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The Humongous Book of Algebra Problems by W. Michael Kelley

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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
dont 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, youre
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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