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**Warm - up 6.4 1. 4x2 – 24x 4x(x – 6) 2. 2x2 + 11x – 21 (2x – 3)(x + 7)**

Factor: 1. 4x2 – 24x 4x(x – 6) 2. 2x2 + 11x – 21 (2x – 3)(x + 7) 3. 4x2 – 36x + 81 (2x – 9)2 Solve: 4. x2 + 10x + 25 = 0 x = -5 5. 6x2 + x = 15 x = 3/2 and -5/3

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**6.4 solving polynomial equations**

Objective – To be able to factor and solve polynomial expressions. CA State Standard - 3.0 Students are adept at operations on polynomials, including long division. - 4.0 Students factor polynomials representing the difference of squares, perfect square trinomials, and the sum and difference of two cubes. by Jason L. Bradbury

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**6.4 solving polynomial equations**

In Ch. 5 we learned how to factor: - A General Trinomial 2x2 – 5x – 12 (2x + 3)(x – 4) - A Perfect Square Trinomial x2 + 10x + 25 (x + 5)(x + 5) = (x +5)2 - The Difference of two Squares 4x2 – 9 (2x)2 – 32 (2x + 3)(2x – 3) - A Common Monomial Factor 6x2 + 15x 3x(2x + 5)

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**(x2 + ?)(x2 – ?) (x2 + 3)(x2 – 9) (x2 + 3)(x – 3)(x + 3)**

Example 1 Factor a) x4 – 6x2 – 27 (x ?)(x2 – ?) (x2 + 3)(x2 – 9) (x2 + 3)(x – 3)(x + 3) b) x4 – 3x2 – 10 (x ?)(x2 – ?) (x2 + 2)(x2 – 5)

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**a = x x3 + 23 ex. x3 + 8 b = 2 (x + 2)(x2 – 2x + 4) a = 2x b = 1**

** Special Factoring Patterns Sum of Two Cubes a3 + b3 = (a + b)(a2 - ab + b2) a = x x3 + 23 ex. x3 + 8 b = 2 (x + 2)(x2 – 2x + 4) Difference of Two Cubes a = 2x a3 – b3 = (a – b)(a2 + ab + b2) b = 1 ex x3 – 1 (2x – 1)(4x2 + 2x + 1) (2x)3 – (1)3 Example 2 x a3 + b3 = (a + b)(a2 - ab + b2) x3 + 53 = (x + 5)(x2 – 5x + 25)

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**Example 3 a) x3 – 27 x3 – 33 = (x – 3)(x2 + 3x + 9) b) 8x3 + 64**

Factor a) x3 – 27 a3 – b3 = (a – b)(a2 + ab + b2) x3 – 33 = (x – 3)(x2 + 3x + 9) b) 8x3 + 64 a3 + b3 = (a + b)(a2 - ab + b2) (2x)3 + (4)3 = (2x + 4)(4x2 – 8x + 16)

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**Extra Example 2 x3 – 2x2 – 9x + 18 x2(x – 2) -9(x – 2)**

Factor by grouping x3 – 2x2 – 9x + 18 x2(x – 2) -9(x – 2) Must be the same (x2 – 9)(x – 2) (x – 3)(x + 3)(x – 2)

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6.4 Homework Page 336 – 337 12 – 14, 21 – 27, and 31

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6.4 Guided Practice Page 336 – 337 12 – 14 and 21 – 24

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