for educational use only © 2010 10.8 factoring using the distributive property brian preston...

For Educational Use Only © 2010

10.8 Factoring Using the

Distributive Property

Brian PrestonAlgebra 1 2009-

2010


How fast do you need to run so this does not happen?

Real World Application


1) Use the distributive property to factor a polynomial.

2) Solve polynomial equations by factoring.

Lesson Objectives


Review

1) Factor the GCF out of 14x4 – 21x2.

Taking the GCF out is always out first step. Sometimes

there will be no GCF.


Review

1) What is the GCF of 14x4 – 21x2.

14x4

21x2

GCF

=

=

=

2 7

3 7

7

x x x x

x x

x = 7x2 x


Review

7x2 (2x2 – 3 )

1) Factor the GCF out of 14x4 – 21x2.


(x)2 2222x

RuleIf you take out a GCF,

always look to factor again.

(x x

2) 3x2 – 12

2)( 2)

= 43(

+ –

= –

x2 – )

3[ ]

3

x


x2

3) 4x3 + 20x2 + 24x

Factor the expression completely.

4 ( )+ 5 + 6

Example

First, factor the GCF out. Then, factor x2 + 5x + 6

normally. Last, make sure the GCF is a part of the answer.

x x


x2 + 5x + 6

– 1 + 6

Example

3)Factors of 6

1 6+ 5

+5=

2 3

Factor the trinomial.

4x( )

1 + 6 7=

6 – 1 5=


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)


4x(x2 + 5x + 63) )

2 + 3

Example

+5=


Factors of 6

1 6

2 32 + 3 5=

3 – 2 1=


1x1x 1x1x

+2

+3+2

+3

4x(x2 + 5x + 63) )

2 + 332( + 3 + 2

Example

+5=


Factors of 6

)( )

x2

1 6

2 34x


4) In the sport of pole-vaulting, the height h (in feet) reached by a pole-vaulter is a function of v, the velocity of the pole-vaulter, as shown in the model below. The constant g is approximately 32 feet per second per second. To reach a height of 9 feet, what is the pole-vaulter’s velocity?velocityvelocity

99

3232

(32)


9v

h =2

2 g


4) 9 =

What is the pole-vaulter’s velocity needed to reach 9 feet high?

=9


v2

2(32)

v2

641

64


64 1

4) 9 =


=9

=


v2

2(32)

v2

64 64

1

276 v2 =24 ft/sec v+-

64 1



24 ft/sec



m2

5) – 7m2 + 28m – 21


– 7 ( )– 4 + 3

Example

First, factor the GCF out. Then, factor x2 + 5x + 6

normally. Last, make sure the GCF is a part of the answer.

m


m2 – 4m + 3

– 1 + – 3

5)Factors of 3

1 3– 4

-4=


– 7( )

Example

1 + 3 4=

3 – 1 2= (–1) to all


1m1m 1m1m

-1

-3

-1

-3

m2 – 4m + 35) – 7( )

– 1 + – 3– 3 – 1( – 3 – 1

-4=


)( )

m2

1 3

– 7

Factors of 3

Example


(3x)23x3x 2222

RuleFactor the expression completely.

(3x 3x

6) 45x4 – 20x2

2)( 2)

= 45 (

+ –

= –

9 – )

5x2 [ ]

5x2

x2 x2


6266(t)2tt

RuleFactor the expression completely.

( t t

7) 4t3 – 144t

6)( 6)

= 364 (

+ –

= –

t2 – )

4t [ ]

4t

t


8) x3 + 2x2 + 3x + 6

Factor the expression by grouping.

Example

xx2( )+ 2

x3 + 2x2 + 3x + 6( ) ( )

x+ 3 ( )+ 2

x( )+ 2 x2( )+ 3

x( )+ 2 x( )+ 2Always try to factor again.

Impossible


– 9x2 )

– 2 – 2– 2 – 2

9) x3 – 2x2 – 9x + 18

Factor the expression by grouping.

Example

xx2( )

x3 – 2x2 – 9x + 18( ) ( )

x– 9 ( )

x( )– 2 (

x( ) x( ) Always try to factor again.(x x3)( 3)+ –(x 2)–


The order to factoring.

First, factor the GCF out.

Then, identify which of the three factoring ways. (Two terms, three terms or four terms.)

Rule

Always try to factor again.


32(2x)2 32x 32x

ExampleSolve.

(2x 2x

10) 8x3 – 18x = 0

3)( 3)

92 (

+ –

–

4 – )

2x [ ]

2x

x x2

=

=

=

0

0

0


2 2

+ 3+ 3

2 2

2 2

(2x – 3)(2x + 3)2x (2x – 3)2x(2x + 3)

– 3– 3

10) = 0= 0= 0 2x – 32x + 3

– 3

( ) ( )+ 3

= 02x( )

2x = – 3 2x = 32

x =2

x =3

2

3

2 –

ExampleSolve.

2

x = 0


3

11) – 6x2 + 34x + 56 = 0


– 2 ( )– 17 – 28

Example

First, factor the GCF out. Then, factor the three-

term way.

xx2 = 0


Example

11) – 2(3x2 – 17x – 28) = 0

Factors of 84

1 84– 17

-17=2 42

Solve the equation by factoring.

3 284 216 147 12

1 + 84 85=

84 – 1 83=


11) – 2(3x2 – 17x – 28) = 0

Example

Factors of 84

1 84

-17=2 42


3 284 216 147 12

2 + 42 44=

42 – 2 40=


11) – 2(3x2 – 17x – 28) = 0

Example

Factors of 84

1 84

-17=2 42


3 284 216 147 12

3 + 28 31=

28 – 3 25=


11) – 2(3x2 – 17x – 28) = 0

4 + – 21

Example

-17=


Factors of 84

1 842 423 284 216 147 12

4 + 21 25=

21 – 4 17= (–1) to all


1x3x 1x3x+4+4

-21-21

(

11) – 2(3x2 – 17x – 28) = 0

4 + – 21– 214( + 4 – 7

Example

-17=


Factors of 84

) )=0

3x2 1 842 423 284 216 147 12

– 2


3 3

– 4– 4 + 7+ 7

(x - 7)(3x + 4)(x - 7)(3x + 4)– 2– 211) = 0= 0= 0 x – 73x + 4

– 4

( ) ( )+ 7

= 0–2( )

3x = – 4 x = 73

x =4

3 –

ExampleSolve.


The order to factoring.

First, factor the GCF out.

Then, identify which of the three factoring ways. (Two terms, three terms or four terms.)

Rule

Always try to factor again.


Key Points & Don’t Forget

1) Step one: take the GCF out.

2) Don’t forget the negative signs.

3) Make sure the expressions or equations are in standard form before factoring.

4) Always try to factor again.


pg. 462-463 #’s 10-37, 40-46 even

The Assignment


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for educational use only © 2010 10.8 factoring using the distributive property brian preston...

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