Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 56



Work Step by Step

Let $z=(c+2)$. Then the given expression, $ (c+2)^2-6(c+2)+5 $, is equivalent to \begin{array}{l} z^2-6z+5 .\end{array} The two numbers whose product is $ac= 1(5)=5 $ and whose sum is $b= -6 $ are $\{ -5,-1 \}$. Using these two numbers to decompose the middle term of the expression, $ z^2-6z+5 $, results to \begin{array}{l} z^2-5z-z+5 \\\\= (z^2-5z)-(z-5) \\\\= z(z-5)-(z-5) \\\\= (z-5)(z-1) .\end{array} Since $z=(c+2)$, then, \begin{array}{l} (z-5)(z-1) \\\\= (c+2-5)(c+2-1) \\\\= (c-3)(c+1) .\end{array}
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