6.4 Factoring and Solving Polynomial Expressions p. 345 Name two special factoring patterns for cubes. Name three ways to factor a polynomial. What is.

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6.4 Factoring and Solving Polynomial Expressions p. 345 Name two special factoring patterns for cubes. Name three ways to factor a polynomial. What is the difference between factoring a polynomial and solving a polynomial?

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

Now we will use Sum of Cubes: a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) x = (x) 3 + (2) 3 = (x + 2)(x 2 – 2x + 4)

Difference of Cubes a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) 8x 3 – 1 = (2x) 3 – 1 3 = (2x – 1)((2x) 2 + 2x* ) (2x – 1)(4x 2 + 2x + 1)

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

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

You try this one! 4x 6 – 20x x 2 = 4x 2 (x 4 - 5x 2 +6) = 4x 2 (x 2 – 2)(x 2 – 3)

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

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

Now, you try one! 2y 5 – 18y = 0 Y=0y=±√3y=±i√3

Name two special factoring patterns for cubes. Sum of two cubes and difference of two cubes. Name three ways to factor a polynomial. Factor by sum or difference of cubes, by grouping or when in quadratic form. What is the difference between factoring a polynomial and solving a polynomial? Solving takes the problem one step further than factoring.

Assignment Page 348, odd