Chapter Twelve — Factoring Polynomials

The Humongous Book of Algebra Problems

270

12.34 Factor the expression: x

6

– 1.

Express the terms of the polynomials as perfect squares: (x

3

)

2

= x

6

and 1

2

= 1.

Apply the difference of perfect squares formula.

The resulting factors are a sum of perfect cubes (x

3

+ 1) and a difference of

perfect cubes (x

3

– 1). Apply the factoring formulas for each, setting a = x and

b = 1.

Factoring Quadratic Trinomials

Turn one trinomial into two binomials

12.35 Factor the expression: x

2

+ 3x + 2.

The quadratic binomial x

2

+ ax + b, once factored, has the form below.

Each of the empty boxes represents a number. Those missing values have sum a

and product b. In this problem, a = 3 and b = 2, so the two missing values must

have a sum of 3 and a product of 2. The correct values are 1 and 2, because

1 + 2 = 3 and 1 · 2 = 2. Substitute 1 and 2 into the boxes of the formula.

x

2

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

The order of the factors is irrelevant, so the answer (x + 2)(x + 1) is equally

correct.

12.36 Factor the expression: x

2

+ 8x + 12.

To factor this quadratic binomial, you must identify two numbers with sum 8

and product 12. Begin by subtracting 1 from the required sum (8 – 1 = 7) to

identify two numbers with the correct sum: 1 + 7 = 8. Unfortunately, 1 and 7 do

not have the required product (1 · 7 ≠ 12).

You could

factor x

6

– 1

using the difference

of perfect cubes

formula, because

(x

2

)

3

= x

6

and 1

3

= 1.

You’ll end up with

(x + 1)(x – 1)(x

4

+ x

2

+ 1).

There’s no easy way

to factor the trinomial

in there, though. The

difference of perfect

squares formula gives

a better answer with

more factors.

In other

words, when

you add the

numbers, you get

the coefcient of

x in the trinomial,

and when you

multiply them, you

get the constant.

This works only when

the coefcient

of x

2

is 1 (like in

this problem).

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