Do Now

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

Quick Check:

1. (x2 – 3x + 5) + (-2x2 + 11x + 1)

2. (8y3 – 7y2 + y) – (9y2 – 5y + 7)

3. -3x2(x3 – 3x2)

4. (2r + 11)(r – 6)

5. (m + 3)(-2m2 + 5m – 1)

6. (5w + 9z)2

Answers:

1. –x2 + 8x +6

2. 8y3 – 16y2 +6y – 7

3. -3x5 + 9x4

4. 2r2 – r – 66

5. -2m3 – m2 +14m – 3

6. 25w2 90wz +81z2

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.

Use the zero-product property

EXAMPLE 1

Solve (x – 4)(x + 2) = 0.

(x – 4)(x + 2) = 0 Write original equation.

x – 4 = 0 x = 4

Zero-product property

Solve for x.

ANSWER

The solutions of the equation are 4 and –2.

oror x + 2 = 0

x = – 2

Use the zero-product propertyEXAMPLE 1

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.

SOLUTION

EXAMPLE 2 Find the greatest common monomial factor

Factor out the greatest common monomial factor.

a. 12x + 42y

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

12x + 42y = 6(2x + 7y)

EXAMPLE 2 Find the greatest common monomial factor

b. The GCF of 4 and 24 is 4. The GCF of x4 and x3 is x3. So, the greatest common monomial factor of the terms is 4x3.

ANSWER

4x4 + 24x3 = 4x3(x + 6)

SOLUTION

Factor out the greatest common monomial factor.

b. 4x4 + 24x3

GUIDED PRACTICE for Example 2

2. Factor out the greatest common monomial factorfrom 14m + 35n.

ANSWER

14m + 35n = 7(2m + 5n)

EXAMPLE 3 Solve an equation by factoring

Solve 2x2 + 8x = 0.

2x2 + 8x = 0

2x(x + 4) = 0

2x = 0

x = 0

or x + 4 = 0

or x = – 4

ANSWER

The solutions of the equation are 0 and – 4.

Solve for x.

Zero-product property

Factor left side.

Write original equation.

EXAMPLE 4 Solve an equation by factoring

Solve 6n2 = 15n.

6n2 – 15n = 0

3n(2n – 5) = 0

3n = 0

n = 0

2n – 5 = 0

n =52

or

or Solve for n.

Zero-product property

Factor left side.

Subtract 15n from each side.

ANSWER

The solutions of the equation are 0 and52 .

GUIDED PRACTICE for Examples 3 and 4

Solve the equation.

ANSWER

0 and – 5

3. a2 + 5a = 0

4. 3s2 – 9s = 0

ANSWER

0 and 3

5. 4x2 = 2x.

ANSWER

0 and 12

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 = -16t2 + 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)

ARMADILLO

EXAMPLE 5 Solve a multi-step problem

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?

SOLUTION

EXAMPLE 5 Solve a multi-step problem

STEP 1

Write a model for the armadillo’s height above the ground.

h = –16t2 + vt + s

h = –16t2 + 14t + 0

h = –16t2 + 14t

Vertical motion model

Substitute 14 for v and 0 for s.

Simplify.

EXAMPLE 5 Solve a multi-step problem

STEP 2Substitute 0 for h. When the armadillo lands, its height above the ground is 0 feet. Solve for t.

0 = –16t2 + 14t

0 = 2t(–8t + 7)

2t = 0

t = 0

–8t + 7 = 0

t = 0.875

or

or Solve for t.

Zero-product property

Factor right side.

Substitute 0 for h.

ANSWER

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. 12y2 + 21y