Precalc0006to0010-page12

Precalc0006to0010-page12 - (Section 0.7: Factoring...

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(Section 0.7: Factoring Polynomials) 0.7.7 e) 4 x 2 + 11 x + 6 = 4 x + 3 () x + 2 Method 1: Trial-and-Error ("Guess") Method 4 xx 2 x 2 x ± ² ³ F =4 x 2 + 1 + 6 + 6 + 1 + 2 + 3 + 3 + 2 ± ² ´ ´ ³ ´ ´ L = 6; need both " + " because of + 11 x Method 2: Factoring by Grouping 4 and 6 are neither prime nor “1,” so we may prefer this method. We want two integers whose product is ac = (4)(6) = 24 and whose sum is b = 11. We want 8 and 3; split the middle term accordingly. 4 x 2 + 11 x + 6 = 4 x 2 + 8 x + 3 x OK to switch ±² ³´ ³ + 6 = 4 x 2 + 8 x + 3 x + 6 ± Group terms = 4 + 2 + 3 x + 2 ± "Local factoring" = 4 x + 3 x + 2 ± "Global factoring" f) 2 x 2 + 10 x + 5 is prime or irreducible over the integers (i.e., it cannot be broken down further using integer coefficients). None of these combinations work:
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This document was uploaded on 12/29/2011.

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