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Mastering Factoring Techniques for Polynomials: Sum, Difference, and Grouping

Dive into factoring techniques including sum and difference of cubes and grouping to solve polynomial equations efficiently. Practice identifying maximums and minimums with examples and real-world scenarios.

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Mastering Factoring Techniques for Polynomials: Sum, Difference, and Grouping

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  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? A. –3 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? A. –3 B. –2 C. 2 D. 4 5-Minute Check 1

  5. Use the table of values for f(x) = x4 – 12x2 + 5. 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. 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. A. 0.5 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. A. 0.5 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. 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 x3 – 400. If the polynomial cannot be factored, write prime. Answer: Example 1

  16. Sum and Difference of Cubes A. Factor the polynomial x3 – 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. Sum and Difference of Cubes B. Factor the polynomial 24x5 + 3x2y3. If the polynomial cannot be factored, write prime. 24x5 + 3x2y3 = 3x2(8x3 + y3) Factor out the GCF. 8x3 and y3 are both perfect cubes, so we can factor the sum of the two cubes. (8x3 + y3) = (2x)3 + (y)3 (2x)3 = 8x3; (y)3 = y3 = (2x + y)[(2x)2 – (2x)(y) + (y)2] Sum of two cubes Example 1

  18. Sum and Difference of Cubes = (2x + y)[4x2 – 2xy + y2] Simplify. 24x5 + 3x2y3 = 3x2(2x + y)[4x2 – 2xy + y2]Replace the GCF. Answer: Example 1

  19. Sum and Difference of Cubes = (2x + y)[4x2 – 2xy + y2] Simplify. 24x5 + 3x2y3 = 3x2(2x + y)[4x2 – 2xy + y2]Replace the GCF. Answer: 3x2(2x + y)(4x2 – 2xy + y2) Example 1

  20. A. B. C. D.prime A. Factor the polynomial 54x5 + 128x2y3. If the polynomial cannot be factored, write prime. Example 1

  21. A. B. C. D.prime A. Factor the polynomial 54x5 + 128x2y3. If the polynomial cannot be factored, write prime. Example 1

  22. A. B. C. D.prime B. Factor the polynomial 64x9 + 27y5. If the polynomial cannot be factored, write prime. Example 1

  23. A. B. C. D.prime B. Factor the polynomial 64x9 + 27y5. If the polynomial cannot be factored, write prime. Example 1

  24. Concept

  25. Factoring by Grouping A. Factor the polynomial x3 + 5x2 – 2x – 10. If the polynomial cannot be factored, write prime. x3 + 5x2 – 2x – 10 Original expression = (x3 + 5x2) + (–2x – 10) Group to find a GCF. = x2(x + 5) – 2(x + 5) Factor the GCF. = (x + 5)(x2 – 2) Distributive Property Answer: Example 2

  26. Factoring by Grouping A. Factor the polynomial x3 + 5x2 – 2x – 10. If the polynomial cannot be factored, write prime. x3 + 5x2 – 2x – 10 Original expression = (x3 + 5x2) + (–2x – 10) Group to find a GCF. = x2(x + 5) – 2(x + 5) Factor the GCF. = (x + 5)(x2 – 2) Distributive Property Answer:(x + 5)(x2 – 2) Example 2

  27. Factoring by Grouping B. Factor the polynomial a2 + 3ay + 2ay2 + 6y3. If the polynomial cannot be factored, write prime. a2 + 3ay + 2ay2 + 6y3 Original expression = (a2 + 3ay) + (2ay2 + 6y3) Group to find a GCF. = a(a + 3y) + 2y2(a + 3y) Factor the GCF. = (a + 3y)(a + 2y2) Distributive Property Answer: Example 2

  28. Factoring by Grouping B. Factor the polynomial a2 + 3ay + 2ay2 + 6y3. If the polynomial cannot be factored, write prime. a2 + 3ay + 2ay2 + 6y3 Original expression = (a2 + 3ay) + (2ay2 + 6y3) Group to find a GCF. = a(a + 3y) + 2y2(a + 3y) Factor the GCF. = (a + 3y)(a + 2y2) Distributive Property Answer:(a + 3y)(a + 2y2) Example 2

  29. A. Factor the polynomial d3 + 2d2 + 4d + 8. If the polynomial cannot be factored, write prime. A. (d + 2)(d2 + 2) B. (d – 2)(d2 – 4) C. (d + 2)(d2 + 4) D. prime Example 2

  30. A. Factor the polynomial d3 + 2d2 + 4d + 8. If the polynomial cannot be factored, write prime. A. (d + 2)(d2 + 2) B. (d – 2)(d2 – 4) C. (d + 2)(d2 + 4) D. prime Example 2

  31. B. Factor the polynomial r2 + 4rs2 + 2sr + 8s3. If the polynomial cannot be factored, write prime. A. (r – 2s)(r + 4s2) B. (r + 2s)(r + 4s2) C. (r + s)(r – 4s2) D. prime Example 2

  32. B. Factor the polynomial r2 + 4rs2 + 2sr + 8s3. If the polynomial cannot be factored, write prime. A. (r – 2s)(r + 4s2) B. (r + 2s)(r + 4s2) C. (r + s)(r – 4s2) D. prime Example 2

  33. Group to find a GCF. Factor the GCF. Combine Cubes and Squares A. Factor the polynomial x2y3 – 3xy3 + 2y3 + x2z3 – 3xz3 + 2z3. If the polynomial cannot be factored, write prime. With six terms, factor by grouping first. Example 3

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

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

  36. Difference of two squares Combine Cubes and Squares B. Factor the polynomial 64x6 – y6. 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. Example 3

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

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

  39. A. B. C. D.prime A. Factor the polynomial r3w2 + 6r3w + 9r3 + w2y3 + 6wy3 + 9y3. If the polynomial cannot be factored, write prime. Example 3

  40. A. B. C. D.prime A. Factor the polynomial r3w2 + 6r3w + 9r3 + w2y3 + 6wy3 + 9y3. If the polynomial cannot be factored, write prime. Example 3

  41. A. B. C. D.prime B. Factor the polynomial 729p6 – k6. If the polynomial cannot be factored, write prime. Example 3

  42. A. B. C. D.prime B. Factor the polynomial 729p6 – k6. If the polynomial cannot be factored, write 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. Volume of object Subtract. Divide. 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. Example 4

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

  46. Subtract 3375 from each side. Difference of cubes Zero Product Property Solve Polynomial Functions by Factoring 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. Quadratic Form A. Write 2x6 – x3 + 9 in quadratic form, if possible. 2x6 – x3 + 9 = 2(x3)2 – (x3) + 9 Answer: Example 5

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