Chapter Twelve — Factoring Polynomials
The Humongous Book of Algebra Problems
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
+ 2x + 10.
To factor by grouping, group together the polynomial terms that have common
factors. In this problem, 3x
have common factors, and 2x and 10 share
a common factor as well.
) + (2x + 10)
The greatest common factor of the left group of terms is 3x
; the greatest
common factor of the right group is 2. Write each group in factored form.
(x + 5) + 2(x + 5)
was 2w + 3yw,
it’d be easy
to tell that w
was the greatest
just plug in (x + 1)
where w is and you
get Problem 12.19.
don’t have to be
monomials. They can
be binomials like
this (or trinomials,
or fractions, or
the list goes on
factor (2y – 1)
out of the
left with 9x in the
rst term and 4 in
the second. Add
those values inside
parentheses (9x + 4)
and multiply by the
(9x + 4)(2y – 1).
you end up grouping
are usually next to