1 / 11

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) =

shelley-hardin
Download Presentation

6.4 Factoring and Solving Polynomial Expressions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related