1. 1 Copyright © Cengage Learning. All rights reserved.

2. 2 6.5 Solving Polynomial Equations by Factoring

3. 3 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.

4. 4 The Zero-Factor Property

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6. 6 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.

7. 7 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.

8. 8 Solving Quadratic Equations by Factoring

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10. 10 Example 2 – Solving Quadratic Equation by Factoring Solve x 2 – 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. x 2 – 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.

11. 11 Solving Quadratic Equations by Factoring

12. 12 Example 4 – A Quadratic Equation with a Repeated Solution Solve x 2 – 2x + 16 = 6x. Solution x 2 – 2x + 16 = 6x x 2 – 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.

13. 13 Example 4 – A Quadratic Equation with a Repeated Solution So, the only solution of the equation is x = 4. This solution is called a repeated solution. cont’d

14. 14 Solving Higher-Degree Equations by Factoring

15. 15 Example 5 – Solving a Polynomial Equation with Three Factors

16. 16 Example 5 – Solving a Polynomial Equation with Three Factors The solutions are x = 0, x = 6, and x = –1. Check these three solutions. cont’d

17. 17 Applications

18. 18 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.

19. 19 Example 7 – Geometry: Dimensions of a Room Solution Labels: Length = x + 4 (feet) Width = x (feet) Area = 192 (square feet) Equation: (x + 4)x = 192 x 2 + 4x – 192 = 0 (x + 16)(x – 12) = 0 x = –16 or x = 12

20. 20 Example 7 – Geometry: Dimensions of a Room 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. cont’d