Do Now

• Pass out calculators. • Work on practice EOC Week # 8.

Quick Check:

1. (x 2 – 3x + 5) + (-2x 2 + 11x + 1) 2. (8y 3 – 7y 2 + y) – (9y 2 – 5y + 7) 3. -3x 2 (x 3 – 3x 2 ) 4. (2r + 11)(r – 6) 5. (m + 3)(-2m 2 + 5m – 1) 6. (5w + 9z) 2

Answers:

1. –x 2 + 8x +6 2. 8y 3 – 16y 2 +6y – 7 3. -3x 5 + 9x 4 4. 2r 2 – r – 66 5. -2m 3 – m 2 +14m – 3 6. 25w 2 90wz +81z 2

Objective:

• To use the zero product property and factor using the greatest common factor.

Zero – Product Property:

• The

zero-product property

is used to solve an equation when one side is zero and the other side is two polynomials being multiplied. • The solutions of an equations like are called

roots

.

EXAMPLE 1 Use the zero-product property Solve

(

x

– 4)(

x

+ 2) = 0.

(

x

– 4 )(

x

+ 2 ) = 0

x

– 4

x

= 4

x

+ 2

x

= 0 = – 2

ANSWER Write original equation.

Zero-product property Solve for x.

The solutions of the equation are

4

and

–2.

EXAMPLE 1 Use the zero-product property

CHECK

Substitute each solution into the original equation to check.

(4  ?

4)(4 + 2) = 0 0  ?

6 = 0 0 = 0 (  2  4)(  ?

2 + 2) = 0  6  ?

0 = 0 0 = 0

GUIDED PRACTICE for Example 1 1. Solve the equation

(

x

– 5)(

x

– 1) = 0.

ANSWER The solutions of the equation are

5

and

1.

EXAMPLE 2 Find the greatest common monomial factor Factor out the greatest common monomial factor.

a.

12

x +

42

y

SOLUTION a.

The GCF of

12

and

42

is

6 . The variables x and y have

no common factor. So, the greatest common monomial factor of the terms is

6.

ANSWER

12

x +

42

y

= 6(2

x +

7

y

)

EXAMPLE 2 Find the greatest common monomial factor Factor out the greatest common monomial factor.

b.

4

x

4

+

24

x

3

SOLUTION b.

The GCF of

4

and

24

is

4 . The GCF of x 4 and x 3 is x 3

. So, the greatest common monomial factor of the terms is

4

x

3 .

ANSWER

4

x

4

+

24

x

3 = 4

x

3 (

x +

6)

GUIDED PRACTICE for Example 2 2. Factor out the greatest common monomial factor from

14

m +

35

n

.

ANSWER

14

m +

35

n

= 7(2

m +

5

n

)

EXAMPLE 3 Solve an equation by factoring Solve

2

x

2

+

8

x

= 0.

2

x

2

+

8

x

= 0 2

x

(

x +

4 )

=

0 2

x =

0

or

x +

4

=

0

x =

0

or

x = –

4

Write original equation.

Factor left side.

Zero-product property Solve for

x.

ANSWER The solutions of the equation are

0

and

–

4.

EXAMPLE

4

Solve an equation by factoring Solve

6

n

2 = 15

n

.

6

n

2

–

15

n =

0

Subtract

15

n

from each side.

3

n

( 2

n –

5 )

=

0

Factor left side.

3

n =

0

or

2

n –

5

=

0

Zero-product property

n =

0

or

5

n =

2

Solve for

n.

ANSWER The solutions of the equation are

0

and

5 .

2

GUIDED PRACTICE for Examples 3 and 4 Solve the equation.

3. a 2

+

5

a

= 0

ANSWER

0

and

–

5

5.

4

x

2 0

and

1 2 = 2

x

.

ANSWER 4.

3

s

2

–

9

s

= 0

ANSWER

0

and

3

Vertical Motion:

• A

projectile

is an object that is propelled into the air but has no power to keep itself in the air. A thrown ball is a projective, but an airplane is not. The height of a projectile can be described by the

vertical motion model

.

• The height h (in feet) of a projectile can be modeled by:

h = -16t 2 + vt + x

t = time (in seconds) the object has been in the air v = initial velocity (in feet per second) s = the initial height (in feet)

EXAMPLE 5 Solve a multi-step problem ARMADILLO A startled armadillo jumps straight into the air with an initial vertical velocity of 14 feet per second. After how many seconds does it land on the ground ?

EXAMPLE 5 Solve a multi-step problem SOLUTION

STEP 1

Write a model for the armadillo

’

s height above the ground.

h = –

16

t

2

+ v t + s h = –

16

t

2

+

14

t +

0

Vertical motion model Substitute 14 for

v

and 0 for s.

h = –

16

t

2 + 14

t

Simplify.

EXAMPLE 5 Solve a multi-step problem

STEP 2

Substitute 0 for

h.

When the armadillo lands, its height above the ground is 0 feet. Solve for

t.

0

= –

16

t

2

+

14

t

0

=

2

t

(

–

8

t +

7 ) 2

t =

0

or

t =

0

ANSWER

or

–

8

t +

7

=

0

Substitute 0 for h.

Factor right side.

Zero-product property

t =

0.875

Solve for t.

The armadillo lands on the ground 0.875

second after the armadillo jumps.

GUIDED PRACTICE for Example 5 6. WHAT IF ? In Example 5, suppose the initial vertical velocity is 12 feet per second.After how many seconds does armadillo land on the ground ?

ANSWER The armadillo lands on the ground 0.75

second after the armadillo jumps.

Exit Ticket

1. Solve (x + 3)(x – 5) = 0 Why does this type of problem have two solutions? 2. Factor out the greatest common monomial factor. a. 8x +12 y b. 12y 2 + 21y