1. Splash Screen

2. Five-Minute Check (over Lesson 5–4) CCSS Then/Now New Vocabulary
Key Concept: Sum and Difference of Cubes
Example 1: Sum and Difference of Cubes
Concept Summary: Factoring Techniques
Example 2: Factoring by Grouping
Example 3: Combine Cubes and Squares
Example 4: Real-World Example: Solve Polynomial Functions by Factoring
Key Concept: Quadratic Form
Example 5: Quadratic Form
Example 6: Solve Equations in Quadratic Form
Lesson Menu

3. Which is not a zero of the function f(x) = x3 – 3x2 – 10x + 24?
B. –2
C. 2
D. 4
5-Minute Check 1

4. Which is not a zero of the function f(x) = x3 – 3x2 – 10x + 24?
B. –2
C. 2
D. 4
5-Minute Check 1

5. Use the table of values for f(x) = x4 – 12x2 + 5
Use the table of values for f(x) = x4 – 12x Estimate the x-coordinates at which any relative maxima and relative minima occur. Which is not a possible relative maximum or relative minimum?
A. x = –2.5
B. x = 0
C. x = 1.5
D. x = 2.5
5-Minute Check 2

6. Use the table of values for f(x) = x4 – 12x2 + 5
Use the table of values for f(x) = x4 – 12x Estimate the x-coordinates at which any relative maxima and relative minima occur. Which is not a possible relative maximum or relative minimum?
A. x = –2.5
B. x = 0
C. x = 1.5
D. x = 2.5
5-Minute Check 2

7. Estimate the x-value at which the relative minimum of f(x) = x4 + x + 2 occurs.
B. 0
C. –0.5
D. –1.5
5-Minute Check 3

8. Estimate the x-value at which the relative minimum of f(x) = x4 + x + 2 occurs.
B. 0
C. –0.5
D. –1.5
5-Minute Check 3

9. For which part(s) of its domain does the function f(x) = x3 – 2x2 – 11x + 12 have negative f(x) values?
A. (–∞, –3), (1, 4)
B. (–4, –3), (1, 3)
C. (–∞, –3), (1, ∞)
D. (–∞, –2), (1, 2)
5-Minute Check 4

10. For which part(s) of its domain does the function f(x) = x3 – 2x2 – 11x + 12 have negative f(x) values?
A. (–∞, –3), (1, 4)
B. (–4, –3), (1, 3)
C. (–∞, –3), (1, ∞)
D. (–∞, –2), (1, 2)
5-Minute Check 4

11. Mathematical Practices 4 Model with mathematics.
Content Standards
A.CED.1 Create equations and inequalities in one variable and use them to solve problems.
Mathematical Practices
4 Model with mathematics.
CCSS

12. You solved quadratic functions by factoring.
Factor polynomials.
Solve polynomial equations by factoring.
Then/Now

13. prime polynomials
quadratic form
Vocabulary

14. Concept

15. Sum and Difference of Cubes
A. Factor the polynomial x 3 – 400. If the polynomial cannot be factored, write prime.
Answer:
Example 1

16. Sum and Difference of Cubes
A. Factor the polynomial x 3 – 400. If the polynomial cannot be factored, write prime.
Answer: The first term is a perfect cube, but the second term is not. It is a prime polynomial.
Example 1

17. 24x 5 + 3x 2y 3 = 3x 2(8x 3 + y 3) Factor out the GCF.
Sum and Difference of Cubes
B. Factor the polynomial 24x 5 + 3x 2y 3. If the polynomial cannot be factored, write prime.
24x 5 + 3x 2y 3 = 3x 2(8x 3 + y 3) Factor out the GCF.
8x 3 and y 3 are both perfect cubes, so we can factor the sum of the two cubes.
(8x 3 + y 3) = (2x)3 + (y)3 (2x)3 = 8x 3; (y)3 = y 3
= (2x + y)[(2x)2 – (2x)(y) + (y)2] Sum of two cubes
Example 1

18. = (2x + y)[4x 2 – 2xy + y 2] Simplify.
Sum and Difference of Cubes
= (2x + y)[4x 2 – 2xy + y 2] Simplify.
24x 5 + 3x 2y 3 = 3x 2(2x + y)[4x 2 – 2xy + y 2] Replace the GCF.
Answer:
Example 1

19. = (2x + y)[4x 2 – 2xy + y 2] Simplify.
Sum and Difference of Cubes
= (2x + y)[4x 2 – 2xy + y 2] Simplify.
24x 5 + 3x 2y 3 = 3x 2(2x + y)[4x 2 – 2xy + y 2] Replace the GCF.
Answer: 3x 2(2x + y)(4x 2 – 2xy + y 2)
Example 1

20. A. Factor the polynomial 54x 5 + 128x 2y 3
A. Factor the polynomial 54x x 2y 3. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 1

21. A. Factor the polynomial 54x 5 + 128x 2y 3
A. Factor the polynomial 54x x 2y 3. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 1

22. B. Factor the polynomial 64x 9 + 27y 5
B. Factor the polynomial 64x y 5. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 1

23. B. Factor the polynomial 64x 9 + 27y 5
B. Factor the polynomial 64x y 5. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 1

24. Concept

25. x 3 + 5x 2 – 2x – 10 Original expression
Factoring by Grouping
A. Factor the polynomial x 3 + 5x 2 – 2x – 10. If the polynomial cannot be factored, write prime.
x 3 + 5x 2 – 2x – 10 Original expression
= (x 3 + 5x 2) + (–2x – 10) Group to find a GCF.
= x 2(x + 5) – 2(x + 5) Factor the GCF.
= (x + 5)(x 2 – 2) Distributive Property
Answer:
Example 2

26. x 3 + 5x 2 – 2x – 10 Original expression
Factoring by Grouping
A. Factor the polynomial x 3 + 5x 2 – 2x – 10. If the polynomial cannot be factored, write prime.
x 3 + 5x 2 – 2x – 10 Original expression
= (x 3 + 5x 2) + (–2x – 10) Group to find a GCF.
= x 2(x + 5) – 2(x + 5) Factor the GCF.
= (x + 5)(x 2 – 2) Distributive Property
Answer: (x + 5)(x 2 – 2)
Example 2

27. a 2 + 3ay + 2ay 2 + 6y 3 Original expression
Factoring by Grouping
B. Factor the polynomial a 2 + 3ay + 2ay 2 + 6y 3. If the polynomial cannot be factored, write prime.
a 2 + 3ay + 2ay 2 + 6y 3 Original expression
= (a 2 + 3ay) + (2ay 2 + 6y 3) Group to find a GCF.
= a(a + 3y) + 2y 2(a + 3y) Factor the GCF.
= (a + 3y)(a + 2y 2) Distributive Property
Answer:
Example 2

28. a 2 + 3ay + 2ay 2 + 6y 3 Original expression
Factoring by Grouping
B. Factor the polynomial a 2 + 3ay + 2ay 2 + 6y 3. If the polynomial cannot be factored, write prime.
a 2 + 3ay + 2ay 2 + 6y 3 Original expression
= (a 2 + 3ay) + (2ay 2 + 6y 3) Group to find a GCF.
= a(a + 3y) + 2y 2(a + 3y) Factor the GCF.
= (a + 3y)(a + 2y 2) Distributive Property
Answer: (a + 3y)(a + 2y 2)
Example 2

29. A. Factor the polynomial d 3 + 2d 2 + 4d + 8
A. Factor the polynomial d 3 + 2d 2 + 4d + 8. If the polynomial cannot be factored, write prime.
A. (d + 2)(d 2 + 2)
B. (d – 2)(d 2 – 4)
C. (d + 2)(d 2 + 4)
D. prime
Example 2

30. A. Factor the polynomial d 3 + 2d 2 + 4d + 8
A. Factor the polynomial d 3 + 2d 2 + 4d + 8. If the polynomial cannot be factored, write prime.
A. (d + 2)(d 2 + 2)
B. (d – 2)(d 2 – 4)
C. (d + 2)(d 2 + 4)
D. prime
Example 2

31. B. Factor the polynomial r 2 + 4rs 2 + 2sr + 8s 3
B. Factor the polynomial r 2 + 4rs 2 + 2sr + 8s 3. If the polynomial cannot be factored, write prime.
A. (r – 2s)(r + 4s 2)
B. (r + 2s)(r + 4s 2)
C. (r + s)(r – 4s 2)
D. prime
Example 2

32. B. Factor the polynomial r 2 + 4rs 2 + 2sr + 8s 3
B. Factor the polynomial r 2 + 4rs 2 + 2sr + 8s 3. If the polynomial cannot be factored, write prime.
A. (r – 2s)(r + 4s 2)
B. (r + 2s)(r + 4s 2)
C. (r + s)(r – 4s 2)
D. prime
Example 2

33. With six terms, factor by grouping first.
Combine Cubes and Squares
A. Factor the polynomial x 2y 3 – 3xy 3 + 2y 3 + x 2z 3 – 3xz 3 + 2z 3. If the polynomial cannot be factored, write prime.
With six terms, factor by grouping first.
Group to find a GCF.
Factor the GCF.
Example 3

34. Distributive Property
Combine Cubes and Squares
Distributive Property
Sum of cubes
Factor.
Example 3

35. Distributive Property
Combine Cubes and Squares
Distributive Property
Sum of cubes
Factor.
Example 3

36. Difference of two squares
Combine Cubes and Squares
B. Factor the polynomial 64x 6 – y 6. If the polynomial cannot be factored, write prime.
This polynomial could be considered the difference of two squares or the difference of two cubes. The difference of two squares should always be done before the difference of two cubes for easier factoring.
Difference of two squares
Example 3

37. Sum and difference of two cubes
Combine Cubes and Squares
Sum and difference of two cubes
Example 3

38. Sum and difference of two cubes
Combine Cubes and Squares
Sum and difference of two cubes
Example 3

39. A. Factor the polynomial r 3w 2 + 6r 3w + 9r 3 + w 2y 3 + 6wy 3 + 9y 3
A. Factor the polynomial r 3w 2 + 6r 3w + 9r 3 + w 2y 3 + 6wy 3 + 9y 3. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 3

40. A. Factor the polynomial r 3w 2 + 6r 3w + 9r 3 + w 2y 3 + 6wy 3 + 9y 3
A. Factor the polynomial r 3w 2 + 6r 3w + 9r 3 + w 2y 3 + 6wy 3 + 9y 3. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 3

41. B. Factor the polynomial 729p 6 – k 6
B. Factor the polynomial 729p 6 – k 6. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 3

42. B. Factor the polynomial 729p 6 – k 6
B. Factor the polynomial 729p 6 – k 6. If the polynomial cannot be factored, write prime. A. B. C.
D. prime
Example 3

43. Solve Polynomial Functions by Factoring
GEOMETRY Determine the dimensions of the cubes below if the length of the smaller cube is one half the length of the larger cube, and the volume of the shaded figure is 23,625 cubic centimeters.
Example 4

44. Solve Polynomial Functions by Factoring
Since the length of the smaller cube is half the length of the larger cube, then their lengths can be represented by x and 2x, respectively. The volume of the object equals the volume of the larger cube minus the volume of the smaller cube.
Volume of object
Subtract.
Divide.
Example 4

45. Subtract 3375 from each side.
Solve Polynomial Functions by Factoring
Subtract 3375 from each side.
Difference of cubes
Zero Product Property
Answer:
Example 4

46. Subtract 3375 from each side.
Solve Polynomial Functions by Factoring
Subtract 3375 from each side.
Difference of cubes
Zero Product Property
Answer: Since 15 is the only real solution, the lengths of the cubes are 15 cm and 30 cm.
Example 4

47. GEOMETRY Determine the dimensions of the cubes below if the length of the smaller cube is one half the length of the larger cube, and the volume of the shaded figure is 5103 cubic centimeters.
A. 7 cm and 14 cm
B. 9 cm and 18 cm
C. 10 cm and 20 cm
D. 12 cm and 24 cm
Example 4

48. GEOMETRY Determine the dimensions of the cubes below if the length of the smaller cube is one half the length of the larger cube, and the volume of the shaded figure is 5103 cubic centimeters.
A. 7 cm and 14 cm
B. 9 cm and 18 cm
C. 10 cm and 20 cm
D. 12 cm and 24 cm
Example 4

49. Concept

50. A. Write 2x 6 – x 3 + 9 in quadratic form, if possible.
2x 6 – x = 2(x 3)2 – (x 3) + 9
Answer:
Example 5

51. A. Write 2x 6 – x 3 + 9 in quadratic form, if possible.
2x 6 – x = 2(x 3)2 – (x 3) + 9
Answer: 2(x 3)2 – (x 3) + 9
Example 5

52. B. Write x 4 – 2x 3 – 1 in quadratic form, if possible.
Answer:
Example 5

53. B. Write x 4 – 2x 3 – 1 in quadratic form, if possible.
Answer: This cannot be written in quadratic form since x 4 ≠ (x 3)2.
Example 5

54. A. Write 6x 10 – 2x 5 – 3 in quadratic form, if possible.
A. 3(2x 5)2 – (2x 5) – 3
B. 6x5(x5) – x5 – 3
C. 6(x 5)2 – 2(x 5) – 3
D. This cannot be written in quadratic form.
Example 5

55. A. Write 6x 10 – 2x 5 – 3 in quadratic form, if possible.
A. 3(2x 5)2 – (2x 5) – 3
B. 6x5(x5) – x5 – 3
C. 6(x 5)2 – 2(x 5) – 3
D. This cannot be written in quadratic form.
Example 5

56. B. Write x 8 – 3x 3 – 11 in quadratic form, if possible.
A. (x 8)2 – 3(x 3) – 11
B. (x 4)2 – 3(x 3) – 11
C. (x 4)2 – 3(x 2) – 11
D. This cannot be written in quadratic form.
Example 5

57. B. Write x 8 – 3x 3 – 11 in quadratic form, if possible.
A. (x 8)2 – 3(x 3) – 11
B. (x 4)2 – 3(x 3) – 11
C. (x 4)2 – 3(x 2) – 11
D. This cannot be written in quadratic form.
Example 5

58. Solve x 4 – 29x 2 + 100 = 0. Original equation Factor.
Solve Equations in Quadratic Form
Solve x 4 – 29x = 0.
Original equation
Factor.
Zero Product Property
Replace u with x 2.
Example 6

59. Take the square root. Answer: Solve Equations in Quadratic Form
Example 6

60. Answer: The solutions of the equation are 5, –5, 2, and –2.
Solve Equations in Quadratic Form
Take the square root.
Answer: The solutions of the equation are 5, –5, 2, and –2.
Example 6

61. Solve x 6 – 35x = 0.
A. 2, 3
B. –2, –3
C. –2, 2, –3, 3
D. no solution
Example 6

62. Solve x 6 – 35x = 0.
A. 2, 3
B. –2, –3
C. –2, 2, –3, 3
D. no solution
Example 6

63. End of the Lesson