1. Splash Screen

2. 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
Lesson Menu

3. Use the related graph of y = x2 – 4 to determine its solutions.
B. 3, –2
C. 2, 0
D. 2, –2
5-Minute Check 1

4. Use the related graph of y = x2 – 4 to determine its solutions.
B. 3, –2
C. 2, 0
D. 2, –2
5-Minute Check 1

5. Use the related graph of y = –x2 – 2x + 3 to determine its solutions.
B. –3, 3
C. –1, 3
D. 3, 1
5-Minute Check 2

6. Use the related graph of y = –x2 – 2x + 3 to determine its solutions.
B. –3, 3
C. –1, 3
D. 3, 1
5-Minute Check 2

7. Solve –2x2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located.
A. 0
B. 0, between 2 and 3
C. between 1 and 2
D. 2, –2
5-Minute Check 3

8. Solve –2x2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located.
A. 0
B. 0, between 2 and 3
C. between 1 and 2
D. 2, –2
5-Minute Check 3

9. Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14.
5-Minute Check 4

10. Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14.
5-Minute Check 4

11. Which term is not another name for a solution to a quadratic equation?
A. zero
B. x-intercept
C. root
D. vertex
5-Minute Check 5

12. Which term is not another name for a solution to a quadratic equation?
A. zero
B. x-intercept
C. root
D. vertex
5-Minute Check 5

13. Mathematical Practices 2 Reason Abstractly and quantitatively.
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.
CCSS

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

15. factored form
FOIL method
Vocabulary

16. Concept

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

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

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

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

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

22. Concept

23. 9y 2 + 3y = 0 Original equation 3y(3y) + 3y(1) = 0 Factor the GCF.
Factor GCF
A. Solve 9y 2 + 3y = 0.
9y 2 + 3y = 0 Original equation
3y(3y) + 3y(1) = 0 Factor the GCF.
3y(3y + 1) = 0 Distributive Property
3y = y + 1 = 0 Zero Product Property
y = 0 Solve each equation.
Answer:
Example 2

24. 9y 2 + 3y = 0 Original equation 3y(3y) + 3y(1) = 0 Factor the GCF.
Factor GCF
A. Solve 9y 2 + 3y = 0.
9y 2 + 3y = 0 Original equation
3y(3y) + 3y(1) = 0 Factor the GCF.
3y(3y + 1) = 0 Distributive Property
3y = y + 1 = 0 Zero Product Property
y = 0 Solve each equation.
Answer:
Example 2

25. 5a 2 – 20a = 0 Original equation 5a(a) – 5a(4) = 0 Factor the GCF.
Factor GCF
B. Solve 5a2 – 20a = 0.
5a 2 – 20a = 0 Original equation
5a(a) – 5a(4) = 0 Factor the GCF.
5a(5a – 4) = 0 Distributive Property
5a = a – 4 = 0 Zero Product Property
a = a = 4 Solve each equation.
Answer:
Example 2

26. 5a 2 – 20a = 0 Original equation 5a(a) – 5a(4) = 0 Factor the GCF.
Factor GCF
B. Solve 5a2 – 20a = 0.
5a 2 – 20a = 0 Original equation
5a(a) – 5a(4) = 0 Factor the GCF.
5a(5a – 4) = 0 Distributive Property
5a = a – 4 = 0 Zero Product Property
a = a = 4 Solve each equation.
Answer: 0, 4
Example 2

27. Solve 12x – 4x2 = 0.
A. 3, 12
B. 3, –4
C. –3, 0
D. 3, 0
Example 2

28. Solve 12x – 4x2 = 0.
A. 3, 12
B. 3, –4
C. –3, 0
D. 3, 0
Example 2

29. x 2 = (x)2; 9 = (3)2 First and last terms are perfect squares.
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 = 0 Original equation
(x – 3)2 = 0 Factor using the pattern.
x – 3 = 0 Take the square root of each side.
x = 3 Add 3 to each side.
Answer:
Example 3

30. x 2 = (x)2; 9 = (3)2 First and last terms are perfect squares.
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 = 0 Original equation
(x – 3)2 = 0 Factor using the pattern.
x – 3 = 0 Take the square root of each side.
x = 3 Add 3 to each side.
Answer: 3
Example 3

31. y2 – 36 = 0 Subtract 36 from each side.
Perfect Squares and Differences of Squares
B. Solve y 2 = 36.
y 2 = 32 Original equation
y2 – 36 = 0 Subtract 36 from each side.
y2 – (6)2 = 0 Write in the form a2 – b2.
(y + 6)(y – 6) = 0 Factor the difference of squares.
y + 6 = 0 y – 6 = 0 Zero Product Property
y = – y = 6 Solve each equation.
Answer:
Example 3

32. y2 – 36 = 0 Subtract 36 from each side.
Perfect Squares and Differences of Squares
B. Solve y 2 = 36.
y 2 = 32 Original equation
y2 – 36 = 0 Subtract 36 from each side.
y2 – (6)2 = 0 Write in the form a2 – b2.
(y + 6)(y – 6) = 0 Factor the difference of squares.
y + 6 = 0 y – 6 = 0 Zero Product Property
y = – y = 6 Solve each equation.
Answer: –6, 6
Example 3

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

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

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

36. x 2 – 2x – 15 = 0 Original equation
Factor Trinomials
x 2 – 2x – 15 = 0 Original equation
x2 + mx + px – 15 = 0 Write the pattern.
x 2 + 3x – 5x – 15 = 0 m = 3 and p = –5
(x 2 + 3x) – (5x + 15) = 0 Group terms with common factors.
x(x + 3) – 5(x + 3) = 0 Factor the GCF from each grouping.
(x – 5)(x + 3) = 0 Distributive Property
x – 5 = 0 x + 3 = 0 Zero Product Property
x = 5 x = –3 Solve each equation.
Answer:
Example 4

37. x 2 – 2x – 15 = 0 Original equation
Factor Trinomials
x 2 – 2x – 15 = 0 Original equation
x2 + mx + px – 15 = 0 Write the pattern.
x 2 + 3x – 5x – 15 = 0 m = 3 and p = –5
(x 2 + 3x) – (5x + 15) = 0 Group terms with common factors.
x(x + 3) – 5(x + 3) = 0 Factor the GCF from each grouping.
(x – 5)(x + 3) = 0 Distributive Property
x – 5 = 0 x + 3 = 0 Zero Product Property
x = 5 x = –3 Solve each equation.
Answer: 5, –3
Example 4

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

39. 5x 2 + 34x + 24 = 0 Original equation
Factor Trinomials
5x x + 24 = 0 Original equation
5x2 + mx + px + 24 = 0 Write the pattern.
5x 2 + 4x + 30x + 24 = 0 m = 4 and p = 30
(5x 2 + 4x) + (30x + 24) = 0 Group terms with common factors.
x(5x + 4) + 6(x + 4) = 0 Factor the GCF from each grouping.
(x + 6)(5x + 4) = 0 Distributive Property
x + 6 = 0 5x + 4 = 0 Zero Product Property
x = –6 Solve each equation.
Example 4

40. Factor Trinomials
Answer:
Example 4

41. Factor Trinomials
Answer:
Example 4

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

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

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

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

46. 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.
Example 5

47. 9 – x 2 = 0 Original expression x 2 – 9 = 0 Multiply both sides by –1.
Solve Equations by Factoring
9 – x 2 = 0 Original expression
x 2 – 9 = 0 Multiply both sides by –1.
(x + 3)(x – 3) = 0 Difference of squares
x + 3 = 0 or x – 3 = 0 Zero Product Property
x = – x = 3 Solve.
Answer:
Example 5

48. 9 – x 2 = 0 Original expression x 2 – 9 = 0 Multiply both sides by –1.
Solve Equations by Factoring
9 – x 2 = 0 Original expression
x 2 – 9 = 0 Multiply both sides by –1.
(x + 3)(x – 3) = 0 Difference of squares
x + 3 = 0 or x – 3 = 0 Zero Product Property
x = – x = 3 Solve.
Answer: The distance between 3 and – 3 is 3 – (–3) or 6 feet.
Example 5

49. Check 9 – x 2 = 0 9 – (3)2 = 0 or 9 – (–3)2 = 0 9 – 9 = 0 9 – 9 = 0
Solve Equations by Factoring
Check 9 – x 2 = 0
9 – (3)2 = 0 or 9 – (–3)2 = 0 ?
9 – 9 = 0 9 – 9 = 0 ?
0 = 0 0 = 0  
Example 5

50. 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?
A. 7 feet
B. 11 feet
C. 14 feet
D. 25 feet
Example 5

51. 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?
A. 7 feet
B. 11 feet
C. 14 feet
D. 25 feet
Example 5

52. End of the Lesson