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For Educational Use Only © 2010
10.5 Factoring x2 + bx + c
Brian PrestonAlgebra 1 2009-
2010
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Real World Application
How wide is the stone border?
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Lesson Objectives
1) Factor a quadratic expression of the form x2 + bx + c.
2) Solve quadratic equations by factoring.
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Now we are going to learn how to do this backwards.
Review
(x + 4)(x + 5)5x 5x 44x x 1)
x2 + 4x+ 5x + 20
x2 + 9x + 20
First Outside Inside Last
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2) x2 + 3x + 2 1
1 + 2
Factors of 2
1 2+ 3
+3=
Factor the trinomial.
Example
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+2+1
+2+1
2) x2 + 3x + 2
1x1x
1
1 + 221( + 2 + 1
+3=
Factor the trinomial.
1 2
Factors of 2
)(1x )
1x2
1x
Example
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3) Mike is building a stone border along two sides of a rectangular Japanese garden that measures 6 yards by 15 yards. His budget limits him to only enough stone to cover 46 square yards. How wide should the border be?
borderborder46
615
(15)(6)Are of border46 (x + 6)
46
6
Real World Application
= Garden area–(x + 15)Total
area
15
border
615
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3)(x + 15)(x + 6) – (15)(6) = 46615 615
– 46– 46
Real World Application
– 46
=+ 21xx2 0– 46
x2 + 15x+ 6x + 90
x2 + 21x + 0
– 90 = 46
= 46
x x xx
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3) x2 + 21x – 46 = 01Factors of 46
1 46+ 21
+21=
2 23
Factor the trinomial.
Real World Application
1 + 46 47=
46 – 1 45=
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3) x2 + 21x – 46 = 01
– 2 + 23 +21=
Factor the trinomial.
1 46
2 23
Factors of 46
Real World Application
2 + 23 25=
23 – 2 21=
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3) x2 + 21x – 46 = 0
+23
-2
+23
-2
11x2
1x 1x1x 1x– 2 + 2323– 2( + 23 – 2
+21=
Factor the trinomial.
1 46
2 23
Factors of 46
)( )
Real World Application
= 0
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– 23– 23 + 2+ 2
(x + 23)(x – 2)(x + 23)(x – 2)
x + 23
ExampleSolve the equation by factoring.
3) = 0
= 0= 0 x – 2
x
– 23
= – 23
( ) ( )
x
+ 2
= 22 yards wide
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Real World Application
How wide is the stone border?
2 yards wide
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What is factoring?
Factoring is another way to solve for variables in a
quadratic equations.
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4) x2 – 5x + 6 1
1 + – 6
Factors of 6
1 6– 5
-5=
2 3
Factor the trinomial.
Example
1 + 6 7=
6 – 1 5= (–1) to all
1 + – 6
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Rule
x2 – 2x – 8
One (-) & One (+)
Patterns for factoring trinomials.
(x + 2)(x - 4)
x2 + 6x + 8(x + 2)(x + 4)
x2 – 6x + 8(x – 2)(x – 4)
Two (+)Two (-)
x2 + 2x – 8(x + 4)(x - 2)
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4) x2 – 5x + 6 1
– 2 + – 3 -5=
Factor the trinomial.
1 6
2 3
Factors of 6
Example
2 + 3 5=
3 – 2 1= (–1) to all
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– 3
-2
– 3
-2
4) x2 – 5x + 6
1x1x
1
– 2 + – 3– 3– 2( – 3 – 2
-5=
Factor the trinomial.
1 6
2 3
Factors of 6
)(1x )
1x2
1x
Example
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5) x2 – 2x – 81Factors of 8
1 8– 2
-2=
2 4
Factor the trinomial.
Example
1 + 8 9=
8 – 1 7=
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5) x2 – 2x – 81
2 + – 4 -2=
Factor the trinomial.
1 8
2 4
Factors of 8
Example
2 + 4 6=
4 – 2 2= (–1) to all
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5) x2 – 2x – 8
– 4– 41x1x
1
+2+2
2 + – 4– 42( – 4 + 2
-2=
Factor the trinomial.
1 8
2 4
Factors of 8
)(1x )
1x2
1x
Example
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16) x2 + 7x – 18Factors of 18
1 18+ 7
+7=
2 9
Factor the trinomial.
Example
3 6
1 + 18 19=
18 – 1 17=
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6) x2 + 7x – 181
– 2 + 9 +7=
Factor the trinomial.
1 18
2 9
Factors of 18
Example
3 6
2 + 9 11=
9 – 2 7=
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+9
-2
+9
-2
6) x2 + 7x – 18
1x1x
1
– 2 + 99– 2( + 9 – 2
+7=
Factor the trinomial.
1 18
2 9
Factors of 18
)(1x )
1x2
1x
Example
3 6
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17) w2 + 13w + 36Factors of 36
+ 13
+13=
Factor the trinomial.
Example
1 36
2 18
3 12
4 96 6
1 + 36 37=
36 – 1 35=
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7) w2 + 13w + 361Factors of 36
+13=
Factor the trinomial.
Example
1 36
2 18
3 12
4 96 6
2 + 18 20=
18 – 2 16=
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7) w2 + 13w + 361Factors of 36
+13=
Factor the trinomial.
Example
1 36
2 18
3 12
4 96 6
3 + 12 15=
12 – 3 9=
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7) w2 + 13w + 361
4 + 9 +13=
Factor the trinomial.Factors of 36
Example
1 36
2 18
3 12
4 96 6
4 + 9 13=
9 – 4 5=
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7) w2 + 13w + 36
+4+4
1w 1w1w 1w
+9+9
1
4 + 994( + 9 + 4
+13=
Factor the trinomial.Factors of 36
)( )
1w2
Example
1 36
2 18
3 12
4 96 6
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8) 32 + 12n + n2
(4 + n)(3 + n)
(n + 1)(n + 2)
ExampleFactor the trinomial.
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8) 32 + 12n + n2
Example
Factors of 321 32+ 12
+12= 2 16
Factor the trinomial.
1
3 12
4 86 6
1 + 32 33=
32 – 1 31=
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8) 32 + 12n + n2
+12=
Factor the trinomial.
Example
1 36
2 18
3 12
4 86 6
Factors of 321
2 + 18 20=
18 – 2 16=
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8) 32 + 12n + n2
+12=
Factor the trinomial.
Example
1 36
2 18
3 12
4 86 6
1Factors of 32
3 + 12 15=
12 – 3 9=
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8) 32 + 12n + n21
4 + 8 +12=
Factor the trinomial.
Example
1 36
2 18
3 12
4 86 6
Factors of 32
4 + 8 12=
8 – 4 4=
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+8+88 +
8) 32 + 12n + n2
1n 1n1n 1n
+4+4
1
4 + 884(4 +
+12=
Factor the trinomial.
)( )
Example
1 36
2 18
3 12
4 86 6
Factors of 321n2
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Example
33
x2 + 3x – 4
– 4– 4
(1) (– 4) a (3)
9) Tell whether the trinomial can be factored.
1ax2 + bx + c = 0a = b =
3
1
1
c =
– 4
b2 – 4ac
2 – 4b c
1
Standard form
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(1) (– 4) (3) 9) 2 – 4
9 + 16
2525 is a perfect square, so Yes
Example
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– 6– 633
x2 + 3x – 6
(1) (– 6) a (3)
10) Tell whether the trinomial can be factored.
1ax2 + bx + c = 0a = b =
3
1
1
c =
– 6
b2 – 4ac
2 – 4b c
1
Standard form
Example
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(1) (– 6) (3) 10) 2 – 4
9 + 24
3333 is not a perfect square, so No
Example
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11) b2 + 7b + 10 = 01Factors of 10
1 10+ 7
+7=
2 5
Solve the equation by factoring.
Example
Standard form
1 + 10 11=
10 – 1 9=
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11) b2 + 7b + 10 = 01
2 + 5 +7=
Solve the equation by factoring.
1 10
2 5
Factors of 10
Example
2 + 5 7=
5 – 2 3=
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+5+51b + 5 + 21b1b 1b
11) b2 + 7b + 10 = 01
+2+2
2 + 552(
+7=
Solve the equation by factoring.
1 10
2 5
Factors of 10
)( )
1b2
Example
= 0
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– 5 – 2– 5 – 2
(b + 5)(b + 2)(b + 5)(b + 2)
b + 5
ExampleSolve the equation by factoring.
11) = 0
= 0= 0 b + 2
b
– 5
= – 5
( ) ( )
b
– 2
= – 2
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12) x2 + 5x – 14 = 01Factors of 14
1 14+ 5
+5=
2 7
Solve the equation by factoring.
Example
Standard form
1 + 14 15=
14 – 1 13=
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12) x2 + 5x – 14 = 01
– 2 + 7 +5=
Solve the equation by factoring.
1 14
2 7
Factors of 14
Example
2 + 7 9=
7 – 2 5=
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+7
-2
+7
-2
1x 1x1x 1x
12) x2 + 5x – 14 = 0
+ 7 – 2
1
– 2 + 77– 2(
+5=
Solve the equation by factoring.
1 14
2 7
Factors of 14
)( )
1x2
Example
= 0
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– 7 + 2– 7 + 2
(x + 7)(x – 2)(x + 7)(x – 2)
x + 7
ExampleSolve the equation by factoring.
12) = 0
= 0= 0 x – 2
x
– 7
= – 7
( ) ( )
x
+ 2
= 2
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+ 4+ 4
Example
13) x2 – 2x – 19 = – 4
Solve the equation by factoring.
+ 4
=– 2xx2 0– 15
Standard form
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13) x2 – 2x – 15 = 01Factors of 15
1 15– 2
-2=
3 5
Solve the equation by factoring.
Example
1 + 15 16=
15 – 1 14=
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13) x2 – 2x – 15 = 01
3 + – 5 -2=
Solve the equation by factoring.
1 15
3 5
Factors of 15
Example
3 + 5 8=
5 – 3 2= (–1) to all
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-5-5+3+3
13) x2 – 2x – 15 = 0
1x 1x1x 1x– 5 + 3
1
3 + – 5– 53(
-2=
Solve the equation by factoring.
1 15
3 5
Factors of 15
)( )
1x2
Example
= 0
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+ 5 – 3+ 5 – 3
(x – 5)(x + 3)(x + 3)
x – 5
ExampleSolve the equation by factoring.
13) = 0
= 0= 0 x + 3
x
+ 5
= 5
( ) ( )
x
– 3
= – 3
(x – 5)
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+ 15+ 15+ 15
Example
14) x2 + 16x = – 15
Solve the equation by factoring.
=+ 16xx2 0+ 15
Standard form
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14) x2 + 16x + 15 = 01
1 + 15
Factors of 15
1 15+ 16
+16=
3 5
Solve the equation by factoring.
Example
1 + 15 16=
15 – 1 14=
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+1+1
+15(
+15
1 + 15
14) x2 + 16x + 15 = 0
1x 1x1x 1x+ 15 + 1
1
151(
+16=
Solve the equation by factoring.
1 15
3 5
Factors of 15
) )
1x2
Example
= 0
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– 1– 1– 15– 15
(x + 15)(x + 1)(x + 15)(x + 1)
x + 15
ExampleSolve the equation by factoring.
14) = 0
= 0= 0 x + 1
x
– 15
= – 15
( ) ( )
x
– 1
= – 1
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1) Don’t forget the negative signs.
2) Make sure the expressions or equations are in standard form before factoring.
Key Points & Don’t Forget
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pg. 462-463 #’s 10-37, 40-46 even
The Assignment
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