Section 6.3
In this section, we are looking at other trinomials.
These trinomials differ from the ones we factored
in the previous section:

They have a leading coefficient that is a number other
than 1

There is no GCF that can be factored out to make the
leading coefficient 1
There are two methods for factoring trinomials of
this type.

Method 1: Trial and Error
List all possible factors until the correct pair of
factors is found.

Method 2: “ac” Method
This method is based on the factoring by
grouping process.
Method 1: Factoring ax2 + bx + c
by trial and error
Factor the trinomial: 2x2 – 5x – 3.
We know that the factors (if they exist) will be a
pair of binomials.
The product of their first terms is 2x2 and the
product of their last term is –3.
Let’s list all the possible factors along with the trinomial
that would result if we were to multiply them together.
Remember, the middle term comes from the product of the
inside terms plus the product of the outside terms.
We can see from the last line that the factors of
2x2 – 5x – 3 are (2x + 1) and (x – 3).
Method 2: Factoring ax2 + bx + c
by
“ac” Method
Factor: 6x2 – 7x + 2
a = 6, b = -7, c = 2, ac = 12
We want to find 2 factors of ac = 12 that combine to b = -7.
Possible factors of 12: Factors of 12
1
12
2
6
3
4
(- 3) ( -4) = 12
(- 3) + ( -4) = - 7
If we rewrite the middle term –7x as –3x – 4x, the trinomial will become
a polynomial with four terms: 6x2 – 3x – 4x + 2
We can factor by grouping!
In a previous lesson, we used factoring by grouping to
factor this polynomial.
For review, here is the complete problem again:
6x2 – 3x – 4x + 2 = (6x2 – 3x) + (– 4x + 2)
= 3x(2x – 1) – 2(2x – 1)
= (2x – 1) (3x – 2)
This means that 6x2 – 7x + 2 can be factored to
(2x – 1)(3x – 2).
To generalize this discussion, here are the steps we
use to factor trinomials by the “ac” Method.
Factor 3x2 – 10x – 8 using these steps.
The trinomial, 3x2 – 10x – 8 has the form ax2 + bx + c,
where a = 3, b = –10, and c = –8.
Step 1: The product ac is 3(–8) = –24.
Step 2: We need two numbers whose product is –24
and whose sum is –10.
Let’s list all the pairs of numbers whose product is –24 to
find the pair whose sum is –10.
Factors of
24
1
24
2
12
( 2 ) (- 12) = - 24
( 2) + (- 12) = - 10
The numbers are 2 and –12.
Step 3: We now rewrite our original trinomial so the
middle term, –10x, is written as the sum of –12x and
2x:
3x2 – 10x – 8 = 3x2 + 2x – 12x – 8
Step 4: Factoring by grouping, we have:
3x2 + 2x – 12x– 8 = (3x2+2x) + (– 12x – 8)
= x(3x + 2) – 4 (3x +2)
= (3x + 2)(x – 4)
We can check our work by multiplying and 3x + 2 and
x – 4 to get 3x2 – 10x – 8.
Factor
Answer:
Factor
Answer:
2 x + 7 x − 15
2
(x+5) (2x-3)
4 x − 11x − 3
2
(4x+1) (x-3)
Factoring Out the Greatest
Common Factor
Factor: 12y3 + 10y2 – 12y
Step 1: Factor out the greatest common factor, 2y:
12y3 + 10y2 – 12y = 2y(6y2 + 5y – 6)
So 2y is one factor. We will now factor the trinomial in
parenthesis to find two additional factors.
Step 2: a = 6, b = 5, c = – 6, ac = 6(–6) = –36
Step 3: We need two numbers whose product is –36
and whose sum is 5.
Factors of 36
1
36
2
18
3
12
4
9
(- 4) ( 9) = - 36
(- 4) + (9) = 5
We now rewrite our original trinomial so the middle
term, 5y, is written as the sum of –4y and 9y:
2y(6y2 +5y– 6) = 2y(6y2 – 4y+9y – 6)
Step 4: Factoring by grouping, we have:
2y(6y2 – 4y+9y – 6) = 2y[2y(3y – 2) + 3(3y – 2)]
= 2y (3y – 2)(2y+3)
The factors are 2y (3y – 2)(2y+3)
We can check our work by multiplying 2y and 3y – 2 and
2y + 3 to get 12y2 +10y– 12.
Section 6.3
Page 443-446
#1, 5, 9, 15, 21, 31, 35