6.4 Factoring and Solving Polynomial Equations. Review of Factoring 2 nd Degree Polynomials x 2 + 9x + 20 = (x+5)(x+4) x 2 - 11x + 30 = (x-6)(x-5) 3x.

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Presentation transcript:

6.4 Factoring and Solving Polynomial Equations

Review of Factoring 2 nd Degree Polynomials x 2 + 9x + 20 = (x+5)(x+4) x x + 30 = (x-6)(x-5) 3x x + 12 = (3x+2)(x+6)

Factoring of higher-degree polynomials x 3 – 2x 2 – 9x + 18 For some polynomials, you can factor by grouping x 3 – 2x 2 – 9x + 18 = x 2 (x-2) - 9(x-2) = (x 2 - 9) (x-2) = (X+3) (X-3) (X-2)

More Factoring of Polynomials 4x 4 – 20x x 2 = 4x 2 (x 2 – 5x + 6) = 4x 2 (x-3) (x-2)

Solving Polynomial Equations If a polynomial is shown as equal to zero, then we can solve it for its solutions. Example from previous slide: 4x 4 – 20x x 2 = 0 4x 2 (x 2 – 5x + 6) = 0 4x 2 (x-3) (x-2) = 0 x = 0, 3, or 2

Another example 2x x = 26x 3 Group all the terms together in standard form 2x x x = 0 Factor out common monomial 2x (x x ) = 0 Factor the trinomial 2x (x 2 – 9) (x 2 – 4) = 0 Factor each binomial 2x(x+3)(x-3)(x+2)(x-2) = 0 x = 0, -3, 3, -2, 2

Practice Examples Text page 348. Problems 27-32

Classwork Text page 349, #42, 44, 48, 50, 52, 68, 74, 78, 84