Chapter Twelve — Factoring Polynomials

The Humongous Book of Algebra Problems

265

12.19 Factor the expression: 2(x + 1) + 3y(x + 1).

This expression consists of two terms, 2(x + 1) and 3y(x + 1). Both terms have

one common factor, the binomial (x + 1), so the greatest common factor of the

expression is x + 1. Divide both terms by that quantity.

Factor the expression by writing it as the product of the greatest common factor

and the preceding quotients.

2(x + 1) + 3y(x + 1) = (x + 1)(2 + 3y)

Factoring by Grouping

You can factor out binomials, too

12.20 Factor the expression: 9x(2y – 1) + 8y – 4.

This expression consists of three terms: 9x(2y – 1) is the ﬁrst term, 8y is the

second, and –4 is the third. Notice that the ﬁrst term is expressed as a product.

The greatest common factor of the second and third terms (8y and –4) is 4.

Factor it out of the expression 8y – 4 to get 4(2y – 1). Rewrite the expression

using the newly factored form of the second and third terms.

The expression contains two terms with a binomial greatest common factor:

2y – 1. Factor the expression using the method described in Problem 12.19.

9x(2y – 1) + 4(2y – 1) = (2y – 1)(9x + 4)

12.21 Factor the expression by grouping: 3x

3

+ 15x

2

+ 2x + 10.

To factor by grouping, group together the polynomial terms that have common

factors. In this problem, 3x

3

and 15x

2

have common factors, and 2x and 10 share

a common factor as well.

(3x

3

+ 15x

2

) + (2x + 10)

The greatest common factor of the left group of terms is 3x

2

; the greatest

common factor of the right group is 2. Write each group in factored form.

3x

2

(x + 5) + 2(x + 5)

If the

expression

was 2w + 3yw,

it’d be easy

to tell that w

was the greatest

common factor;

just plug in (x + 1)

where w is and you

get Problem 12.19.

Common factors

don’t have to be

monomials. They can

be binomials like

this (or trinomials,

or fractions, or

square roots…

the list goes on

and on).

When you

factor (2y – 1)

out of the

expression, you’re

left with 9x in the

rst term and 4 in

the second. Add

those values inside

parentheses (9x + 4)

and multiply by the

greatest common

factor:

(9x + 4)(2y – 1).

The terms

you end up grouping

are usually next to

each other.

Chapter Twelve — Factoring Polynomials

The Humongous Book of Algebra Problems

266

The expression now contains two terms with a common factor: x + 5.

Factor it out.

(x + 5)(3x

2

+ 2)

Therefore, the factored form of 3x

3

+ 15x

2

+ 2x + 10 is (x + 5)(3x

2

+ 2).

12.22 Factor by grouping: 14xy + 21x + 8y + 12.

The ﬁrst two terms share the common factor 7x; the last two terms share the

common factor 4.

(14xy + 21x) + (8y + 12)

Factor each group.

7x(2y + 3) + 4(2y + 3)

Factor the binomial 2y + 3 out of both terms.

(2y + 3)(7x + 4)

The factored form of 14xy + 21x + 8y + 12 is (2y + 3)(7x + 4).

12.23 Factor by grouping: 3x

3

– 12x

2

+ x – 4.

The ﬁrst two terms have a greatest common factor of 3x

2

; group them together.

(3x

3

– 12x

2

) + x – 4

Factor the grouped expression.

3x

2

(x – 4) + x – 4

The only factor common to the last two terms of the expression (x and –4) is 1.

Factor it out.

3x

2

(x – 4) + 1(x – 4)

Factor the binomial x – 4 out of both terms.

(x – 4)(3x

2

+ 1)

Therefore, the factored form of 3x

3

– 12x

2

+ x – 4 is (x – 4)(3x

2

+ 1).

12.24 Factor the expression by grouping: 4x

5

– 8x

3

– 5x

2

+ 10.

The ﬁrst two terms share common factor 4x

3

and the last two terms share

common factor 5.

Unlike Problems 12.20–12.23, once factored, these groups do not contain a

common binomial term. In fact, the binomials x

2

– 2 and –x

2

+ 2 are opposites.

To apply the factoring by grouping technique, the binomials must be equal.

Satisfy this requirement by factoring –1 out of the second group.

You can tell the

binomials are opposites

because the signs of

the corresponding terms

are opposites: x

2

and –x

2

are opposites, as are –2

and +2

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