Aim: How do we factor completely and by grouping?
Objective: use common factoring to factor expressions by grouping and to factor expressions completely.
Lesson Development: Factoring is undoing multiplication and it can produce more than just two factors.
For example: 2( x  4)( x  7) and 4 x(3x  2)(3x  2)
To completely factor an expression means to write it as a product which includes binomials that contain no
greatest common factors (gcf’s). Whether we factor, we always factor completely. It is always easiest to
completely factor by looking for a gcf of the expression first. Once removed, the factoring then either consists
of the difference of perfect squares or standard trinomial techniques.
EX:
a) 4 x 2  12 x  40
b) 6 x 2  24
4 x 2  12 x  40
6 x 2  24
4( x 2  3x  10)
6( x 2  4)
4( x  5)( x  2)
6( x  2)( x  2)
EX1: Write each of the following in its completely factored form.
a) 10 x 2  55x  105
b) 75  3x 2
5(2 x 2  11x  21)
5(2 x  3)( x  7)
75  3x 2
c) 12 x 2  57 x  15
However if the gcf = 1, then we go straight to
factoring.
12 x 2  57 x  15
3(4 x 2  19 x  5)
3(4 x  1)( x  5)
3(25  x 2 )
3(5  x )(5  x )
EX: 4 x 2  5x  6  (4 x  3)( x  2)
You now have essentially three types of factoring: (1) greatest common factor, (2) difference of perfect squares,
and (3) trinomials. We can combine gcf factoring with the other two to completely factor quadratic
expressions. Today we will introduce a new type of factoring known as factoring by grouping. This technique
requires you to see a factor to as a binomial or trinomial.
EX:
a) x(2 x  1)  7(2 x  1)
b) 5x( x  2)  4( x  2)
GCF: (2x+1)
GCF = (x – 2)
x(2 x  1)  7(2 x  1)
5x( x  2)  4( x  2)
(2 x  1)( x  7)
( x  2)(5x  4)
c) ( x  5)( x  7)  ( x  7)( x  1)
GCF = (x – 7)
( x  5)( x  7)  ( x  7)( x  1)
( x  7)( x  5  x  1)
( x  7)(2 x  6) don't forget to factor completely
2( x  7)( x  3)
When an expression contains four terms, we can try factoring by grouping them into two pairs.
a) 3x3  2 x 2  27 x  18
b) 18x3  9 x 2  2 x  1
gcf of first two terms: x 2
gcf of last two terms: 9
18 x 3  9 x 2  2 x  1
3x 3  2 x 2  27 x  18
(2 x  1)(9 x 2  1)
(2 x  1)(3x  1)(3x  1)
9 x 2 (2 x  1)  1(2 x  1)
x 2 (3x  2)  9(3x  2)
(3x  2)( x 2  9)
(3x  2)( x  3)( x  3)
Don’t forget to factor completely
c) x 2  ab  ax  bx
Be careful when you use factoring by grouping. Don't force
the method when it does not apply. This can lead to errors.
Since gcf = 1 as it’s arranged, we will try rearranging to get a gcf  1 .
x 2  ab  ax  bx
x 2  ax  ab  bx
x ( x  a )  b( a  x )
x ( x  a )  b( x  a )
( x  b)( x  a )
HW#12: See worksheet below
Solutions:
1) a) (x+7)(x+5)
e) (2x-1)(3x-1)
b) (4x-3)(x-2)
f) (x+5)(x+11)
c) (x-3)(2x+15)
d) (x+4)(3x-5)
2) ( x  3)(3x  5)
3) a) (2x+7)(5x+3)
4) a) (x +2)(x – 2)(5x+2)
5) a) (x+a)(x-c)
6) x = -3, 3, 5
b) (3x – 5)(4x + 1)
b) (3x-1)(3x+1)(2x-3)
b) (y+b)(a+x)
c) (x+2)(x-5)(x+5)
d) (2 x 2  3)(4 x  5)