Download 6.5 Solving Polynomial Equations by Factoring

6.5 Solving Polynomial Equations by Factoring
What You Will Learn
 Use the Zero-Factor Property to solve
equations.
 Solve quadratic equations by factoring.
 Solve higher-degree polynomial equations by
factoring.
 Solve application problems by factoring.
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The Zero-Factor Property
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The Zero-Factor Property
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Example 1 – Using the Zero-Factor Property
The Zero-Factor Property is the primary property for solving
equations in algebra. For instance, to solve the equation
you can use the Zero-Factor Property to conclude that either
(x – 1) or (x + 2) equals 0. Setting the first factor equal to 0
implies that x = 1 is a solution.
Similarly, setting the second factor equal to 0 implies that
x = –2 is a solution.
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Example 1 – Using the Zero-Factor Property
So, the equation (x – 1)(x + 2) = 0 has exactly two solutions:
x = 1 and x = –2.
Check these solutions by substituting them into the original
equation.
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Solving Quadratic Equations by
Factoring
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Solving Quadratic Equations by Factoring
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Example 2 – Solving Quadratic Equation by Factoring
Solve x2 – x – 6 = 0.
Solution
First, make sure that the right side of the equation is zero.
Next, factor the left side of the equation.
Finally, apply the Zero-Factor Property to find the solutions.
x2 – x – 6 = 0
(x + 2) (x – 3) = 0
x+2=0
x = –2
x–3=0
x=3
The solutions are x = –2 and x = 3.
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Solving Quadratic Equations by Factoring
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Example 4 – A Quadratic Equation with a Repeated Solution
Solve x2 – 2x + 16 = 6x.
Solution
x2 – 2x + 16 = 6x
x2 – 8x + 16 = 0
(x – 4)2 = 0
x – 4 = 0 or x – 4 = 0
x=4
Note that even though the left side of this equation has two
factors, the factors are the same.
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Example 4 – A Quadratic Equation with a Repeated Solution
cont’d
So, the only solution of the equation is x = 4.
This solution is called a repeated solution.
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Solving Higher-Degree Equations by
Factoring
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Example 5 – Solving a Polynomial Equation with Three Factors
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Example 5 – Solving a Polynomial Equation with Three Factors
cont’d
The solutions are x = 0, x = 6, and x = –1. Check these
three solutions.
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Applications
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Example 7 – Geometry: Dimensions of a Room
A rectangular room has an area of 192 square feet. The
length of the room is 4 feet more than its width, as shown
below. Find the dimensions of the room.
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Example 7 – Geometry: Dimensions of a Room
Solution
Labels:
Length = x + 4
Width = x
Area = 192
(feet)
(feet)
(square feet)
Equation:
(x + 4)x = 192
x2 + 4x – 192 = 0
(x + 16)(x – 12) = 0
x = –16 or x = 12
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Example 7 – Geometry: Dimensions of a Room
cont’d
Because the negative solution does not make sense,
choose the positive solution x = 12.
When the width of the room is 12 feet, the length of the
room is
Length = x + 4 = 12 + 4 = 16 feet.
So, the dimensions of the room are 12 feet by 16 feet.
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Homework:
Page 304 # 7, 11, 14
Page 305 # 17 – 23 odd
Page 306 # 27 – 33 odd
Page 307 # 41 – 47 odd
Page 308 # 55
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