,
and show that the product simplifies to
the second expression.
Rewrite the subtraction as addition to avoid
sign errors.
[x 1 (2y)](x2 1 xy 1 y2)
2
Apply the distributive property.
Multiply each term in the first set of
parentheses by each term in the second
set of parentheses.
x(x2 1 xy 1 y2) 1 (2y)(x2 1 xy 1 y2)
3
2
2
2
2
3
5 x 1 x y 1 xy 1 (2x y) 1 (2xy ) 1 (2y )
3
Simplify.
The terms x2y and (2x2y) are opposites.
The terms xy2 and (2xy2) are opposites.
simplified expression is equal to
▸ The
3
3
3
3
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x 1 (2y ), or x 2 y .
TRY
Show that the expression
(a2 1 b2) (a4 2 a2b2 1 b4) is equivalent
to a6 1 b6.
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EXAMPLE A Factor: 125x3 1 64y3
1
Examine the polynomial and see if it fits
an identity.
Both of the variables are raised to the
third power.
The coefficients are also perfect cubes.
Use the sum of cubes identity to factor the
expression.
2
Rewrite the expression to show the sum
of cubes.
125x3 1 64y3 5 (5x)3 1 (4y)3
3
Use the identity to factor.
Recall the sum of cubes identity.
x3 1 y3 5 (x 1 y)(x2 2 xy 1 y2)
Substitute 5x for x and 4y for y.
(5x)3 1 (4y)3
2
2
5 (5x 1 4y)[(5x) 2 (5x)(4y) 1 (4y) ]
Simplify.
is
▸ The factored expression
2
(5x 1 4y)(25x 2 20xy 1 16y2).
TRY
Factor the polynomial x4 2 7x2 1 12 by
inspection. (Hint: Think about how you
would factor x2 2 7x 1 12.)
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4
Unit 1: Polynomial, Rational, and Radical Relationships
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EXAMPLE B Factor: x4 1 5x2 2 36
1
Treat the expression as a quadratic in
which the variable is x2.
Let z 5 x2, and rewrite the expression as
a quadratic.
z2 1 5z 2 36
2
Factor the quadratic by inspection.
The numbers 4 and 9 have a difference of 5
and a product of 36.
z2 1 5z 2 36 5 (z 2 4)(z 1 9)
3
2
Substitute x2 back into the expression.
2
The expression (x 2 4)(x 1 9) is partly
factored, but it can be factored further.
x4 1 5x2 2 36 5 (x2 2 4)(x2 1 9)
The first term, x2 2 4, is a difference of
squares.
It is equal to (x 1 2)(x 2 2).
The second term, x2 1 9, is a sum of
squares.
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It can be rewritten as a difference of
squares and factored with imaginary
numbers.
x2 1 9 5 x2 2 (29)
2
2
5 x 2 (3i)
5 (x 1 3i)(x 2 3i)
4
Write the polynomial in fully
factored form.
▸ x
4
SC U S S
DI
2
1 5x 2 36
5 (x 1 2)(x 2 2)(x 1 3i)(x 2 3i)
Could a polynomial have complex roots
that have both a real and an imaginary
part, such as 1 1 2i?
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EXAMPLE C Factor: n4 1 4n3 1 6n2 1 4n 1 1
1
Examine the polynomial, and see if it fits
an identity.
It does not appear to fit a quartic or
quadratic identity.
Look for a GCF.
There is no GCF among the terms.
Look for any other pattern that may help.
The terms are in descending order with a
term for each power of x.
The coefficients of the terms are 1, 4, 6,
4, 1. These numbers correspond to row 4
of Pascal’s Triangle. This is most likely an
expansion of a binomial.
2
Determine the binomial that was
expanded.
Since the coefficients are the same as
those in Pascal’s Triangle, they have not
been multiplied by any other factors.
The binomial that was expanded must be
(n 1 1). Since the leading term is n4, it is
raised to the 4th power.
▸ n
4
TRY
Factor the binomial expansion.
x4 1 8 x3 1 24 x2 1 32 x 1 16
38 3
2
4
1 4n 1 6n 1 4n 1 1 5 (n 1 1)
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There is only one variable, n. All of the
terms are added. So, the expanded
binomial must be in the form (n 1 k).
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Problem Solving
READ
Find the zeroes of the function f (x) 5 3x4 2 24x.
PLAN
Factor out the GCF. Then, look for an identity or other pattern that can be used to factor the
expression fully. Once the function is factored, the zeros can be found.
SOLVE
Factor out the GCF:
f (x) 5 3x(
)
.
The expression inside parentheses is a difference of
Factor the expression in parentheses.
.
The function, in fully factored form, is f (x) 5
Find the zero associated with each factor.
For the factor 3x, the zero is
For the factor (x 2
.
), the zero is
.
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To find the zeros for the quadratic factor, use the quadratic formula.
The zeros are
and
▸ The zeros of the function are
.
,
,
, and
.
CHECK
Substitute each zero for x into the original function, and show that f (x) 5 0.
f (
)5
f (
)5
f (
)5
f (
)5
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Practice
Factor each quadratic polynomial by inspection.
x2 1 x 2 56
2.
x2 2 8x 1 12
3.2x2 1 5x 1 3
HI
NT
1.
The constant term, 3,
is a prime number.
Factor each polynomial by grouping.
5.
4
3
2
6.2x 1 2x 1 14x 1 14x
3
2
7.2x 1 16x 1 x 1 8
HI
NT
4.4x3 1 3x2 1 12x 1 9
x3 1 2x2 2 5x 2 10
Factor out the
GCF first.
Factor each polynomial completely. Look for identities and expanded binomials.
8.1,000a3 1 273
4
3
2 2
3
10. x 1 3x y 1 3x y 1 xy
11. z4 2 625y4
REMEMBER Look for common
factors first.
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9.20a3 2 45a
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Factor each polynomial. Use real and imaginary numbers.
12.2a2 1 72
13. b4 2 81
Find all zeros of each polynomial.
14. f (x) 5 x2 2 144
15. g(x) 5 x4 2 16
16. h(x) 5 x4 2 4x3 1 6x2 2 4x 1 1
Solve.
17.
18.
DERIVE Rewrite x4 2 y4 as a difference of squares. Then, derive the difference of
quartics identity.
SYNTHESIZE Factor x6 2 y6 as a difference of cubes.
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Now, factor x6 2 y6 as a difference of squares.
Use the results to write an identity for (x4 1 x2y2 1 y4).
(x4 1 x2y2 1 y4) 5
19.
APPLY Use the identity you created in the last item to factor a4 1 9a2 1 81.
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