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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.
Turn one trinomial into two binomials
12.35 Factor the expression: x
2
+ 3x + 2.
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).
Theres no easy way
to factor the trinomial
in there, though. The
difference of perfect
squares formula gives
more factors.
In other
words, when
numbers, you get
the coefcient of
x in the trinomial,
and when you
multiply them, you
get the constant.
This works only when
the coefcient
of x
2
is 1 (like in
this problem).

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