1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 4–2) CCSS Then/Now New Vocabulary Key Concept: FOIL Method for Multiplying Binomials Example 1:Translate Sentences into Equations Concept Summary: Zero Product Property Example 2:Factor GCF Example 3:Perfect Squares and Differences of Squares Example 4:Factor Trinomials Example 5:Real-World Example: Solve Equations by Factoring

3. Over Lesson 4–2 5-Minute Check 1 A.4, –4 B.3, –2 C.2, 0 D.2, –2 Use the related graph of y = x 2 – 4 to determine its solutions.

4. Over Lesson 4–2 5-Minute Check 2 A.–3, 1 B.–3, 3 C.–1, 3 D.3, 1 Use the related graph of y = –x 2 – 2x + 3 to determine its solutions.

5. Over Lesson 4–2 5-Minute Check 3 A.0 B.0, between 2 and 3 C.between 1 and 2 D.2, –2 Solve –2x 2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located.

6. Over Lesson 4–2 5-Minute Check 4 A.10, –4 B.5, –1 C.–2, 7 D.–5, 2 Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14.

7. Over Lesson 4–2 5-Minute Check 5 A.zero B.x-intercept C.root D.vertex Which term is not another name for a solution to a quadratic equation?

8. CCSS Content Standards A.SSE.2 Use the structure of an expression to identify ways to rewrite it. F.IF.8.a Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. Mathematical Practices 2 Reason Abstractly and quantitatively.

9. Then/Now You found the greatest common factors of sets of numbers. Write quadratic equations in intercept form. Solve quadratic equations by factoring.

10. Vocabulary factored form FOIL method

11. Concept

12. Example 1 Translate Sentences into Equations (x – p)(x – q)=0Write the pattern. Simplify. Replace p with and q with –5. Use FOIL.

13. Example 1 Multiply each side by 2 so b and c are integers. Answer: Translate Sentences into Equations

14. Example 1 A.ans B.ans C.ans D.ans

15. Concept

16. Example 2 Factor GCF A. Solve 9y 2 + 3y = 0. 9y 2 + 3y= 0Original equation 3y(3y) + 3y(1) = 0Factor the GCF. 3y(3y + 1)= 0Distributive Property 3y = 0 3y + 1 = 0Zero Product Property y = 0 Solve each equation. Answer:

17. Example 2 Factor GCF B. Solve 5a 2 – 20a = 0. 5a 2 – 20a= 0Original equation 5a(a) – 5a(4)= 0Factor the GCF. 5a(5a – 4)= 0Distributive Property 5a = 0 a – 4 = 0Zero Product Property a = 0 a = 4Solve each equation. Answer: 0, 4

18. Example 2 A.3, 12 B.3, –4 C.–3, 0 D.3, 0 Solve 12x – 4x 2 = 0.

19. Example 3 Perfect Squares and Differences of Squares A. Solve x 2 – 6x + 9 = 0. x 2 = (x) 2 ; 9 = (3) 2 First and last terms are perfect squares. 6x = 2(x)(3)Middle term equals 2ab. x 2 – 6x + 9 is a perfect square trinomial. x 2 + 6x + 9 = 0Original equation (x – 3) 2 = 0 Factor using the pattern. x – 3= 0Take the square root of each side. x= 3Add 3 to each side. Answer: 3

20. Example 3 Perfect Squares and Differences of Squares B. Solve y 2 = 36. y 2 = 32Original equation y 2 – 36= 0Subtract 36 from each side. y 2 – (6) 2 = 0Write in the form a 2 – b 2. (y + 6)(y – 6) = 0Factor the difference of squares. y + 6 = 0y – 6 = 0Zero Product Property y = –6 y = 6Solve each equation. Answer: –6, 6

21. Example 3 A.8, –8 B.8, 0 C.8 D.–8 Solve x 2 – 16x + 64 = 0.

22. Example 4 Factor Trinomials A. Solve x 2 – 2x – 15 = 0. ac =–15a = 1, c = –15

23. Example 4 Factor Trinomials x 2 – 2x – 15= 0Original equation Answer: 5, –3 x 2 + mx + px – 15= 0Write the pattern. x 2 + 3x – 5x – 15= 0m = 3 and p = –5 (x 2 + 3x) – (5x + 15) = 0Group terms with common factors. x(x + 3) – 5(x + 3)= 0Factor the GCF from each grouping. (x – 5)(x + 3)= 0Distributive Property x – 5 = 0 x + 3= 0Zero Product Property x = 5 x= –3Solve each equation.

24. Example 4 Factor Trinomials B. Solve 5x 2 + 34x + 24 = 0. ac =120a = 5, c = 24

25. Example 4 Factor Trinomials 5x 2 + 34x + 24= 0Original equation 5x 2 + mx + px + 24= 0Write the pattern. 5x 2 + 4x + 30x + 24= 0m = 4 and p = 30 (5x 2 + 4x) + (30x + 24) = 0Group terms with common factors. x(5x + 4) + 6(x + 4)= 0Factor the GCF from each grouping. (x + 6)(5x + 4)= 0Distributive Property x + 6 = 0 5x + 4= 0Zero Product Property x = –6 Solve each equation.

26. Example 4 Factor Trinomials Answer:

27. Example 4 Solve 6x 2 – 5x – 4 = 0. A. B. C. D.

28. Example 4 A.(3s + 1)(s – 4) B.(s + 1)(3s – 4) C.(3s + 4)(s – 1) D.(s – 1)(3s + 4) B. Factor 3s 2 – 11s – 4.

29. Example 5 Solve Equations by Factoring ARCHITECTURE The entrance to an office building is an arch in the shape of a parabola whose vertex is the height of the arch. The height of the arch is given by h = 9 – x 2, where x is the horizontal distance from the center of the arch. Both h and x are measured in feet. How wide is the arch at ground level? To find the width of the arch at ground level, find the distance between the two zeros.

30. Example 5 Solve Equations by Factoring 9 – x 2 =0Original expression x 2 – 9 =0Multiply both sides by –1. (x + 3)(x – 3) =0Difference of squares x + 3 = 0 or x – 3 =0Zero Product Property x = –3 x =3Solve. Answer:The distance between 3 and – 3 is 3 – (–3) or 6 feet.

31. Example 5 Solve Equations by Factoring Check9 – x 2 =0 9 – (3) 2 =0or9 – (–3) 2 =0 ?? 9 – 9 =09 – 9=0 ?? 0 =00=0

32. Example 5 A.7 feet B.11 feet C.14 feet D.25 feet TENNIS During a match, Andre hit a lob right off the court with the ball traveling in the shape of a parabola whose vertex was the height of the shot. The height of the shot is given by h = 49 – x 2, where x is the horizontal distance from the center of the shot. Both h and x are measured in feet. How far was the lob hit?

33. End of the Lesson