10.4 Solving Polynomial Equations in Factored Form

Objective:• I will use the zero-product property

to find solutions to polynomial equations that are factored into the product of linear equations.

Vocabulary1. Zero-product property - for all real

numbers, if ab = 0, then a = 0 or b = 0

2. Repeated factor - polynomial with more than one of the same factor

Using the Zero-Product Property

Solve the equation (x - 4)(x + 3) = 0

a = x - 4, b = x + 3

Using the zero-product property:

x - 4 = 0 or x + 3 = 0

+4 +4 - 3 - 3

x = 4 x = -3

The solutions are -3 and 4.

Guided Practice

1. Solve: (x - 8)(x + 4) = 0

2. Solve: (x - 14)(x - 23) = 0

Solving a Repeated-Factor Equation

Solve (x - 4)2 = 0

This equation has a repeated factor because (x - 4)2 = (x - 4)(x - 4)

So x - 4 = 0

+4 +4

x = 4

The solution is 4.

Same factor

Guided Practice

1. Solve: (x + 5)2 = 0

2. Solve: (x - 9)3 = 0

Solving a Factored Cubic Equation

Solve (4x + 3)(2x - 1)(x - 5) = 0• a polynomial equation has as many solutions

as it does linear factors

4x + 3 = 0 2x - 1 = 0 x - 5 = 0

- 3 - 3 + 1 + 1 + 5 + 5

4x = -3 2x = 1 x = 5

4 4 2 2

x = x =

The solution is , , and 5.

34

34

1

2

1

2

Guided Practice

1. Solve: (2x + 4)(3x - 9)(x - 1) = 0

2. Solve: (5x - 30)(4x - 8)(x + 1)2 = 0

Factoring a Polynomial, then Solving

• Factor and Solve: x2 - 5x + 4 = 0

Factor: (x - 1)(x - 4) = 0

Solve: x - 1 = 0 x - 4 = 0

+1 +1 +4 +4

x = 1 x = 4

Solution: x = 1, 4

Guided Practice

1. Factor and Solve: x2 + 3x + 2 = 0

2. Factor and Solve: x2 - 2x - 63 = 0

Independent Practice

Solve.

1. (x - 3)(x + 4) = 0

2. (x + 8)2 = 0

3. (5x + 10)(2x - 16)(x - 4) = 0

4. x3 + 2x2 + 2x + 4 = 0

Factors, Solutions, and x-intercepts

For any quadratic polynomial ax2 + bx + c = 0:• If (x - p) is a factor of ax2 + bx + c = 0• Then x = p is a solution of ax2 + bx + c = 0• Then p is an x-intercept of the graph of

ax2 + bx + c = 0

Sketching Graphs of Quadratic Equations Using Factors and x-intercepts

Sketch a graph of y = (x + 2)(x - 4)

First find the x-intercepts

x + 2 = 0 x - 4 = 0

- 2 - 2 + 4 + 4

x = -2 x = 4

x-intercepts: (-2,0) and (4,0)

Sketching Graphs of Quadratic Equations Using Factors and x-intercepts Continued…

Second find the vertex• The x-value of the vertex is the average of

the two x-interceptsx = 4 + (-2) = 2 = 1

2 2Substitute x into the equation to find yy = (x + 2)(x - 4) = (1 + 2)(1 - 4) = (3)(-3)y = -9Vertex: (1,-9)

Sketching Graphs of Quadratic Equations Using Factors and x-intercepts Continued…

Sketch the graph

-20

-10

0

10

20

30

40

50

60

0 5 10 15 20

Series1

-7 55-6 40-5 27-4 16-3 7-2 0-1 -50 -81 -92 -83 -54 05 76 167 27

x y

Using a Quadratic Model

An arch is modeled by the equation

y = -0.25(x - 6)(x + 6), with x and y measured in feet. How wide is the arch at the base? How high is the arch?

Using a Quadratic ModelSolutionx-intercepts (where y = 0):x - 6 = 0 x + 6 = 0 + 6 + 6 - 6 - 6 x = 6 x = -

6

-6, 6

Vertex (maximum height):

x = 0 (average of x-values)

y = -0.25(0 - 6)(0 + 6) = -0.25(-6)(6) = 9

Maximum height = 9 ft

Using a Quadratic Model Continued…

Arch Width (horizontal distance where y = 0):

x2 - x1 = 6 - (-6) = 6 + 6 = 12

Arch Width = 12 ft

-4

-2

0

2

4

6

8

10

-10 -5 0 5 10

Width (ft)

Heig

ht

(ft)

Series1

-6 0-5 2.75-4 5-3 6.75-2 8-1 8.750 91 8.752 83 6.754 55 2.756 0

x y