1. Ch 10: Polynomials F) Factoring ax2 + bx + c
Objective:
To factor polynomials when a ≠ 1

2. Definitions Polynomial in Standard Form: ax2 + bx + c = 0
A polynomial written in descending
order based on the exponents.
Polynomial in Factored Form: (a1x + c1)(a2x + c2) = 0
A polynomial written as the product
of two (or more) binomials.

3. Table Method a1 c1 a2 c2 b = a1 c2 + c1 a2
Arrange the polynomial in Standard Form everything on one side and zero on the other
Create a 5 column table listing all of the factors of a in columns 1 and 3
List all of the factors of c in columns 2 & 4
Multiply the values in columns 1 & 4 and Multiply the values in columns 2 & 3. Place the sum of the products in column 5
Look for the value of b in column 5 and place the values from columns 1 & 2 and 3 & 4 into the Factored Form.
(a1x + )(a2x + ) = 0 c1 c2 a1 c1 a2 c2
b = a1 c2 + c1 a2

4. Table Method ( ) ( ) Example 1 2x2 + 5x + 3 = 0 12 1 3 = 3 1 −1−3
Reverse Order
12
1 3
= 3 1
−1−3
= −3−1 a1 c2 c1 a2 a1 c1 a2 c2
b = +
( )
( )
1 3 + 1 1 2 3
5 =
12 1 −1 2 −3
1−3 +
−12 1 3 2 1
1 1 +
32
1−1 +
−32 1 −3 2 −1
( x + )( x + ) = 0 a1 c1 a2 c2
(1x + 1)
(2x + 3)
= 0
(x + 1)(2x + 3) = 0

5. Table Method ( ) ( ) Example 2 2x2 − 9x + 7 = 0 12 1 7 = 7 1 −1−7
Reverse Order
12
1 7
= 7 1
−1−7
= −7−1 a1 c2 c1 a2 a1 c1 a2 c2
b = +
1 7 + 1 1 2 7
9 =
12
( )
( ) 1 −1 2 −7
-9 =
1−7 +
−12 1 7 2 1
1 1 +
72
1−1 +
−72 1 −7 2 −1
( x + )( x + ) = 0 a1 c1 a2 c2
(1x +−1)
(2x +−7)
= 0
(x − 1)(2x − 7) = 0

6. Table Method ( ) ( ) −1 8 = 8−1 Example 3 1−8 = −8 1 17 −2 4
Reverse Order
−1 8
= 8−1
Example 3
1−8
= −8 1
17
−2 4
= 4−2
7x2 − 55x − 8 = 0
2−4
= −4 2 a1 c2 c1 a2 a1 c1 a2 c2
b = + + 1 −1 7 8 1=
1 8
−17 1 1 7 −8
−1=
1−8 +
17 1 −2 7 4
−10 =
1 4 +
−27
10 =
1−4 +
27 1 2 7 −4 1 8 −1
55 =
1−1 +
87 7
( )
( )
−55 =
1 1 +
−87 1 −8 7 1 + 1 4 7 −2
1−2
47 1 −4 7 2
1 2 +
−47
( x + )( x + ) = 0 a1 c1 a2 c2
(1x +−8)
(7x + 1)
= 0
(x − 8)(7x + 1) = 0

7. Table Method ( ) ( ) 14 Example 4 −1 3 = 3−1 22 1−3 = −3 1
Reverse Order
14
Example 4
−1 3
= 3−1
22
1−3
= −3 1
4x2 + 4x – 3 = 0 a1 c2 c1 a2 a1 c1 a2 c2
b = + + 1 −1 4 3
−1=
1 3
−14 1 1 4 −3 1=
1−3 +
14 1 4
11=
1−1 +
34 3 −1
−11=
1 1 +
−34 1 −3 4 1
( )
( ) −1 3
4 =
2 3 +
−12 2 2
2−3 +
12 2 1 2 −3 + 2 3 2 −1
2−1
32 2 −3 2 1
2 1 +
−32
( x + )( x + ) = 0 a1 c1 a2 c2
(2x +−1)
(2x + 3)
= 0
(2x − 1)(2x + 3) = 0

8. Algebra Tiles x + 1 2x + 3 + + = (2x + 3) (x + 1) = 0
Arrange the polynomial in Standard Form everything on one side and zero on the other
Lay out the tiles that represent the polynomial
2x2 + 5x + 3 = 0
3) Rearrange the tiles so they form a rectangle
4) The tiles across the bottom and down the right side represent the binomials in Factored Form.
x + 1
2x + 3 + + =
(2x + 3)
(x + 1)
= 0
2x2 5x 3

9. 1) 2) 3) 4) 2x2 + 7x + 3 = 0 2x2 − 19x + 24 = 0 (2x + 1)(x + 3) = 0
Classwork 1) 2)
2x2 + 7x + 3 = 0
2x2 − 19x + 24 = 0
(2x + 1)(x + 3) = 0
(2x − 3)(x − 8) = 0 3) 4)
5x2 − 29x + 20 = 0
2x2 − 7x − 49 = 0
(5x − 4)(x − 5) = 0
(2x + 7)(x − 7) = 0

10. 5) 6) 7) 8) 4x2 + 4x + 1 = 0 4x2 − 25 = 0 + 0x (2x + 1)(2x + 1) = 0