Factoring - (Section 0.7: Factoring Polynomials) 0.7.5...

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(Section 0.7: Factoring Polynomials) 0.7.5 Example Set 5 (Factoring Polynomials) Factor the following polynomials over the integers. a) x 2 + 9 x + 20 b) x 2 ± 20 x + 100 (Hint: This is a Perfect Square Trinomial (PST).) c) x 2 ± 4 x ± 12 d) 3 x 2 ± 20 x ± 7 e) 4 x 2 + 11 x + 6 f) 2 x 2 + 10 x + 5 g) ± 3 x 2 + 6 x ± 3 h) x 4 ± 16 i) a 3 ± 3 a + 2 a 2 b ± 6 b (Hint: Use Factoring by Grouping . This is when we group terms and factor each group “locally” before we factor the entire expression “globally” by factoring out the GCF.) j) 4 x 2 + 9 y 2 k) x 3 + 125 y 3 l) x 3 ± 125 y 3
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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?
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This note was uploaded on 09/08/2011 for the course MATH 141 taught by Professor Staff during the Fall '11 term at Mesa CC.

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Factoring - (Section 0.7: Factoring Polynomials) 0.7.5...

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