1. Factoring x2 + bx + c Objective:
Students will use factor trinomials of the
form x2 + bx + c.

2. Algebra Standards:
11.0 Students apply basic factoring techniques to second
and simple third degree polynomials. These techniques
include finding a common factor to all of the terms in a
polynomial and recognizing the difference of two squares,
and recognizing perfect squares of binomials.
14.0 Students solve a quadratic equation by factoring or
completing the square.

3. (x + 3) (x + 2) First Outer Inner Last x2 +2x +3x +6 x2 + 5x + 6 6
Vocabulary
Multiply – Foil
(x + 3) (x + 2)
First
Outer
Inner
Last x2
+2x
+3x +6 x2
+ 5x
+ 6 6
Factor
( )( ) x + 2 x + 3
1 • 6
2 • 3

4. 48 a) x2 + 8x + 12 12 d) x2 – 16x + 48 ( )( ) x + 2 x + 6 ( )( ) x – 4
#1 Factor 48 a) x2
+ 8x
+ 12 12 d) x2
– 16x
+ 48
1 • 48
1 • 12
2• 24
( )( ) x + 2 x + 6
2• 6
( )( ) x – 4 x – 12
3 • 16
3 • 4
4 • 12
6 • 8 24 b) x2
+ 10x
+ 24 e) x2
+11x
+ 24 24
1 • 24
2 • 12
1 • 24
( )( ) x + 4 x + 6
3 • 8
( )( ) x + 3 x + 8
2• 12
4 • 6
3 • 8
4 • 6 c) x2
– 12x
+ 36 36 f) x2
– 6x
+ 21 21
1 • 36
1 • 21
( )( ) x – 6 x – 6
2• 18
( )( ) x x
3• 7
3 • 12
4 • 9
Not Factorable
6 • 6

5. a) x2 – 4x – 12 12 d) x2 + 9x – 36 36 ( )( ) x + 2 x – 6 ( )( ) x – 3
#2 Factor a) x2
– 4x
– 12 12 d) x2
+ 9x
– 36 36
1 • 12
1 • 36
( )( ) x + 2 x – 6
2• 6
( )( ) x – 3 x + 12
2• 18
3 • 4
3 • 12
4 • 9 b) x2
+ 18x
– 40 40
6 • 6 e) x2
– 3x
– 10 10
1 • 40
( )( ) x – 2 x + 20
2• 20
1 • 10
4 • 10
( )( ) x + 2 x – 5
2• 5
5• 8 c) x2
– 3x
– 18 18 f) x2
+ 4x
– 9 9
1 • 18
1 • 9
( )( ) x + 3 x – 6
2• 9
( )( ) x x
3• 3
3 • 6
Not Factorable

6. 14 10 1) x2 – 5x – 14 5) x2 – 7x + 10 ( )( ) x + 2 x – 7 ( )( ) x – 2
Check Point – Factor: 14 10 1) x2
– 5x
– 14 5) x2
– 7x
+ 10
1 • 14
2• 7
1 • 10
( )( ) x + 2 x – 7
2 • 5
( )( ) x – 2 x – 5 36 2) x2
+ 9x
+ 20 20 6) x2
– 5x
– 36
1 • 36
2 • 18
1 • 20
3 • 12
( )( ) x + 4 x + 5
2• 10
( )( ) x + 4 x – 9
4 • 9
4 • 5
6 • 6 45 3) x2
+ 4x
– 45 7) z2
– z
+ 12 12
1 • 45
( )( ) z 3 z 4
1 • 12
3 • 15
( )( ) x
2• 6 – 5 x + 9
5 • 9
Not Factorable
3• 4 9 8) a2
– 4ab
– 12b2 12 4) x2
+ 6xy
+ 9y2
1 • 12
1 • 9
( )( ) x 3 y x + 3 y
( )( ) a + 2 b a – 6 b +
2 • 6
3 • 3
3 • 4

7. The solutions are -2 and -7 .
#3 Solve a Quadratic Equation
Solve x2
+ 9x =
-14
by factoring.
x2 + 9x = -14 14
+14
+14
1 • 14
2• 7 x2
+ 9x
+ 14 =
( )( ) x + 2 x + 7 =
x + 2 = 0 or
x + 7 = 0 -7 -7 -2 -2 x = -7 x = -2
The solutions are -2 and -7 .

8. Solve x2 – 8x = 20 by factoring. x2 – 8x = 20 20 -20 -20 x2 – 8x – 20
Check Point:
Solve x2
– 8x = 20
by factoring.
x2 – 8x = 20 20
-20
-20
1 • 20
2 • 10 x2
– 8x
– 20 =
4 • 5
( )( ) x + 2 x – 10 =
x + 2 = 0 or
x – 10 = 0
+10
+10 -2 -2 x = 10 x = -2
The solutions are -2 and 10 .

9. by factoring and the Quadratic Formula. Factoring:
Check Point:
Solve x2
– 5x = –4
by factoring and the Quadratic
Formula.
Factoring:
x2 – 5x = –4 +4 +4 x2
– 5x
+ 4 = 4
( )( ) x – 1 x – 4 =
1 • 4
2 • 2
x – 1 = 0 or
x – 4 = 0 +4 +4 +1 +1 x = 4 x = 1
The solutions are 1and 4 .

10. = = = = Solve x2 – 5x = –4 by factoring and the Quadratic Formula.
Check Point:
Solve x2
– 5x = –4
by factoring and the Quadratic
Formula.
Quadratic Formula:
x2 – 5x = –4 2
(-5) – -
(-5) 4
(1)
(4) +4 +4 = 2
(1) x2
– 5x
+ 4 = 25
– 16 5 = 2 5 5 3 = = 2 2
a = 1,
b = -5,
c = 4 5 – 3 5 + 3 2
The solutions are 1and 4 . 2
x = and 4 1

11. Sketch the graph of y = (x – 1)(x – 4) The solutions are 1 and 4
y- intercept =
(-1)
(-4)
The solutions are 1 and 4 x y
Find the x-coordinate
of the vertex 1 + 4 5
x = = =
2.5 2 2
Find the y-coordinate
of the vertex
y = ( – 1)( – 4)
2.5
2.5
y =
(1.5)
(-1.5) =
-2.25
Vertex = (2.5, -2.25)

12. Assignment
Book Pg. 599 – 600
#13 – 35 odd, 36, 37, 38