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2. CLO and Warm Up I will be able to solve a quadratic equation through factoring and applying the zero product property by discussing the process with my partner. Warm Up: Factor. You may use X-Box or Trial and Error. 1. x 2 + 5x + 62. 2x 2 + 11x + 15

3. 5.8 Solving Equations by Factoring and Problem Solving

4. Polynomial Equations Equations that set 2 polynomials equal to each other. Standard form has a 0 on one side of the equation. Quadratic Equations Polynomial equations of degree 2.

5. Zero-Factor Property If a and b are real numbers and if ab = 0, then a = 0 or b = 0. This property is true for three or more factors also.

6. Solve: (x – 4)(x + 2) = 0 x – 4 = 0 or x + 2 = 0 x = 4 or x = –2 You can check the solutions on your own. Example

7. Solve: x 2 – 5x = 24 First write the quadratic equation in standard form. x 2 – 5x – 24 = 0 Now factor. x 2 – 5x – 24 = (x – 8)(x + 3) = 0 Set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = –3 Example Continued

8. Check both possible answers in the original equation. 8 2 – 5(8) = 64 – 40 = 24 True (–3) 2 – 5(–3) = 9 – (–15) = 24 True So the solutions are 8 or –3. Example (cont) ? ?

9. Solving Polynomial Equations by Factoring 1)Write the equation in standard form so that one side of the equation is 0. 2)Factor the polynomial completely. 3)Set each factor containing a variable equal to 0. 4)Solve the resulting equations. 5)Check each solution in the original equation. Solving Polynomial Equations

10. Solve: 4x(8x + 9) = 5 4x(8x + 9) = 5 32x 2 + 36x = 5 32x 2 + 36x – 5 = 0 (8x – 1)(4x + 5) = 0 8x – 1 = 0 or 4x + 5 = 0 8x = 1 4x = – 5 Example Continued and The solutions are

11. Check both possible answers in the original equation. True ? ? andThe solutions are Example (cont)

12. Solve: Replace x with each solution in the original equation. The solutions all check. Example

13. Strategy for Problem Solving General Strategy for Problem Solving 1)Understand the problem Read and reread the problem Choose a variable to represent the unknown Construct a drawing, whenever possible Propose a solution and check 2)Translate the problem into an equation 3)Solve the equation 4)Interpret the result Check proposed solution in problem State your conclusion

14. Example The product of two consecutive positive integers is 132. Find the two integers. 1. UNDERSTAND Read and reread the problem. If we let x = one of the unknown positive integers, then x + 1 = the next consecutive positive integer. Continued

15. 2. TRANSLATE Continued two consecutive positive integers x(x + 1) is = 132 The product of Example (cont)

16. 3. SOLVE Continued x(x + 1) = 132 x 2 + x = 132 x 2 + x – 132 = 0 Write in standard form. (x + 12)(x – 11) = 0 x + 12 = 0 or x – 11 = 0 Set each factor equal to 0. x = – 12 or x = 11 Solve. Factor. Apply the distributive property. Example (cont)

17. 4.INTERPRET Check: Remember that x is suppose to represent a positive integer. So, although x = -12 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 11, then x + 1 = 12. The product of the two numbers is 11 · 12 = 132, our desired result. State: The two positive integers are 11 and 12. Example (cont)

18. Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. (leg a) 2 + (leg b) 2 = (hypotenuse) 2 The Pythagorean Theorem

19. Find the length of the shorter leg of a right triangle if the longer leg is 10 miles more than the shorter leg and the hypotenuse is 10 miles less than twice the shorter leg. Example Continued 1.) Understand Read and reread the problem. If we let x = the length of the shorter leg, then x + 10 = the length of the longer leg and 2x – 10 = the length of the hypotenuse.

20. Example (cont) 2.) Translate Continued By the Pythagorean Theorem, (leg a) 2 + (leg b) 2 = (hypotenuse) 2 x 2 + (x + 10) 2 = (2x – 10) 2 3.) Solve x 2 + (x + 10) 2 = (2x – 10) 2 x 2 + x 2 + 20x + 100 = 4x 2 – 40x + 100Multiply the binomials. 2x 2 + 20x + 100 = 4x 2 – 40x + 100Simplify the left side. x = 0 or x = 30Set each factor = 0 and solve, 0 = 2x(x – 30)Factor the right side. 0 = 2x 2 – 60xSubtract 2x 2 + 20x + 100 from both sides.

21. 4.) Interpret Check: Remember that x is suppose to represent the length of the shorter side. So, although x = 0 satisfies our equation, it cannot be a solution for the problem we were presented. If we let x = 30, then x + 10 = 40 and 2x – 10 = 50. Since 302 + 402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem checks out. State: The length of the shorter leg is 30 miles. (Remember that is all we were asked for in this problem.) Example (cont)