# Algebra 2 Section 4-6

J
Jimbo LambMath Teacher and Technology Coach um Annville-Cleona Secondary School
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### Algebra 2 Section 4-6

• 2. Essential Questions • How do you factor polynomials? • How do you solve polynomial equations by factoring?
• 4. Vocabulary 1. Prime Polynomials: Polynomials that cannot be factored 2. Quadratic Form:
• 5. Vocabulary 1. Prime Polynomials: Polynomials that cannot be factored 2. Quadratic Form: Rewriting a polynomial so that it ﬁts the pattern of a standard form quadratic; occurs with the degree of the ﬁrst term is twice that of the second, with a constant third term
• 6. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping
• 7. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3)
• 8. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b)
• 9. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b) a3 + b3 = (a + b)(a2 − ab + b2 )
• 10. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b) a3 + b3 = (a + b)(a2 − ab + b2 ) a3 − b3 = (a − b)(a2 + ab + b2 )
• 11. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b) a3 + b3 = (a + b)(a2 − ab + b2 ) a3 − b3 = (a − b)(a2 + ab + b2 ) a2 + 2ab + b2 = (a + b)2
• 12. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b) a3 + b3 = (a + b)(a2 − ab + b2 ) a3 − b3 = (a − b)(a2 + ab + b2 ) a2 + 2ab + b2 = (a + b)2 a2 − 2ab + b2 = (a − b)2
• 13. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b) a3 + b3 = (a + b)(a2 − ab + b2 ) a3 − b3 = (a − b)(a2 + ab + b2 ) a2 + 2ab + b2 = (a + b)2 a2 − 2ab + b2 = (a − b)2 acx 2 + (ad + bc)x + bd = (ax + b)(cx + d )
• 14. Number of Terms Factoring Technique General Case Any Greatest Common Factor (GCF) 2 Difference of Squares 2 Sum of Cubes 2 Difference of Cubes 3 Perfect Square Trinomial 3 General Trinomials 4+ Grouping 4a5 b7 +12ab = 4ab(a4 b6 + 3) a2 − b2 = (a + b)(a − b) a3 + b3 = (a + b)(a2 − ab + b2 ) a3 − b3 = (a − b)(a2 + ab + b2 ) a2 + 2ab + b2 = (a + b)2 a2 − 2ab + b2 = (a − b)2 acx 2 + (ad + bc)x + bd = (ax + b)(cx + d ) ax + bx + ay + by = (a + b)(x + y )
• 15. Example 1 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 −1000 b. 24x 5 + 3x 2 y 3 c. x 3 + 400
• 16. Example 1 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 −1000 (x −10)(x 2 +10x +100) b. 24x 5 + 3x 2 y 3 c. x 3 + 400
• 17. Example 1 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 −1000 (x −10)(x 2 +10x +100) b. 24x 5 + 3x 2 y 3 3x 2 (8x 3 + y 3 ) c. x 3 + 400
• 18. Example 1 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 −1000 (x −10)(x 2 +10x +100) b. 24x 5 + 3x 2 y 3 3x 2 (8x 3 + y 3 ) 3x 2 (2x + y )(4x 2 − 2xy + y 2 ) c. x 3 + 400
• 19. Example 1 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 −1000 (x −10)(x 2 +10x +100) b. 24x 5 + 3x 2 y 3 3x 2 (8x 3 + y 3 ) 3x 2 (2x + y )(4x 2 − 2xy + y 2 ) c. x 3 + 400 Prime
• 20. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 + 5x 2 − 2x −10
• 21. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 + 5x 2 − 2x −10 (x 3 + 5x 2 )+ (−2x −10)
• 22. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 + 5x 2 − 2x −10 (x 3 + 5x 2 )+ (−2x −10) x 2 (x + 5)− 2(x + 5)
• 23. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 3 + 5x 2 − 2x −10 (x 3 + 5x 2 )+ (−2x −10) x 2 (x + 5)− 2(x + 5) (x + 5)(x 2 − 2)
• 24. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. b. a2 + 3ay + 2ay 2 + 6y 3
• 25. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. b. a2 + 3ay + 2ay 2 + 6y 3 (a2 + 3ay )+ (2ay 2 + 6y 3 )
• 26. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. b. a2 + 3ay + 2ay 2 + 6y 3 (a2 + 3ay )+ (2ay 2 + 6y 3 ) a(a + 3y )+ 2y 2 (a + 3y )
• 27. Example 2 Factor each polynomial. If the polynomial cannot be factored, write prime. b. a2 + 3ay + 2ay 2 + 6y 3 (a2 + 3ay )+ (2ay 2 + 6y 3 ) a(a + 3y )+ 2y 2 (a + 3y ) (a + 3y )(a + 2y 2 )
• 28. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 2 y 3 − 3xy 3 + 2y 3 + x 2 z3 − 3xz3 + 2z3
• 29. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 2 y 3 − 3xy 3 + 2y 3 + x 2 z3 − 3xz3 + 2z3 (x 2 y 3 − 3xy 3 + 2y 3 )+ (x 2 z3 − 3xz3 + 2z3 )
• 30. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 2 y 3 − 3xy 3 + 2y 3 + x 2 z3 − 3xz3 + 2z3 (x 2 y 3 − 3xy 3 + 2y 3 )+ (x 2 z3 − 3xz3 + 2z3 ) y 3 (x 2 − 3x + 2)+ z3 (x 2 − 3x + 2)
• 31. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 2 y 3 − 3xy 3 + 2y 3 + x 2 z3 − 3xz3 + 2z3 (x 2 y 3 − 3xy 3 + 2y 3 )+ (x 2 z3 − 3xz3 + 2z3 ) y 3 (x 2 − 3x + 2)+ z3 (x 2 − 3x + 2) (x 2 − 3x + 2)(y 3 + z3 )
• 32. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. a. x 2 y 3 − 3xy 3 + 2y 3 + x 2 z3 − 3xz3 + 2z3 (x 2 y 3 − 3xy 3 + 2y 3 )+ (x 2 z3 − 3xz3 + 2z3 ) y 3 (x 2 − 3x + 2)+ z3 (x 2 − 3x + 2) (x 2 − 3x + 2)(y 3 + z3 ) (x − 2)(x −1)(y + z)(y 2 − yz + z2 )
• 33. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. b. 64x 6 − y 6
• 34. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. b. 64x 6 − y 6 (8x 3 + y 3 )(8x 3 − y 3 )
• 35. Example 3 Factor each polynomial. If the polynomial cannot be factored, write prime. b. 64x 6 − y 6 (8x 3 + y 3 )(8x 3 − y 3 ) (2x + y )(4x 2 − 2xy + y 2 )(2x − y )(4x 2 + 2xy + y 2 )
• 36. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters.
• 37. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x x x
• 38. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x
• 39. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x x 3 − 1 8 x 3 = 23625
• 40. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x x 3 − 1 8 x 3 = 23625 7 8 x 3 = 23625
• 41. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x x 3 − 1 8 x 3 = 23625 7 8 x 3 = 23625 x 3 = 27000
• 42. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x x 3 − 1 8 x 3 = 23625 7 8 x 3 = 23625 x 3 = 27000 x 33 = 270003
• 43. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x x 3 − 1 8 x 3 = 23625 7 8 x 3 = 23625 x 3 = 27000 x 33 = 270003 x = 30
• 44. Example 4 A small cube is cut from a larger cube. Determine the dimensions of the cubes if the length of the smaller cube is is one half the length of the larger cube, and the volume of the object is 23,625 cubic centimeters. x 3 − x 2 ⎛ ⎝⎜ ⎞ ⎠⎟ 3 = 23625 x x x x 3 − 1 8 x 3 = 23625 7 8 x 3 = 23625 x 3 = 27000 x 33 = 270003 x = 30 The larger cube has a length of 30cm. The smaller cube has a length of 15 cm.
• 45. Example 5 Write each expression in quadratic form, if possible. a. 2x 6 − x 3 + 9 b. x 4 + 2x 3 −1
• 46. Example 5 Write each expression in quadratic form, if possible. a. 2x 6 − x 3 + 9 2(x 3 )2 − (x 3 )+ 9 b. x 4 + 2x 3 −1
• 47. Example 5 Write each expression in quadratic form, if possible. a. 2x 6 − x 3 + 9 2(x 3 )2 − (x 3 )+ 9 Let u = x 3 b. x 4 + 2x 3 −1
• 48. Example 5 Write each expression in quadratic form, if possible. a. 2x 6 − x 3 + 9 2(x 3 )2 − (x 3 )+ 9 2u2 −u + 9 Let u = x 3 b. x 4 + 2x 3 −1
• 49. Example 5 Write each expression in quadratic form, if possible. a. 2x 6 − x 3 + 9 2(x 3 )2 − (x 3 )+ 9 2u2 −u + 9 Let u = x 3 b. x 4 + 2x 3 −1 Not possible
• 50. Example 6 Solve. x 4 − 29x 2 +100 = 0
• 51. Example 6 Solve. x 4 − 29x 2 +100 = 0 Let u = x 2
• 52. Example 6 Solve. x 4 − 29x 2 +100 = 0 Let u = x 2 u2 − 29u +100 = 0
• 53. Example 6 Solve. x 4 − 29x 2 +100 = 0 Let u = x 2 u2 − 29u +100 = 0 (u − 25)(u − 4) = 0
• 54. Example 6 Solve. x 4 − 29x 2 +100 = 0 Let u = x 2 u2 − 29u +100 = 0 (u − 25)(u − 4) = 0 (x 2 − 25)(x 2 − 4) = 0
• 55. Example 6 Solve. x 4 − 29x 2 +100 = 0 Let u = x 2 u2 − 29u +100 = 0 (u − 25)(u − 4) = 0 (x 2 − 25)(x 2 − 4) = 0 (x + 5)(x − 5)(x + 2)(x − 2) = 0
• 56. Example 6 Solve. x 4 − 29x 2 +100 = 0 Let u = x 2 u2 − 29u +100 = 0 (u − 25)(u − 4) = 0 (x 2 − 25)(x 2 − 4) = 0 (x + 5)(x − 5)(x + 2)(x − 2) = 0 x = −5,x = 5,x = −2,x = 2
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