Precalc0006to0010-page11

Precalc0006to0010-page11 - (Section 0.7: Factoring...

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(Section 0.7: Factoring Polynomials) 0.7.6 § Solution a) ax 2 + bx + c = x 2 + 9 x + 20 = x + 5 () x + 4 We want 5 and 4, because they have product = c = 20 and (since this is the a = 1 case) sum = b = 9. We can rearrange the factors: x + 4 x + 5 . b) x 2 x 2 ± ± 20 x + 100 10 2 ± Guess that this is a PST for now. ²³ ´´ µ ´´ = x ± 10 Check: 2 x ± 10 = ± 20 x ²³ ´µ ´ 2 ,o r x ± 10 x ± 10 = x ± 10 2 c) x 2 ± 4 x ± 12 = x ± 6 x + 2 How do we know we need ± 6 and + 2? The constant term, c , is negative, so use opposite signs: one "+" and one " ± ." The middle coefficient, b , is negative, so the negative number must be higher in absolute value than the positive number; it “carries more weight.” 2-factorizations of –12 (which is c ) Think: What? • What?? = –12. Sum = b = ± 4? ( a = 1 case ) ± 12, +1 No ± 6, +2 Yes ± Can stop ± 4, +3 No d) F + O + I + L = 3 x 2 ± 20 x ± 7 = 3 x + 1 x ± 7 F = First product (product of the First terms)
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