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B. Common Binomial Factors and Factoring by Grouping

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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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



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B. Common Binomial Factors and Factoring by Grouping

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