B. Common Binomial Factors and Factoring by Grouping
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Section R.4 Factoring Polynomials and Solving Polynomial Equations by Factoring
B. You’ve just seen how
we can factor common
binomial factors and factor by
grouping
41
When asked to factor an expression, first look for common factors. The resulting
expression will be easier to work with and help ensure the final answer is written in
completely factored form. If a four-term polynomial cannot be factored as written, try
rearranging the terms to find a combination that enables factoring by grouping.
C. Factoring Quadratic Polynomials
A quadratic polynomial is one that can be written in the form ax2 ϩ bx ϩ c, where
a, b, c ʦ ޒand a 0. One common form of factoring involves quadratic trinomials
such as x2 ϩ 7x ϩ 10 and 2x2 Ϫ 13x ϩ 15. While we know 1x ϩ 521x ϩ 22 ϭ
x2 ϩ 7x ϩ 10 and 12x Ϫ 321x Ϫ 52 ϭ 2x2 Ϫ 13x ϩ 15 using F-O-I-L, how can we
factor these trinomials without seeing the original expression in advance? First, it helps
to place the trinomials in two families—those with a leading coefficient of 1 and those
with a leading coefficient other than 1.
ax 2 ؉ bx ؉ c, where a ؍1
When a ϭ 1, the only factor pair for x2 (other than 1 # x2 2 is x # x and the first term in
each binomial will be x: (x
)(x
). The following observation helps guide us to
the complete factorization. Consider the product 1x ϩ b2 1x ϩ a2:
1x ϩ b21x ϩ a2 ϭ x2 ϩ ax ϩ bx ϩ ab
ϭ x ϩ 1a ϩ b2x ϩ ab
2
F-O-I-L
distributive property
Note the last term is the product ab (the lasts), while the coefficient of the middle
term is a ϩ b (the sum of the outers and inners). Since the last term of x2 Ϫ 8x ϩ 7 is
7 and the coefficient of the middle term is Ϫ8, we are seeking two numbers with a
product of positive 7 and a sum of negative 8. The numbers are Ϫ7 and Ϫ1, so the factored form is 1x Ϫ 721x Ϫ 12. It is also helpful to note that if the constant term is positive, the binomials will have like signs, since only the product of like signs is positive.
If the constant term is negative, the binomials will have unlike signs, since only the
product of unlike signs is negative. This means we can use the sign of the linear term
(the term with degree 1) to guide our choice of factors.
Factoring Trinomials with a Leading Coefficient of 1
If the constant term is positive, the binomials will have like signs:
1x ϩ 2 1x ϩ 2 or 1x Ϫ 2 1x Ϫ 2 ,
to match the sign of the linear (middle) term.
If the constant term is negative, the binomials will have unlike signs:
1x ϩ 2 1x Ϫ 2,
with the larger factor placed in the binomial
whose sign matches the linear (middle) term.
EXAMPLE 4
ᮣ
Factoring Trinomials
Factor these expressions:
a. Ϫx2 ϩ 11x Ϫ 24
Solution
ᮣ
b. x2 Ϫ 10 Ϫ 3x
a. First rewrite the trinomial in standard form as Ϫ11x2 Ϫ 11x ϩ 242. For
x2 Ϫ 11x ϩ 24, the constant term is positive so the binomials will have like
signs. Since the linear term is negative,
Ϫ11x2 Ϫ 11x ϩ 242 ϭ Ϫ11x Ϫ 21x Ϫ 2
ϭ Ϫ11x Ϫ 821x Ϫ 32
like signs, both negative
1Ϫ82 1Ϫ32 ϭ 24; Ϫ8 ϩ 1Ϫ32 ϭ Ϫ11
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b. First rewrite the trinomial in standard form as x2 Ϫ 3x Ϫ 10. The constant
term is negative so the binomials will have unlike signs. Since the linear term
is negative,
x2 Ϫ 3x Ϫ 10 ϭ 1x ϩ 2 1x Ϫ 2
ϭ 1x ϩ 22 1x Ϫ 52
unlike signs, one positive and one negative
5 7 2, 5 is placed in the second binomial;
122 1Ϫ52 ϭ Ϫ10; 2 ϩ 1Ϫ52 ϭ Ϫ3
Now try Exercises 13 and 14
ᮣ
Sometimes we encounter prime polynomials, or polynomials that cannot be factored. For x2 ϩ 9x ϩ 15, the factor pairs of 15 are 1 # 15 and 3 # 5, with neither pair
having a sum of ϩ9. We conclude that x2 ϩ 9x ϩ 15 is prime.
ax 2 ؉ bx ؉ c, where a
1
If the leading coefficient is not one, the possible combinations of outers and inners are
more numerous. Furthermore, the sum of the outer and inner products will change depending on the position of the possible factors. Note that 12x ϩ 321x ϩ 92 ϭ
2x2 ϩ 21x ϩ 27 and 12x ϩ 921x ϩ 32 ϭ 2x2 ϩ 15x ϩ 27 result in a different middle
term, even though identical numbers were used.
To factor 2x2 Ϫ 13x ϩ 15, note the constant term is positive so the binomials must
have like signs. The negative linear term indicates these signs will be negative. We then
list possible factors for the first and last terms of each binomial, then sum the outer and
inner products.
Possible First and Last Terms
for 2x2 and 15
1. 12x Ϫ 121x Ϫ 152
2. 12x Ϫ 1521x Ϫ 12
3. 12x Ϫ 321x Ϫ 52
4. 12x Ϫ 521x Ϫ 32
WORTHY OF NOTE
The number of trials needed to
factor a polynomial can also be
reduced by noting that the two
terms in any binomial cannot share
a common factor (all common
factors are removed in a preliminary
step).
EXAMPLE 5
ᮣ
Sum of
Outers and Inners
Ϫ30x Ϫ 1x ϭ Ϫ31x
Ϫ2x Ϫ 15x ϭ Ϫ17x
Ϫ10x Ϫ 3x ϭ Ϫ13x
d
Ϫ6x Ϫ 5x ϭ Ϫ11x
As you can see, only possibility 3 yields a linear term of Ϫ13x, and the correct factorization is then 12x Ϫ 321x Ϫ 52. With practice, this trial-and-error process can be
completed very quickly.
If the constant term is negative, the number of possibilities can be reduced by finding a factor pair with a sum or difference equal to the absolute value of the linear coefficient, as we can then arrange the sign in each binomial to obtain the needed result
as shown in Example 5.
Factoring a Trinomial Using Trial and Error
Factor 6z2 Ϫ 11z Ϫ 35.
Solution
ᮣ
Note the constant term is negative (binomials will have unlike signs) and ͿϪ11Ϳ ϭ 11.
The factors of 35 are 1 # 35 and 5 # 7. Two possible first terms are: (6z
)(z
)
and (3z )(2z
), and we begin with 5 and 7 as factors of 35.
(6z
Outer and Inner
Products
)(z
)
Sum
Difference
1. (6z
5)(z
7)
42z ϩ 5z
47z
42z Ϫ 5z
37z
2. (6z
7)(z
5)
30z ϩ 7z
37z
30z Ϫ 7z
23z
(3z
Outer and Inner
Products
)(2z
)
Sum
Difference
3. (3z
5)(2z
7)
21z ϩ 10z
31z
21z Ϫ 10z
11z
4. (3z
7)(2z
5)
15z ϩ 14z
29z
15z Ϫ 14z
1z
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43
Since possibility 3 yields a linear term of 11z, we need not consider other factors
of 35 and write the factored form as 6z2 Ϫ 11z Ϫ 35 ϭ 13z 5212z 72. The
signs can then be arranged to obtain a middle term of Ϫ11z: 13z ϩ 5212z Ϫ 72,
Ϫ21z ϩ 10z ϭ Ϫ11z ✓.
C. You’ve just seen how
we can factor quadratic
polynomials
Now try Exercises 15 and 16
ᮣ
D. Factoring Special Forms and Quadratic Forms
Next we consider methods to factor each of the special products we encountered in
Section R.2.
The Difference of Two Squares
WORTHY OF NOTE
In an attempt to factor a sum of
two perfect squares, say v2 ϩ 49,
let’s list all possible binomial
factors. These are (1) 1v ϩ 721v ϩ 72,
(2) 1v Ϫ 721v Ϫ 72, and
(3) 1v ϩ 721v Ϫ 72. Note that (1) and
(2) are the binomial squares
1v ϩ 72 2 and 1v Ϫ 72 2, with each
product resulting in a “middle”
term, whereas (3) is a binomial
times its conjugate, resulting in a
difference of squares: v2 Ϫ 49. With
all possibilities exhausted, we
conclude that the sum of two
squares is prime!
EXAMPLE 6
ᮣ
Multiplying and factoring are inverse processes. Since 1x Ϫ 721x ϩ 72 ϭ x2 Ϫ 49, we
know that x2 Ϫ 49 ϭ 1x Ϫ 721x ϩ 72. In words, the difference of two squares will
factor into a binomial and its conjugate. To find the terms of the factored form, rewrite
each term in the original expression as a square: 1 2 2.
Factoring the Difference of Two Perfect Squares
Given any expression that can be written in the form A2 Ϫ B2,
A2 Ϫ B2 ϭ 1A ϩ B21A Ϫ B2
Note that the sum of two perfect squares A2 ϩ B2 cannot be factored using real numbers (the expression is prime). As a reminder, always check for a common factor first
and be sure to write all results in completely factored form. See Example 6(c).
Factoring the Difference of Two Perfect Squares
Factor each expression completely.
a. 4w2 Ϫ 81
b. v2 ϩ 49
c. Ϫ3n2 ϩ 48
Solution
ᮣ
a. 4w2 Ϫ 81 ϭ 12w2 2 Ϫ 92
ϭ 12w ϩ 92 12w Ϫ 92
b. v2 ϩ 49 is prime.
c. Ϫ3n2 ϩ 48 ϭ Ϫ31n2 Ϫ 162
ϭ Ϫ33 n2 Ϫ 142 2 4
ϭ Ϫ31n ϩ 42 1n Ϫ 42
1
d. z4 Ϫ 81
ϭ 1z2 2 2 Ϫ 1 19 2 2
ϭ 1z2 ϩ 19 2 1z2 Ϫ 19 2
ϭ 1z2 ϩ 19 2 3 z2 Ϫ 1 13 2 2 4
ϭ 1z2 ϩ 19 2 1z ϩ 13 2 1z Ϫ 13 2
2
e. x Ϫ 7 ϭ 1x2 2 Ϫ 1 172 2
ϭ 1x ϩ 1721x Ϫ 172
1
d. z4 Ϫ 81
e. x2 Ϫ 7
write as a difference of squares
A 2 Ϫ B 2 ϭ 1A ϩ B21A Ϫ B2
factor out Ϫ3
write as a difference of squares
A 2 Ϫ B 2 ϭ 1A ϩ B21A Ϫ B2
write as a difference of squares
A 2 Ϫ B 2 ϭ 1A ϩ B21A Ϫ B2
write as a difference of squares (z 2 ϩ 19 is prime)
result
write as a difference of squares
A 2 Ϫ B 2 ϭ 1A ϩ B21A Ϫ B2
Now try Exercises 17 and 18
Perfect Square Trinomials
ᮣ
Since 1x ϩ 72 2 ϭ x2 ϩ 14x ϩ 49, we know that x2 ϩ 14x ϩ 49 ϭ 1x ϩ 72 2. In words,
a perfect square trinomial will factor into a binomial square. To use this idea effectively,
we must learn to identify perfect square trinomials. Note that the first and last terms of
x2 ϩ 14x ϩ 49 are the squares of x and 7, and the middle term is twice the product of
these two terms: 217x2 ϭ 14x. These are the characteristics of a perfect square trinomial.
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Factoring Perfect Square Trinomials
Given any expression that can be written in the form A2 Ϯ 2AB ϩ B2,
1. A2 ϩ 2AB ϩ B2 ϭ 1A ϩ B2 2
2. A2 Ϫ 2AB ϩ B2 ϭ 1A Ϫ B2 2
EXAMPLE 7
ᮣ
Factoring a Perfect Square Trinomial
Factor 12m3 Ϫ 12m2 ϩ 3m.
Solution
12m3 Ϫ 12m2 ϩ 3m
ϭ 3m14m2 Ϫ 4m ϩ 12
ᮣ
check for common factors: GCF ϭ 3m
factor out 3m
For the remaining trinomial 4m2 Ϫ 4m ϩ 1 p
1. Are the first and last terms perfect squares?
4m2 ϭ 12m2 2 and 1 ϭ 112 2 ✓ Yes.
2. Is the linear term twice the product of 2m and 1?
2 # 2m # 1 ϭ 4m ✓ Yes.
Factor as a binomial square: 4m2 Ϫ 4m ϩ 1 ϭ 12m Ϫ 12 2
This shows 12m3 Ϫ 12m2 ϩ 3m ϭ 3m12m Ϫ 12 2.
Now try Exercises 19 and 20
CAUTION
ᮣ
ᮣ
As shown in Example 7, be sure to include the GCF in your final answer. It is a common
error to “leave the GCF behind.”
In actual practice, these calculations can be performed mentally, making the
process much more efficient.
Sum or Difference of Two Perfect Cubes
Recall that the difference of two perfect squares is factorable, but the sum of two perfect
squares is prime. In contrast, both the sum and difference of two perfect cubes are factorable. For either A3 ϩ B3 or A3 Ϫ B3 we have the following:
1. Each will factor into the product of a binomial
and a trinomial:
2. The terms of the binomial are the quantities
being cubed:
3. The terms of the trinomial are the square of A,
the product AB, and the square of B, respectively:
4. The binomial takes the same sign as the original
expression
5. The middle term of the trinomial takes the
opposite sign of the original expression
(the last term is always positive):
(
)(
binomial
)
trinomial
(A
B)(
)
(A
B)(A2
AB
B 2)
(A Ϯ B)(A2
AB
B 2)
(A Ϯ B)(A2 ϯ AB ϩ B 2)
Factoring the Sum or Difference of Two Perfect Cubes: A3 Ϯ B3
1. A3 ϩ B3 ϭ 1A ϩ B2 1A2 Ϫ AB ϩ B2 2
2. A3 Ϫ B3 ϭ 1A Ϫ B21A2 ϩ AB ϩ B2 2