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: 64



Work Step by Step

Let $z=(x+3)$. Then the given expression, $ 3(x+3)^2+2(x+3)-5 $, is equivalent to $ 3z^2+2z-5 $.\\ The two numbers whose product is $ac= 3(-5)=-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, $ 3z^2+2z-5 $, is \begin{array}{l}\require{cancel} 3z^2+5z-3z-5 \\\\= (3z^2+5z)-(3z+5) \\\\= z(3z+5)-(3z+5) \\\\= (3z+5)(z-1) .\end{array} Since $z=(x+3)$, then, \begin{array}{l} (3z+5)(z-1) \\\\= (3(x+3)+5)((x+3)-1) \\\\= (3x+9+5)(x+3-1) \\\\= (3x+14)(x+2) .\end{array}
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