1. Slide 6- 1 Copyright © 2012 Pearson Education, Inc.

2. 6.1 Introduction to Polynomial Factorizations and Equations ■ Graphical Solutions ■ The Principle of Zero Products ■ Terms with Common Factors ■ Factoring by Grouping ■ Factoring and Equations

3. Slide 6- 3 Copyright © 2012 Pearson Education, Inc. Graphical Solutions Whenever two polynomials are set equal to each other, the result is a polynomial equation. In this section we learn how to solve polynomial equations both graphically and algebraically.

4. Slide 6- 4 Copyright © 2012 Pearson Education, Inc. Example Solve: x 2 = 4x. Solution Intersect Method: We can find the real-number solutions of a polynomial equation by finding the points of intersection of two graphs.

5. Slide 6- 5 Copyright © 2012 Pearson Education, Inc. continued Zero Method: Rewrite the equation so that one side is 0 and then find the x-intercepts of one graph or the zeros of the function. Rewrite the equation so that one side is 0: x 2 = 4x x 2 – 4x = 0

6. Slide 6- 6 Copyright © 2012 Pearson Education, Inc. Zeros and Roots The x-values for which a function f(x) is 0 are called the zeros of the function. The x-values for which an equation such as f(x) = 0 is true are called the roots of the equation.

7. Slide 6- 7 Copyright © 2012 Pearson Education, Inc. Example Find the zeros of the function given by f(x) = x 3 – 2x 2 – 5x + 6. Solution Graph the equation, choosing a window that shows the x-intercepts of the graph. This may require several attempts. To find the zeros use the ZERO option from the CALC menu.

8. Slide 6- 8 Copyright © 2012 Pearson Education, Inc. continued f(x) = x 3 – 2x 2 – 5x + 6 To find the zeros use the ZERO option from the CALC menu. Use the same procedure for the other two zeros. The zeros are  2, 1 and 3.

9. Slide 6- 9 Copyright © 2012 Pearson Education, Inc. The Principle of Zero Products For any real numbers a and b: If ab = 0, then a = 0 or b = 0. If a = 0 or b = 0, then ab = 0.

10. Slide 6- 10 Copyright © 2012 Pearson Education, Inc. Example Solve: (x – 4)(x + 3) = 0. Solution According to the principle of zero products, at least one factor must be 0. x – 4 = 0 or x + 3 = 0 x = 4 or x =  3 For 4:For  3: (x – 4)(x + 3) = 0 (4 – 4)(4 + 3) = 0 (  3 – 4)(  3 + 3) = 0 0(7) = 0 0(  7) = 0 0 = 0 TRUE 0 = 0 TRUE (  3, 0) (4, 0)

11. Slide 6- 11 Copyright © 2012 Pearson Education, Inc. Factoring To factor a polynomial is to find an equivalent expression that is a product. An equivalent expression of this type is called a factorization of the polynomial.

12. Slide 6- 12 Copyright © 2012 Pearson Education, Inc. Terms with Common Factors When factoring a polynomial, we look for factors common to every term and then use the distributive law. MultiplyFactor 4x(x 2 + 3x  4) 4x 3 + 12x 2  16x = 4x  x 2 + 4x  3x  4x  4= 4x  x 2 + 4x  3x  4x  4 = 4x 3 + 12x 2  16x= 4x(x 2 + 3x  4)

13. Slide 6- 13 Copyright © 2012 Pearson Education, Inc. Example Factor: 28x 6 + 32x 3. Solution The prime factorization of 28x 6 is 2  2  7  x  x  x  x  x  x The prime factorization of 32x 3 is 2  2  2  2  2  x  x  x The largest common factor is 2  2  x  x  x or 4x 3. 28x 6 + 32x 3 = 4x 3  7x 3 + 4x 3  8 = 4x 3 (7x 3 + 8)

14. Slide 6- 14 Copyright © 2012 Pearson Education, Inc. Example Factor: 12x 5  21x 4 + 24x 3 Solution The prime factorization of 12x 5 is 2  2  3  x  x  x  x  x The prime factorization of 21x 4 is 3  7  x  x  x  x The prime factorization of 24x 3 is 2  2  2  3  x  x  x The largest common factor is 3  x  x  x or 3x 3. 12x 5  21x 4 + 24x 3 = 3x 3  4x 2  3x 3  7x + 3x 3  8 = 3x 3 (4x 2  7x + 8)

15. Slide 5- 15 Copyright © 2012 Pearson Education, Inc. Tips for Factoring 1. Factor out the largest common factor, if one exists. 2. The common factor multiplies a polynomial with the same number of terms as the original polynomial. 3. Factoring can always be checked by multiplying. Multiplication should yield the original polynomial.

16. Slide 6- 16 Copyright © 2012 Pearson Education, Inc. Factoring by Grouping Sometimes algebraic expressions contain a common factor with two or more terms. Example Factor x 2 (x + 2) + 3(x + 2). Solution The binomial (x + 2) is a factor of both x 2 (x + 2) and 3(x + 2). Thus, x + 2 is a common factor. x 2 (x + 2) + 3(x + 2) = (x + 2)x 2 + (x + 2)3 = (x + 2)(x 2 + 3)

17. Slide 6- 17 Copyright © 2012 Pearson Education, Inc. Example Write an equivalent expression by factoring. a) 3x 3 + 9x 2 + x + 3 b) 9x 4 + 6x  27x 3  18 Solution a) 3x 3 + 9x 2 + x + 3 = (3x 3 + 9x 2 ) + (x + 3) = 3x 2 (x + 3) + 1(x + 3) = (x + 3)(3x 2 + 1) Don’t forget to include the 1.

18. Slide 6- 18 Copyright © 2012 Pearson Education, Inc. Example continued b) 9x 4 + 6x  27x 3  18 = (9x 4 + 6x) + (  27x 3  18) = 3x(3x 3 + 2) + (  9)(3x 3 + 2) = (3x 3 + 2)(3x  9) = (3x 3 + 2)3(x  3) = 3(3x 3 + 2)(x  3)

19. Slide 5- 19 Copyright © 2012 Pearson Education, Inc. Factoring out ‒ 1 b – a = ‒ 1(a – b) = ‒ (a – b)

20. Slide 6- 20 Copyright © 2012 Pearson Education, Inc. Example Factor: ax – bx + by – ay. Solution ax – bx + by – ay = (ax – bx) + (by – ay) = x(a – b) + y(b – a) = x(a – b) + y( ‒ 1)(a – b) = x(a – b) – y(a – b) = (a – b)(x – y)

21. Slide 6- 21 Copyright © 2012 Pearson Education, Inc. Factoring and Equations Example Solve: 7x 2 = 35x. Solution Use the principle of zero products if there is a 0 on one side of the equation and the other side is in factored form. 7x 2 = 35x 7x 2 – 35x = 0 Subtracting 35x. One side is now 0. 7x(x – 5) = 0 Factoring 7x = 0 or x – 5 = 0 Use the principle of zero products x = 0 or x = 5 We check by substitution or graphically.

22. Slide 6- 22 Copyright © 2012 Pearson Education, Inc. To Use the Principle of Zero Products 1. Write an equivalent equation with 0 on one side, using the addition principle. 2. Factor the nonzero side of the equation. 3.Set each factor that is not a constant equal to 0. 4.Solve the resulting equations.