Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 303: 41



Work Step by Step

Let $z=(a+5)$. Then the given expression, $ (a+5)^2-5(a+5)-24 $, is equivalent to $ z^2-5z-24 $. The two numbers whose product is $ac= 1(-24)=-24 $ and whose sum is $b= -5 $ are $\{ -8,3 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ z^2-5z-24 $, is \begin{array}{l}\require{cancel} z^2-8z+3z-24 \\\\= (z^2-8z)+(3z-24) \\\\= z(z-8)+3(z-8) \\\\= (z-8)(z+3) .\end{array} Since $z=(a+5)$, then, \begin{array}{l} (a+5-8)(a+5+3) \\\\= (a-3)(a+8) .\end{array}
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