Factoring Polynomials
Factor by Grouping

When polynomials contain four terms, it
is sometimes easier to group like terms in
order to factor.
 Your goal is to create a common factor.
 You can also move terms around in the
polynomial to create a common factor.
 Practice makes you better in recognizing
common factors.
Factoring Four Term
Polynomials
Factor by Grouping
Example 1:
FACTOR: 3xy - 21y + 5x – 35
 Factor the first two terms:
3xy - 21y = 3y (x – 7)
 Factor the last two terms:
+ 5x - 35 = 5 (x – 7)
 The green parentheses are the same so
it’s the common factor
Now you have a common factor
(x - 7) (3y + 5)

Factor by Grouping
Example 2:
FACTOR: 6mx – 4m + 3rx – 2r
 Factor the first two terms:
6mx – 4m = 2m (3x - 2)
 Factor the last two terms:
+ 3rx – 2r = r (3x - 2)
 The green parentheses are the same so
it’s the common factor
Now you have a common factor
(3x - 2) (2m + r)

Factor by Grouping
Example 3:





FACTOR: 15x – 3xy + 4y –20
Factor the first two terms:
15x – 3xy = 3x (5 – y)
Factor the last two terms:
+ 4y –20 = 4 (y – 5)
The green parentheses are opposites so change
the sign on the 4
- 4 (-y + 5) or – 4 (5 - y)
Now you have a common factor
(5 – y) (3x – 4)
Examples:
1. 6 x  9 x  4 x  6
3
2
The GCF of
6 x  9 x is 3x .
3
2
2
6 x3  9 x 2 4 x  6
3x 2  2 x  3 2  2 x  3
The GCF of
4 x  6 is 2.
  2 x  3  3 x 2  2 
These two terms must be the same.
2. x  x  x  1
3
The GCF of
x3  x 2 is x 2 .
2
x3  x 2  x  1
x2  x  1 1 x  1
The GCF of
x  1 is 1.
These two terms must be the same.
  x  1  x 2  1
Exercise: 1. Factor 2xy + 3y – 4x – 6.
2. Factor 2a2 + 3bc – 2ab – 3ac.
3. x3  2 x 2  x  2
4. x 2 y 2  ay 2  ab  bx 2
https://quizizz.com/admin/quiz/58c006b92933d574185484c0/factoring
Try These:Factor by grouping.
a. 8 x 3  2 x 2  12 x  3
b. 4 x 3  6 x 2  6 x  9
c. x 3  x 2  x  1
d. 3a  6b  5a  10ab
2
Factoring Trinomials
Factoring Trinominals
1. When trinomials have a degree of “2”,
they are known as quadratics.
2. We learned earlier to use the
“diamond” to factor trinomials that had
a “1” in front of the squared term.
x2 + 12x + 35
(x + 7)(x + 5)
Factoring these trinomials is based on reversing the FOIL process.
Example: Factor x2 + 3x + 2. Express the trinomial as a product of
two binomials with leading term x and
2
x + 3x + 2 = (x + a)(x + b)
unknown constant terms a and b.
F O I
L
= x2 + bx + ax + ba Apply FOIL to multiply the binomials.
= x2 + (b + a)x + ba
= x2 + (1 + 2)x + 1 · 2
Since ab = 2 and a + b = 3, it follows
that a = 1 and b = 2.
(Product-sum method)
Therefore, x2 + 3x + 2 = (x + 1)(x + 2).
Copyright © by Houghton Mifflin Company,
Inc. All rights reserved.
12
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Example: Factor x2 – 8x + 15.
x2 – 8x + 15 = (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a + b = -8 and ab = 15.
It follows that both a and b are negative.
Negative Factors of 15
Sum
- 1, - 15
-15
-3, - 5
-8
x2 – 8x + 15 = (x – 3)(x – 5).
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
13 + 36
Example: Factor x2 + 13x
36.
x2 + 13x + 36 = (x + a)(x + b)
= x2 + (a + b)x + ab
Therefore a and b are:
Positive Factors of 36
Sum
two positive factors of 36
1, 36
37
whose sum is 13.
2, 18
3, 12
20
15
4, 9
6, 6
13
12
x2 + 13x + 36 = (x + 4)(x + 9)
Factor
1)
x2 + 15x + 56
2)
x2 + 10x + 9
3)
x2 + 2x – 24
4) x2 + 5x - 14
More Factoring Trinomials
3. When there is a coefficient larger than
“1” in front of the squared term, we can
use a modified diamond or square to
find the factors.
3. Always remember to look for a GCF
before you do ANY other factoring.
More Factoring Trinomials
5. Let’s try this example
3x2 + 13x + 4
Make a box
Write the factors of the first term.
Write the factors of the last term.
Multiply on the diagonal and add to see if
you get the middle term of the
trinomial. If so, you’re done!
Copyright © by Houghton Mifflin Company, Inc. All rights reserved.
Factoring complex trinomials of the form ax2 + bx + c, (a  1)
can be done by decomposition or cross-check method.
Example: Factor 3x2 + 8x + 4.
3  4 = 12
Decomposition Method
1. Find the product of
first and last terms
3. Rewrite the middle
term decomposed
into the two numbers
4. Factor by grouping
in pairs
2. We need to find factors of 12
whose sum is 8
3x2 + 2x + 6x + 4
= (3x2 + 2x) + (6x + 4)
= x(3x + 2) + 2(3x + 2)
= (3x + 2) (x + 2)
3x2 + 8x + 4 = (3x + 2) (x + 2)
1, 12
2, 6
3, 4
Example: Factor 4x2 + 8x – 5.
4  5 = 20
We need to find factors of 20
whose difference is 8
Rewrite the middle term
decomposed into the
two numbers
Factor by grouping
in pairs
4x2 – 2x + 10x – 5
= (4x2 – 2x) + (10x – 5)
= 2x(2x – 1) + 5(2x – 1)
= (2x – 1) (2x + 5)
4x2 + 8x – 5 = (2x –1)(2x – 5)
1, 20
2, 10
4, 5
Factor Completely
1)
2x2 + 7x + 3
2)
3m2 – m – 4
3)
6t2 + 23t + 7
4)
5k2 + k –18
Difference of Squares
Difference of Squares

When factoring using a difference of
squares, look for the following three
things:
– only 2 terms
– minus sign between them
– both terms must be perfect squares
 If all 3 of the above are true, write two
( ), one with a + sign and one with a – sign
: ( + ) ( - ).
Try These
 1.
 2.
 3.
 4.
 5.
 6.
a2 – 16
x2 – 25
4y2 – 16
9y2 – 25
3r2 – 81
2a2 + 16
Perfect Square Trinomials
Perfect Square Trinomials

When factoring using perfect square
trinomials, look for the following three
things:
– 3 terms
– last term must be positive
– first and last terms must be perfect
squares
 If all three of the above are true, write one (
)2 using the sign of the middle term.
Try These
 1.
 2.
 3.
 4.
 5.
 6.
a2 – 8a + 16
x2 + 10x + 25
4y2 + 16y + 16
9y2 + 30y + 25
3r2 – 18r + 27
2a2 + 8a - 8