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6.4 Factoring and Solving Polynomial Expressions

6.4 Factoring and Solving Polynomial Expressions. p. 345. Types of Factoring:. From Chapter 5 we did factoring of: GCF : 6x 2 + 15x = 3x (2x + 5) PTS : x 2 + 10x + 25 = (x + 5) 2 DOS : 4x 2 – 9 = (2x + 3)(2x – 3) Bustin’ da B = 2x 2 – 5x – 12 = (2x 2 - 8x) + (3x – 12) =

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6.4 Factoring and Solving Polynomial Expressions

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  1. 6.4 Factoring and Solving Polynomial Expressions p. 345

  2. Types of Factoring: • From Chapter 5 we did factoring of: • GCF : 6x2 + 15x = 3x (2x + 5) • PTS : x2 + 10x + 25 = (x + 5)2 • DOS : 4x2 – 9 = (2x + 3)(2x – 3) • Bustin’ da B = 2x2 – 5x – 12 = • (2x2 - 8x) + (3x – 12) = • 2x(x – 4) + 3(x – 4)= • (x – 4)(2x + 3)

  3. Now we will use Sum of Cubes: • a3 + b3 = (a + b)(a2 – ab + b2) • x3 + 8 = • (x)3 + (2)3 = • (x + 2)(x2 – 2x + 4)

  4. Difference of Cubes • a3 – b3 = (a – b)(a2 + ab + b2) • 8x3 – 1 = • (2x)3 – 13 = • (2x – 1)((2x)2 + 2x*1 + 12) • (2x – 1)(4x2 + 2x + 1)

  5. When there are more than 3 terms – use GROUPING • x3 – 2x2 – 9x + 18 = • (x3 – 2x2) + (-9x + 18) = Group in two’s • with a ‘+’ in the middle • x2(x – 2) - 9(x – 2) = GCF each group • (x – 2)(x2 – 9) = • (x – 2)(x + 3)(x – 3) Factor all that can be • factored

  6. Factoring in Quad form: • 81x4 – 16 = • (9x2)2 – 42 = • (9x2 + 4)(9x2 – 4)= Can anything be • factored still??? • (9x2 + 4)(3x – 2)(3x +2) • Keep factoring ‘till you can’t factor any more!!

  7. You try this one! • 4x6 – 20x4 + 24x2 = • 4x2 (x4 - 5x2 +6) = • 4x2 (x2 – 2)(x2 – 3)

  8. In Chapter 5, we used the zero property. (when multiplying 2 numbers together to get 0 – one must be zero)The also works with higher degree polynomials

  9. Solve: • 2x5 + 24x = 14x3 • 2x5 - 14x3 + 24x = 0 Put in standard form • 2x (x4 – 7x2 +12) = 0 GCF • 2x (x2 – 3)(x2 – 4) = 0 Bustin’ da ‘b’ • 2x (x2 – 3)(x + 2)(x – 2) = 0 Factor • everything • 2x=0 x2-3=0 x+2=0 x-2=0 set all • factors to 0 • X=0 x=±√3 x=-2 x=2

  10. Now, you try one! • 2y5 – 18y = 0 • Y=0 y=±√3 y=±i√3

  11. Assignment

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