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 304: 78



Work Step by Step

Factoring the $GCF=3$, then the given expression, $ 3x^2+6x-45 $, is equivalent to \begin{array}{l} 3(x^2+2x-15) .\end{array} The two numbers whose product is $ac= 1(-15)=-15 $ and whose sum is $b= 2 $ are $\{ 5,-3 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 3(x^2+2x-15) $, is \begin{array}{l}\require{cancel} 3(x^2+5x-3x-15) \\\\= 3[(x^2+5x)-(3x+15)] \\\\= 3[x(x+5)-3(x+5)] \\\\= 3[(x+5)(x-3)] \\\\= 3(x+5)(x-3) .\end{array}
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