5.2 Factoring
Trinomials
Factoring Trinomials
Using the FOIL method, we see that the product of the binomial k − 3 and k +1
is
(k − 3)(k + 1) = k2 − 2k − 3.
Multiplying
Suppose instead that we are given the polynomial k2 − 2k − 3 and want to
rewrite it as the product (k − 3)(k + 1). That is,
k2 − 2k − 3 = (k − 3)(k + 1).
Factoring
Recall from Section 5.1 that this process is called factoring the polynomial.
Factoring reverses or “undoes” multiplying.
Slide 5.2-3
Objective 1
Factor trinomials with a coefficient of 1
for the second-degree term.
Slide 5.2-4
Factor trinomials with a coefficient of 1 for the second-degree
term.
When factoring polynomials with integer coefficients, we use only integers in
the factors. For example, we can factor x2 + 5x + 6 by finding integers m and n
such that
x2 + 5x + 6 = (x + m)(x + n).
To find these integers m and n, we first use FOIL to multiply the two
binomials on the right side of the equation:
 x  m x  n  x22  nx  mx  mn.
 x   n  m x  mn.
Comparing this result with x2 + 5x + 6 shows that we must find integers m
and n having a sum of 5 and a product of 6.
Product of m and n is 6.
x2  5x  6  x2   n  m x  mn.
Sum of m and n is 5.
Slide 5.2-5
Factor trinomials with a coefficient of 1 for the seconddegree term. (cont’d)
Since many pairs of integers have a sum of 5, it is best to begin by listing
those pairs of integers whose product is 6. Both 5 and 6 are positive, so
consider only pairs in which both integers are positive.
Both
pairs have a product of 6, but only the pair 2 and 3 has a sum of 5. So 2 and
3 are the required integers, and
x2 + 5x + 6 = (x + 2)(x + 3).
Check by using the FOIL method to multiply the binomials. Make sure that
the sum of the outer and inner products produces the correct middle term.
Slide 5.2-6
CLASSROOM
EXAMPLE 1
Factoring a Trinomial with All Positive Terms
Factor y2+ 12y + 20.
Solution:
Factors of 20
1, 20
2, 10
4, 5
Sums of
Factors
1 + 20 = 21
2 + 10 = 12
4+5 =9
  y  10 y  2
You can check your factoring by graphing both the unfactored and factored
forms of polynomials on your graphing calculators.
Slide 5.2-7
CLASSROOM
EXAMPLE 2
Factoring a Trinomial with a Negative Middle Term
Factor y2 − 10y + 24.
Solution:
Factors of 24
− 1 , −24
−2 , −12
−3 , −8
−4 , −6
Sums of Factors
−1 + (−24) = −25
−2 + (−12) = −14
−3 + (−8) = −11
−4 + (−6) = −10
  y  6 y  4
Slide 5.2-8
CLASSROOM
EXAMPLE 3
Factoring a Trinomial with a Negative Last (Constant) Term
Factor z2 + z − 30.
Solution:
Factors of − 30 Sums of Factors
− 1 , 30
−1 + (30) = 29
1 , − 30
1 + (−30) = −29
5,−6
5 + (− 6) = −1
−5 , 6
−5 + (6) = 1
  z  6 z  5
Slide 5.2-9
CLASSROOM
EXAMPLE 4
Factoring a Trinomial with Two Negative Terms
Factor a2 − 9a − 22.
Solution:
Factors of −22
Sums of Factors
−1 , 22
1, −22
−2 , 11
2 , −11
−1 + 22 = 21
1 + (−22) = −21
−2 + 11 = 9
2 + (−11) = −9
  a  11 a  2
Slide 5.2-10
Factor trinomials with a coefficient of 1 for the seconddegree term. (cont’d)
Some trinomials cannot be factored by using only integers. We call such
trinomials prime polynomials.
Summarize the signs of the binomials when factoring a trinomial whose
leading coefficient is positive.
1. If the last term of the trinomial is positive, both binomials will
have the same “middle” sign as the second term.
2. If the last term of the trinomial is negative, the binomials will have
one plus and one minus “middle” sign.
Slide 5.2-11
CLASSROOM
EXAMPLE 5
Factor if possible.
m  8m  14
2
Prime
Deciding Whether Polynomials Are Prime
Solution:
Factors of 14
Sums of
Factors
−1 , −14
−1 + (−14) =
−15
−2 + (−7) = −9
−2 , −7
y  y2
2
Prime
Factors of 2 Sums of Factors
1, 2
1+2=3
Slide 5.2-12
Factor trinomials with a coefficient of 1 for the seconddegree term. (cont’d)
Guidelines for Factoring x2 + bx + c
Find two integers whose product is c and whose sum is b.
1. Both integers must be positive if b and c are positive.
2. Both integers must be negative if c is positive and b is negative.
3. One integer must be positive and one must be negative if c is negative.
Slide 5.2-13
CLASSROOM
EXAMPLE 6
Factoring a Trinomial with Two Variables
Factor r2 − 6rs + 8s2.
Solution:
Factors of 8s2
−1s , −8s
−2s , −4s
Sums of Factors
− 1s + (−8s) =
−9s
−2s + (−4s) = −6s
  r  4s  r  2s 
Slide 5.2-14
Objective 2
Factor such trinomials after factoring out
the greatest common factor.
Slide 5.2-15
Factor such trinomials after factoring out the greatest
common factor.
If a trinomial has a common factor, first factor it out.
Slide 5.2-16
CLASSROOM
EXAMPLE 7
Factoring a Trinomial with a Common Factor
Factor 3x4 − 15x3 + 18x2.
Solution:
 3x 2  x 2  5 x  6 
 3x2  x  3 x  2
When factoring, always look for a common factor first. Remember to include the
common factor as part of the answer. As a check, multiplying out the factored form
should always give the original polynomial.
Slide 5.2-17