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Factoring Polynomials Objective: To factor polynomials by GCF, special products and grouping terms.

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Presentation on theme: "Factoring Polynomials Objective: To factor polynomials by GCF, special products and grouping terms."— Presentation transcript:

1 Factoring Polynomials Objective: To factor polynomials by GCF, special products and grouping terms

2 By GCF Factor each Polynomial 3x 4 – 12x 3 + 6x 2 The GCF of 3x 4, -12x 3 and 6x 2 is 3x 2 3x 4 – 12x 3 + 6x 2 = 3x 2 (x 2 – 4x + 2) 8xy 2 – 10x 2 y The GCF of 8xy 2 and -10x 2 y is 2xy 8xy 2 – 10x 2 y = 2xy(4y – 5x)

3 Perfect Square Trinomials a 2 + 2ab + b 2 = (a + b) 2 a 2 – 2ab + b 2 = (a – b) 2 Example: factor 4x 2 – 20x + 25 (2x) 2 – 2(2x)(5) + (5) 2 (2x – 5) 2 Factor x 2 + 12x + 36 (x) 2 + 2(x)(6) + (6) 2 (x + 6) 2

4 Difference Of Two Squares a 2 – b 2 = (a + b)(a – b) Example: factor x 2 – 49 (x) 2 – (7) 2 = (x + 7)(x – 7) Factor: 25h 2 – 9k 2 (5h) 2 – (3k) 2 = (5h + 3k)(5h – 3k)

5 Sum And Difference Of Cubes a 3 + b 3 = (a + b)(a 2 – ab + b 2 ) a 3 – b 3 = (a – b)(a 2 + ab + b 2 ) Example: factor x 3 – 8 (x) 3 – (2) 3 = (x – 2)(x 2 + (x)(2) + (2) 2 ) = (x – 2)(x 2 + 2x + 4) Factor x 3 + 27y 3 (x) 3 + (3y) 3 = (x +3y)(x 2 – x(3y) + (3y) 2 ) = (x + 3y)(x 2 – 3xy +9y 2 )

6 Factoring By Grouping In many instances a polynomial will not be a special product, but may be able to be factored by rearranging its terms. Factor 2xy – 3 – x + 6y 2xy – x + 6y – 3 x(2y – 1) + 3(2y – 1) (2y – 1)(x + 3) Note: this process often requires some trial and error.

7 Try These! Factor each Polynomial 1.12x 3 y 2 – 6x 2 y +9xy 3 2.x 2 – 8x + 16 3.x 2 – 64y 2 4.8x 3 + 125y 3 5.8x 2 y + 12x + 3y + 2xy 2 3xy(4x 2 y – 2x + 3y 2 ) (x – 4) 2 (x – 8y)(x + 8y) (2x + 5y)(4x 2 – 10xy + 25y 2 ) (4x + y)(2xy + 3)

8 Try These! All these problems will require more than one type of factorization 1.x 4 – 16 2.x 6 – y 6 3.2x 2 y – 8y 3 4.6x 2 y + 3xy 2 – 6xy – 3y 2 (x 2 + 4)(x + 2)(x – 2) solution solution (x + y)(x – y)(x 2 + xy + y 2 )(x 2 – xy + y 2 ) solutionsolution 2y(x + 2y)(x – 2y) solutionsolution 3y(2x + y)(x – 1) solutionsolution

9 x 4 – 16 (x 2 ) 2 – 4 2 ((x 2 ) + 4)((x 2 ) – 4) (x 2 + 4)((x) 2 – (2) 2 ) (x 2 + 4)(x + 2)(x – 2)

10 x 6 – y 6 (x 3 ) 2 – (y 3 ) 2 (x 3 + y 3 )(x 3 – y 3 ) (x + y)(x 2 – xy + y 2 )(x – y)(x 2 + xy +y 2 )

11 2x 2 y – 8y 3 2x 2 y – 2 3 y 3 2y(x 2 – 2 2 y 2 ) 2y((x) 2 – (2y) 2 ) 2y(x + 2y)(x – 2y)

12 6x 2 y + 3xy 2 – 6xy – 3y 2 3xy(2x + y) – 3y(2x + y) (2x + y)(3xy – 3y) (3xy – 3y)(2x + y) 3y(x – 1)(2x + y)


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