1. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 2 Factoring and Applications Chapter 6

3. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 3 6.7 Solving Quadratic Equations by Factoring

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 4 Objectives 1.Solve quadratic equations by factoring. 2.Solve other equations by factoring. 6.7 Solving Quadratic Equations by Factoring

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 5 Quadratic Equations A quadratic equation is an equation that can be written in the form ax 2 + bx + c = 0, where a, b, and c, are real numbers, with a ≠ 0. The given form is called standard form. 6.7 Solving Quadratic Equations by Factoring

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 6 6.7 Solving Quadratic Equations by Factoring Zero-Factor Property If a and b are real numbers and ab = 0, then a = 0 or b = 0. In words, if the product of two numbers is 0, then at least one of the numbers must be 0. One number must be 0, but both may be 0. Solving Quadratic Equations by Factoring

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 7 (2y – 3)(y + 1) = 0 Example 1 Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring By the zero-factor property, the only way that the product of these two factors can be 0 is if at least one of the factors equals 0. 2y – 3 = 0 or y + 1 = 0 2y = 3y = –1 The product is equal to 0. Solve each equation. The solution set is

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 8 x 2 – 3x = 4 Example 2 Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring Subtract 4. (x – 4)(x + 1) = 0 Zero-factor propertyx – 4 = 0 or x + 1 = 0 x = 4x = –1 x 2 – 3x – 4 = 0 Factor the trinomial. Solve each equation. First, write the equation in standard form. The solution set is

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 9 6.7 Solving Quadratic Equations by Factoring Solving Quadratic Equations by Factoring Solving a Quadratic Equation by Factoring Step 1Write the equation in standard form, that is, with all terms on one side of the equals sign in descending powers of the variable and 0 on the other side. Step 2Factor completely. Step 3Use the zero-factor property to set each factor with a variable equal to 0. Step 4Solve the resulting equations. Step 5 Check each solution in the original equation.

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 10 Divide each side by 2. 2x 2 + 30 = –16x Example 3 Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring Standard form 2(x 2 + 8x + 15) = 0 Factor. (x + 3)(x + 5) = 0 Zero-factor propertyx + 3 = 0 or x + 5 = 0 x = –3x = –5 2x 2 + 16x + 30 = 0 Factor out the GCF, 2. x 2 + 8x + 15 = 0 Solve each equation.

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 11 2(–3) 2 + 30 = –16(–3) Example 3 (continued) Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring Check that the solution set is {–3, –5} by substituting in the original equation. 18 + 30 = 48 ? 48 = 48 ? 2(–5) 2 + 30 = –16(–5) 50 + 30 = 80 ? 80 = 80 ? CAUTION A common error is to include the common factor 2 as a solution. Only factors containing variables lead to solutions.

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 12 y(2y + 5) = 42 Example 4(a) Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring 2y 2 + 5y – 42 = 0 (2y – 7)(y + 6) = 0 2y – 7 = 0 or y + 6 = 0 2y = 7y = –6 2y 2 + 5y = 42 We need to write this equation in standard form. Multiply. Subtract 42. Factor. Zero-factor property The solution set is

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 13 x 2 = 4x Example 4(b) Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring x(x – 4) = 0 x = 0 or x – 4 = 0 x = 4 x 2 – 4x = 0Standard form Factor. Zero-factor property The solution set is {0, 4}. x = 0

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 14 CAUTION In Example 4(a), the zero-factor property could not be used to solve the equation y(2y + 5) = 42 in its given form because of the 42 on the right. The zero-factor property applies only to a product that equals 0. In Example 4(b), it is tempting to begin by dividing each side of the equation x 2 = 4x by x to get x = 4. Note that we do not get the other solution, 0, if we divide by a variable. (We may divide each side of an equation by a nonzero real number, however. For instance, in Example 3, we divided each side by 2.) Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 15 x 2 + 64 = –16x Example 5 Solve the equation. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring Standard form Factor. (x + 8) 2 = 0 Zero-factor propertyx + 8 = 0 or x + 8 = 0 x = – 8 x 2 + 16x + 64 = 0 Solve each equation. The solution set is {– 8}. This is called a double solution.

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 16 Note Not all quadratic equations can be solved by factoring. Solving Quadratic Equations by Factoring 6.7 Solving Quadratic Equations by Factoring

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.7 – Slide 17 Example 6 Solve the equation. 3a 3 – 48a = 0 Solving Other Equations by Factoring 6.7 Solving Quadratic Equations by Factoring 3a(a 2 – 16) = 0 3a(a + 4)(a – 4) = 0 a = –4 Zero-product property Factor out 3a. Factor a 2 – 16. 3a = 0 or a + 4 = 0 or a – 4 = 0 a = 0 a = 4 The solution set is {0, – 4, 4}.