Chapter 6
Section 3
6.3
1
2
More on Factoring Trinomials
Factor trinomials by grouping when the
coefficient of the squared term is not 1.
Factor trinomials by using the FOIL
method.
More on Factoring Trinomials
Trinomials such as 2x2 + 7x + 6, in which the
coefficient of the squared term is not 1, are factored
with extensions of the methods from the previous
sections. One such method uses factoring by grouping
from Section 6.1.
Slide 6.3 - 3
Objective 1
Factor trinomials by grouping
when the coefficient of the squared
term is not 1.
Slide 6.3 - 4
Factor trinomials by grouping when the
coefficient of the squared term is not 1.
Recall that a trinomial such as m2 + 3m + 2 is factored by
finding two numbers whose product is 2 and whose sum is 3. To
factor 2x2 + 7x + 6, we look for two integers whose product is
2 · 6 = 12 and whose sum is 7.
Sum
2 x2  7 x  6
Product is 2 · 6 = 12
Slide 6.3 - 5
Factor trinomials by grouping when the
coefficient of the squared term is not 1. (cont’d)
By considering pairs of positive integers whose product is 12,
we find the necessary integers to be 3 and 4. We use these
integers to write the middle term, 7x, as 7x = 3x + 4x. The
trinomial 2x2 + 7x + 6 becomes
2 x 2  7 x  6  2 x 2  3x  4 x  6.
  2 x 2  3x    4 x  6 
 x  2x  3  2  2x  3
  x  2 2x  3
Slide 6.3 - 6
EXAMPLE 1
Factoring Trinomials by
Grouping
Factor.
Solution:
3 p2  4 p  1
 3 p2  3 p 1 p  1
  3 p 2  3 p    1 p  1
 3 p  p 1 1 p 1
  3 p 1 p 1
12 z  16 z  3  12 z  18z  2 z  3
 12 z 2  18 z    2 z  3
2
2
 6z  2z  3 1(2z  3)   6z 1 2z  3

 
8r 2  6rt  5t 2  8r 2  10rt  4rt  5t 2  8r 2  10rt  4rt  5t 2

 2r  4r  5t  1t  4r  5t    2r  t  4r  5t 
Slide 6.3 - 7
EXAMPLE 2
Factoring a Trinomial with a
Common Factor by Grouping
Factor 6p4 + 21p3 + 9p2.
Solution:
 3 p 2  2 p 2  7 p  3
 3 p  2 p  6 p  1 p  3
2
2
2

 3 p  2 p  6 p   1  p  3
 3 p 2  2 p  p  3  1 p  3 
2
 3 p2  2 p  1 p  3
Slide 6.3 - 8
Objective 2
Factor trinomials by using the
FOIL method.
Slide 6.3 - 9
Factor trinomials by using the FOIL method.
To factor 2x2 + 7x + 6, again using an alternate method
explained in Section 6.2, we use the FOIL method in reverse.
We want to write the equation 2x2 + 7x + 6 as the product of
two binomials.
2 x2  7 x  6  


The product of the two first terms of the binomials is 2x2.
The possible factors of 2x2 are 2x and x or −2x and −x. Since all
terms of the trinomial are positive, we consider only positive
factors. Thus, we have
2 x2  7 x  6   2 x
 x
.
Slide 6.3 - 10
Factor trinomials by using the FOIL
method. (cont’d)
The product of the two last terms, 6, can be factored as 1 · 6, 6 · 1,
3 · 2, or 3 · 2. Try each pair to find the pair that gives the correct
middle term, 7x.
 2x 1 x  6
 2x  6 x 1
x
6x
12x
2x
Incorrect
Incorrect
13x
8x
If the terms of the original polynomial have greatest common factor 1,
then all of that polynomials binomial factors also have GCF 1.
Now try the number 2 and 3 as factors of 6. Because of the common
factor 2 in 2x + 2, (2x + 2)(x + 3) will not work, so we try (2x + 3)(x + 2).
 2x  3 x  2
3x
4x
Correct
7x
Slide 6.3 - 11
EXAMPLE 3
Factoring a Trinomial with All
Positive Terms by Using FOIL
Factor 6p2 + 19p + 10.
Solution:
6 p  21p  5
 6 p 101p 1
2p
10 p
30 p
6p
Incorrect
Incorrect
32 p
16 p
 2 p 103 p 1
3 p  2 2 p  5
13p
4p
2p
15 p
Incorrect
Correct
15 p
19 p
Slide 6.3 - 12
EXAMPLE 4
Factoring a Trinomial with a
Negative Middle Term by
Using FOIL
Factor 10m2 – 23m + 12.
Solution:
 2m 125m 1
60m
2m
62m
10m  21m  6
Incorrect
2m
60m
62m
Incorrect
 2m  35m  4
Correct
15m
8m
23m
Slide 6.3 - 13
EXAMPLE 5
Factoring a Trinomial with
a Negative Last Term by
Using FOIL
Factor 5p2 + 13p – 6.
Solution:
5 p  3 p  2
Incorrect
3p
15p
12 p
5 p  2 p  3
2 p
15p
13p
Correct
Slide 6.3 - 14
EXAMPLE 6
Factoring a Trinomial with
Two Variables
Factor 6m2 + 11mn – 10n2.
Solution:
 6m 10n m 1n
3m  2n 2m  5n
4mn
15mn
11mn
10mn
6mn
4mn
Incorrect
Correct
Slide 6.3 - 15
EXAMPLE 7
Factor.
28x4  58x3  30 x2
Solution:
 2 x 14 x  29 x  15 
2
2
 2x2  7x  3 2x  5
Factoring Trinomials with
Common Factors
24x3  32x2 y  6xy 2
 2 x 12 x 2  16 xy  3 y 2 
 2x  6x  y  2x  3 y 
6x
35x
29 x
2xy
18xy
16xy
Slide 6.3 - 16