1. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1 Section 5.7 Polynomial Equations and Their Applications Copyright © 2013, 2009, 2006 Pearson Education, Inc. 1

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3. 3 Solving Polynomial Equations We have spent much time on learning how to factor polynomials. Now we will look at one important use of factoring. In this section, we will use factoring to solve equations of degree 2 and higher. Up to this point, we have only looked at solving equations of degree one.

4. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 4 Solving Polynomial Equations Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form where a, b, and c are real numbers, with. A quadratic equation in x is also called a second-degree polynomial equation in x. The Zero-Product Rule If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. If AB = 0, then A = 0 or B = 0.

5. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 5 Solving Polynomial Equations Solving a Quadratic Equation by Factoring 1) If necessary, rewrite the equation in the standard form, moving all terms to one side, thereby obtaining zero on the other side. 2) Factor completely. 3) Apply the zero-product principle, setting each factor containing a variable equal to zero. 4) Solve the equations in step 3. 5) Check the solutions in the original equation.

6. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 6 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and obtain zero on the other side. Subtract 45 from both sides and write the equation in standard form. Subtract 45 from both sides Simplify 2) Factor. Factor

7. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 7 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. ? ? ? ?

8. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 8 Solving Polynomial EquationsCONTINUED ?? The graph of is shown at right. true

9. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 9 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: 1) Move all terms to one side and obtain zero on the other side. Subtract 4x from both sides and write the equation in standard form. Note: Do NOT divide both sides by x. We could lose a potential solution!!! Subtract 4x from both sides Simplify 2) Factor. Factor

10. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 10 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. or CONTINUED 5) Check the solutions in the original equation. Check 0: Check 4: ?? true

11. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 11 Solving Polynomial EquationsCONTINUED The solutions are 0 and 4. The solution set is {0,4}. The graph of Is shown at right.

12. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 12 Solving Polynomial EquationsEXAMPLE SOLUTION Solve: Be careful! Although the left side of the original equation is factored, we cannot use the zero-product principle since the right side of the equation is NOT ZERO!! 1) Move all terms to one side and obtain zero on the other side. Subtract 14 from both sides and write the equation in standard form. Simplify Subtract 14 from both sides

13. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 13 Solving Polynomial EquationsCONTINUED 2) Factor. Before we can factor the equation, we must simplify it. FOIL Simplify Now we can factor the polynomial equation.

14. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 14 Solving Polynomial EquationsCONTINUED 3) & 4) Set each factor equal to zero and solve the resulting equations. or 5) Check the solutions in the original equation. ?? ??

15. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 15 Solving Polynomial EquationsCONTINUED true The graph of Is shown at right.

16. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 16 Objective #1: Example

17. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 17 Objective #1: Example

18. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 18 Objective #1: Example

19. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 19 Objective #1: Example

20. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 20 Objective #1: Example

21. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 21 Objective #1: Example

22. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 22 Objective #1: ExampleCONTINUED

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24. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 24 Solving Polynomial Equations A polynomial equation is the result of setting two polynomials equal to each other. The equation is in standard form if one side is 0 and the polynomial on the other side is in standard form, that is, in descending powers of the variable. The degree of a polynomial equation is the same as the highest degree of any term in the equation. Some polynomials equations of degree 3 or higher can be solved by moving all terms to one side, thereby obtaining 0 on the other side. Once the equation is in standard form, factor and then set each factor equal to 0.

25. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 25 Solving Polynomial EquationsEXAMPLE SOLUTION Solve by factoring: 1) Move all terms to one side and obtain zero on the other side. This is already done. + 2) Factor. Use factoring by grouping. Group terms that have a common factor. Common factor is Common factor is -1.

26. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 26 Solving Polynomial EquationsCONTINUED Factor out the common binomial, x – 2, from each term Factor completely by factoring as the difference of two squares

27. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 27 Solving Polynomial Equations 3) & 4) Set each factor equal to zero and solve the resulting equations. CONTINUED or The graph of is shown at right.

28. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 28 Objective #2: Example

29. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 29 Objective #2: Example

30. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 30 Objective #2: ExampleCONTINUED

31. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 31 Objective #2: ExampleCONTINUED

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33. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 33 Polynomial Equations in ApplicationEXAMPLE A gymnast dismounts the uneven parallel bars at a height of 8 feet with an initial upward velocity of 8 feet per second. The function describes the height of the gymnast’s feet above the ground, s (t), in feet, t seconds after dismounting. The graph of the function is shown below.

34. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 34 Polynomial Equations in ApplicationSOLUTION When will the gymnast be 8 feet above the ground? Identify the solution(s) as one or more points on the graph. We note that the graph of the equation passes through the line y = 8 twice. Once when x = 0 and once when x = 0.5. This can be verified by determining when y = s (t) = 8. That is, CONTINUED Original equation Replace s (t) with 8 Subtract 8 from both sides Factor

35. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 35 Polynomial Equations in Application Now we set each factor equal to zero. CONTINUED We have just verified the information we deduced from the graph. That is, the gymnast will indeed be 8 feet off the ground at t = 0 seconds and at t = 0.5 seconds. These solutions are identified by the dots on the graph on the next page. or

36. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 36 Polynomial Equations in ApplicationCONTINUED

37. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 37 The Pythagorean Theorem The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse. If the legs have lengths a and b, and the hypotenuse has length c, then A C B c a b Hypotenuse has length c. The two legs have lengths of a and b.

38. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 38 The Pythagorean TheoremEXAMPLE SOLUTION A tree is supported by a wire anchored in the ground 5 feet from its base. The wire is 1 foot longer than the height that it reaches on the tree. Find the length of the wire. Draw a diagram. Tree 5 feet

39. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 39 The Pythagorean Theorem Tree 5 feet CONTINUED x – 1 x

40. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 40 The Pythagorean Theorem We can now use the Pythagorean Theorem to solve for x, the length of the wire. CONTINUED This is the equation arising from the Pythagorean Theorem Square x – 1 and 5 Add 1 and 25 Subtract from both sides Add 2x to both sides Divide both sides by 2 Therefore, the solution is x = 13 feet.

41. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 41 Objective #3: Example

42. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 42 Objective #3: ExampleCONTINUED

43. Copyright © 2013, 2009, 2006 Pearson Education, Inc. 43 Objective #3: ExampleCONTINUED