For Educational Use Only © 2010 10.5 Factoring x 2 + bx + c Brian Preston Algebra 1 2009-2010.

For Educational Use Only © 2010

10.5 Factoring x2 + bx + c

Brian PrestonAlgebra 1 2009-

2010


Real World Application

How wide is the stone border?


Lesson Objectives

1) Factor a quadratic expression of the form x2 + bx + c.

2) Solve quadratic equations by factoring.


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


2) x2 + 3x + 2 1

1 + 2

Factors of 2

1 2+ 3

+3=

Factor the trinomial.

Example


+2+1

+2+1

2) x2 + 3x + 2

1x1x

1

1 + 221( + 2 + 1

+3=


1 2

Factors of 2

)(1x )

1x2

1x

Example


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


= Garden area–(x + 15)Total

area

15

border

615


3)(x + 15)(x + 6) – (15)(6) = 46615 615

– 46– 46


– 46

=+ 21xx2 0– 46

x2 + 15x+ 6x + 90

x2 + 21x + 0

– 90 = 46

= 46

x x xx


3) x2 + 21x – 46 = 01Factors of 46

1 46+ 21

+21=

2 23



1 + 46 47=

46 – 1 45=


3) x2 + 21x – 46 = 01

– 2 + 23 +21=


1 46

2 23

Factors of 46


2 + 23 25=

23 – 2 21=


3) x2 + 21x – 46 = 0

+23

-2

+23

-2

11x2

1x 1x1x 1x– 2 + 2323– 2( + 23 – 2

+21=


1 46

2 23

Factors of 46

)( )


= 0


– 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



How wide is the stone border?

2 yards wide


What is factoring?

Factoring is another way to solve for variables in a

quadratic equations.


4) x2 – 5x + 6 1

1 + – 6

Factors of 6

1 6– 5

-5=

2 3


Example

1 + 6 7=

6 – 1 5= (–1) to all

1 + – 6


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)


4) x2 – 5x + 6 1

– 2 + – 3 -5=


1 6

2 3

Factors of 6

Example

2 + 3 5=

3 – 2 1= (–1) to all


– 3

-2

– 3

-2

4) x2 – 5x + 6

1x1x

1

– 2 + – 3– 3– 2( – 3 – 2

-5=


1 6

2 3

Factors of 6

)(1x )

1x2

1x

Example


5) x2 – 2x – 81Factors of 8

1 8– 2

-2=

2 4


Example

1 + 8 9=

8 – 1 7=


5) x2 – 2x – 81

2 + – 4 -2=


1 8

2 4

Factors of 8

Example

2 + 4 6=

4 – 2 2= (–1) to all


5) x2 – 2x – 8

– 4– 41x1x

1

+2+2

2 + – 4– 42( – 4 + 2

-2=


1 8

2 4

Factors of 8

)(1x )

1x2

1x

Example


16) x2 + 7x – 18Factors of 18

1 18+ 7

+7=

2 9


Example

3 6

1 + 18 19=

18 – 1 17=


6) x2 + 7x – 181

– 2 + 9 +7=


1 18

2 9

Factors of 18

Example

3 6

2 + 9 11=

9 – 2 7=


+9

-2

+9

-2

6) x2 + 7x – 18

1x1x

1

– 2 + 99– 2( + 9 – 2

+7=


1 18

2 9

Factors of 18

)(1x )

1x2

1x

Example

3 6


17) w2 + 13w + 36Factors of 36

+ 13

+13=


Example

1 36

2 18

3 12

4 96 6

1 + 36 37=

36 – 1 35=


7) w2 + 13w + 361Factors of 36

+13=


Example

1 36

2 18

3 12

4 96 6

2 + 18 20=

18 – 2 16=


7) w2 + 13w + 361Factors of 36

+13=


Example

1 36

2 18

3 12

4 96 6

3 + 12 15=

12 – 3 9=


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=


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


8) 32 + 12n + n2

(4 + n)(3 + n)

(n + 1)(n + 2)

ExampleFactor the trinomial.


8) 32 + 12n + n2

Example

Factors of 321 32+ 12

+12= 2 16


1

3 12

4 86 6

1 + 32 33=

32 – 1 31=


8) 32 + 12n + n2

+12=


Example

1 36

2 18

3 12

4 86 6

Factors of 321

2 + 18 20=

18 – 2 16=


8) 32 + 12n + n2

+12=


Example

1 36

2 18

3 12

4 86 6

1Factors of 32

3 + 12 15=

12 – 3 9=


8) 32 + 12n + n21

4 + 8 +12=


Example

1 36

2 18

3 12

4 86 6

Factors of 32

4 + 8 12=

8 – 4 4=


+8+88 +

8) 32 + 12n + n2

1n 1n1n 1n

+4+4

1

4 + 884(4 +

+12=


)( )

Example

1 36

2 18

3 12

4 86 6

Factors of 321n2


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


(1) (– 4) (3) 9) 2 – 4

9 + 16

2525 is a perfect square, so Yes

Example


– 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


(1) (– 6) (3) 10) 2 – 4

9 + 24

3333 is not a perfect square, so No

Example


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=


11) b2 + 7b + 10 = 01

2 + 5 +7=


1 10

2 5

Factors of 10

Example

2 + 5 7=

5 – 2 3=


+5+51b + 5 + 21b1b 1b

11) b2 + 7b + 10 = 01

+2+2

2 + 552(

+7=


1 10

2 5

Factors of 10

)( )

1b2

Example

= 0


– 5 – 2– 5 – 2

(b + 5)(b + 2)(b + 5)(b + 2)

b + 5


11) = 0

= 0= 0 b + 2

b

– 5

= – 5

( ) ( )

b

– 2

= – 2


12) x2 + 5x – 14 = 01Factors of 14

1 14+ 5

+5=

2 7


Example

Standard form

1 + 14 15=

14 – 1 13=


12) x2 + 5x – 14 = 01

– 2 + 7 +5=


1 14

2 7

Factors of 14

Example

2 + 7 9=

7 – 2 5=


+7

-2

+7

-2

1x 1x1x 1x

12) x2 + 5x – 14 = 0

+ 7 – 2

1

– 2 + 77– 2(

+5=


1 14

2 7

Factors of 14

)( )

1x2

Example

= 0


– 7 + 2– 7 + 2

(x + 7)(x – 2)(x + 7)(x – 2)

x + 7


12) = 0

= 0= 0 x – 2

x

– 7

= – 7

( ) ( )

x

+ 2

= 2


+ 4+ 4

Example

13) x2 – 2x – 19 = – 4


+ 4

=– 2xx2 0– 15

Standard form


13) x2 – 2x – 15 = 01Factors of 15

1 15– 2

-2=

3 5


Example

1 + 15 16=

15 – 1 14=


13) x2 – 2x – 15 = 01

3 + – 5 -2=


1 15

3 5

Factors of 15

Example

3 + 5 8=

5 – 3 2= (–1) to all


-5-5+3+3

13) x2 – 2x – 15 = 0

1x 1x1x 1x– 5 + 3

1

3 + – 5– 53(

-2=


1 15

3 5

Factors of 15

)( )

1x2

Example

= 0


+ 5 – 3+ 5 – 3

(x – 5)(x + 3)(x + 3)

x – 5


13) = 0

= 0= 0 x + 3

x

+ 5

= 5

( ) ( )

x

– 3

= – 3

(x – 5)


+ 15+ 15+ 15

Example

14) x2 + 16x = – 15


=+ 16xx2 0+ 15

Standard form


14) x2 + 16x + 15 = 01

1 + 15

Factors of 15

1 15+ 16

+16=

3 5


Example

1 + 15 16=

15 – 1 14=


+1+1

+15(

+15

1 + 15

14) x2 + 16x + 15 = 0

1x 1x1x 1x+ 15 + 1

1

151(

+16=


1 15

3 5

Factors of 15

) )

1x2

Example

= 0


– 1– 1– 15– 15

(x + 15)(x + 1)(x + 15)(x + 1)

x + 15


14) = 0

= 0= 0 x + 1

x

– 15

= – 15

( ) ( )

x

– 1

= – 1


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


pg. 462-463 #’s 10-37, 40-46 even

The Assignment


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Bibliography

For Educational Use Only © 2010 10.5 Factoring x 2 + bx + c Brian Preston Algebra 1 2009-2010.

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