For Educational Use Only © 2010 10.4 Solving Polynomial Equations in Factored Form Brian Preston Algebra 1 2009-2010

For Educational Use Only © 2010

10.4 Solving Polynomial

Equations in Factored Form

Brian PrestonAlgebra 1 2009-

2010


Real World Application

How high is the tallest arch in the world?


Lesson Objectives

1) Solve a polynomial equation in factored form.

2) Relate factors and x-intercepts.


Review

(3x + 4)(x + 5)5x 5x 443x 3x 1)

3x2 + 4x+ 15x + 203x2 + 19x + 20

First Outside Inside Last


(z – 4)

What does this mean?

Review

(z – 4)2(z – 4) There are two (z- 4)


– 3– 3+ 2+ 2

(x – 2)(x – 2)(x + 3)2)

Definition

Starting with 2 or more binomials and solving for the variable.

= 0(x + 3)

= 0= 0 x + 3x – 2

x

+ 2

= 2

( ) ( )

x

– 3

= – 3


aa b

Rule

Zero-Product Property = the product of two factors is zero only when at

least one of the factors is zero.

= 0b

= 0= 0 baa = 0

( ) ( )b = 0or


3) An arch is modeled by the equations y = – 0.00635(x – 315)(x + 315) with x & y measured in feet. How wide is the base of the arch? How high is the arch?

widehigh

y = – 0.00635(x – 315)(x + 315)



– 0.00635(x – 315)(x + 315)

–315–315+315+315

All you need to graph are the x-intercepts & the vertex.

(x – 315)(x + 315)– 0.00635

Example

= 0= 0 x + 315x - 315

x

+315

= 315

( ) ( )

x

–315

= –315

= 0–0.00635( )

y =3)


Definition


x-intercepts = 315 & – 315

vertex =(Average of the x-int, #)

– 0.00635(x – 315)(x + 315)= 0y =3)


00 630

– 0.00635(x – 315)(x + 315)3)

Definition

= 0y =

x-intercepts = 315 & – 315


315 + – 315

2= 0,( ? )

– 0.00635( – 315)( + 315)= 630y =

– 315315

0 0x x


x-intercepts

3) y = – 0.00635(x – 315)(x + 315)

-200

-600

-1000

-200-600

200

600

1000

200

600 1000-1000

(-315,0) (315,0)

(0,630)

Review

vertex

315 & – 315

(0,630)



How high is the tallest arch in the world? 630 feet


(x + 5)(x + 5)(x + 5)4)

Example

Solve.

= 02


– 5– 5– 5– 5

(x + 5)(x + 5)(x + 5)(x + 5)4)

Example

Solve.

= 0

= 0= 0 x + 5x + 5

x

– 5

= – 5

( ) ( )

x

– 5

= – 5


(4c – 8)(7c + 21)(4c – 8)(7c + 21)

7 7

– 21– 21

4 4

+ 8+ 8

5)

Example

Solve.= 0

= 0= 0 7c + 214c – 8

4c

+ 8

= 8

( ) ( )

7c

– 21

= – 21+ 0 + 04

c = 27

c = – 3


– 1– 4 – 1– 4

(x + 1)(x + 4)(x + 4)(x + 1)776)

Example

Solve.

= 0

= 0= 0 x + 1x + 4

x

– 4

= – 4

( ) ( )

x

– 1

= – 1

= 07( )


(b + 6)(b + 6)(b + 6)7)

Example

Solve.

= 028(b – 9)


– 6 – 6– 6 – 6+ 9+ 9

(b + 6)(b + 6)(b – 9)(b + 6)(b + 6)(b – 9)

Example

Solve.

= 0

= 0b – 9

b

+ 9

= 9

( )

7) 8

= 08( ) = 0b + 6

b

– 6

= – 6

( ) = 0b + 6

b

– 6

= – 6

( )

8


(x – 2)(x + 3)

Definition

These equations can be graphed.

= 0y =

The variable solutions are x-intercepts. They allow

you to graph more easily.


111

11

8) Sketch the graph of y = 2 – 2 – 3.

b2(1)

– (– 2)

a= = 2

2

– 211 xx (1)(1)

0 2 3– 1x

y

Review

Before, you had to

do this to graph.


– 4

8) Sketch the graph of y = 1(1)2 – 2(1) – 3.

= y 1 – 2= y =

– 31 – 5

– 4

0 1 2 3– 1

– 4

x

y

(1)

Review


– 3– 3222


= y 1 – 4= y =

– 34 – 7

x(2) x(2)1

– 4 – 3

0 1 3– 1x

y

(4)

Review


03

02 33


= y 1 – 6= y =

– 39 – 9

x(3) x(3)1

– 4 – 3 0

0 1– 1x

y

(9)

Review


000

– 3– 33x

y

2


= y 1 – 0= y =

– 30 – 3

x(0) x(0)1

– 3 – 4 – 3 0

1– 1

(0)

Review


– 1– 1– 1

00

03x

y

2


= y 1 + 2= y =

– 31 – 1

x(– 1) x(– 1)1

0 – 3 – 4 – 3 0

1

(1)

Review


-1

-3

-5

-1-3

1

3

5

1 3 5-5

8) Sketch the graph of y = x2 – 2x – 3.

0 – 3 – 4 – 3 0

x

y

– 1 0 1 2 3– 1

0

(-1,0)

(0,-3)(2,-3)

(3,0)

(1,-4)0

– 3

1

– 4

2

– 3

3

0

Review


– 5– 5 + 3+ 3

(x – 3)(x – 3)(x + 5)(x + 5)9)

Definition

= 0

= 0= 0 x – 3 x + 5

x

– 5

= – 5

( ) ( )

x

+ 3

= 3



(x + 5)(x – 3)9)

Definition


= 0y =

x-intercepts = – 5 & 3



– 16

(x + 5)(x – 3)9)

Definition

= 0y =x-intercepts = – 5 & 3


– 5 + 3

2= – 1,( ? )

( + 5)( – 3) = – 16y =

– 1

3– 5

– 1

– 1 – 1 – 16x x


-4

-12

-20

-4-12

4

12

20

412 20-20

9) y = (x + 5)(x – 3)

(-5,0) (3,0)

(– 1,-16)

Review

x-intercepts

vertex

– 5 & 3

(– 1,– 16)

For Educational Use Only © 2010– 1 – 1

1

– 2– 2 + 6+ 6

(x – 6)(– x + 2)(x – 6)(– x + 2)10)

Definition

= 0= 0= 0 x – 6 – x + 2( ) ( )

x

+ 6

= 6


– x

– 2

= – 2+ 0– 1 x = 2


(– x + 2)(x – 6)10)

Definition


= 0y =

x-intercepts = 2 & 6



4

= 44

44

(– x + 2)(x – 6)10)

Definition

= 0y =x-intercepts = 2 & 6


2 + 6

2= 4 ,( ? )

(– + 2)( – 6)y =

62

4 4x x


-2

-6

-10

-2-6

2

6

10

2 6 10-10

10) y = (– x + 2)(x – 6)

(2,0)

(6,0)

(4,4)

Review

x-intercepts

vertex

2 & 6

(4, 4)


1) Don’t forget the negative signs.

2) You can sketch quadratic equations.

Key Points & Don’t Forget


pg. 407-408 #’s 11-18,23-26, 31, 40, 41-49

odd

The Assignment


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For Educational Use Only © 2010 10.4 Solving Polynomial Equations in Factored Form Brian Preston Algebra 1 2009-2010