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Factoring and Solving Polynomial Equations Section 6.4

Factoring and Solving Polynomial Equations Section 6.4. Special Factoring Patterns. Sum of two cubes a 3 + b 3 = (a + b) (a 2 − ab + b 2 ) x 3 + 8 = (x + 2) (x 2 − 2x + 4) Difference of two cubes a 3 − b 3 = (a − b) (a 2 + ab + b 2 ) 8x 3 − 1 = (2x − 1) (4x 2 + 2x + 1).

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Factoring and Solving Polynomial Equations Section 6.4

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  1. Factoring and Solving Polynomial EquationsSection 6.4

  2. Special Factoring Patterns Sum of two cubes a3 + b3 = (a + b) (a2− ab + b2) x3 + 8 = (x + 2) (x2 − 2x + 4) Difference of two cubes a3− b3 = (a − b) (a2 + ab + b2) 8x3− 1 = (2x − 1) (4x2 + 2x + 1)

  3. Factoring the Sum of Difference of Cubes Factor each polynomial • x3+ 27 • 16u5− 250u2

  4. Factoring by Grouping Factor the polynomial x3− 2x2 − 9x + 18 Group the factors x3− 2x2 − 9x + 18 Take GFC of each x2(x − 2) −9(x − 2) Reverse distributive (x − 2)(x2 −9) Factor More??? (x − 2)(x −3)(x + 3) Factor More??? No—Done!!!

  5. Factoring Polynomials in Quadratic Form Factor each polynomial 81x4 − 16 4x6 − 20x4 + 24x2 =(9x2)2 −42 =(9x2 +4)(9x2−4) =(9x2 +4)(3x +2)(3x− 2) =4x2(x4−5x2 +6) =4x2(x2−2)(x2−3)

  6. Solving a Polynomial Equation Solve 2x5 + 24x = 14x3 2x5−14x3 + 24x = 0 2x(x4 −7x2 +12) = 0 2x(x2 −3)(x2 −4) = 0 2x(x2 −3)(x −2)(x + 2) = 0 2x = 0, x2 −3 = 0, x − 2 = 0, x + 2 = 0 x = 0, x = +√3, x = − √3, x = 2, x = −2

  7. In 1980 archeologists at the ruins of Caesara discovered a huge hydraulic concrete block with a volume of 330 cubic yards. The block’s dimensions are x yards high by 13x −15 yards wide. What is the height? Volume = Height = Length= Width= 330 x 13x −11 13x − 15 330 = x(13x −11)(13x −15) 0 = 169x3 −338x2 + 165x −330 0 = 169x2(x −2)+ 165(x − 2) 0 + (169x2 +165)(x − 2) x = 2 only real solution. So, 13x −15 = 11, 13x − 11 = 15

  8. Assignment 6.4 Page 348, 34-84 even

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