Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - 4-4 Factoring Quadratic Expressions - Practice and Problem-Solving Exercises - Page 222: 88

Answer

$2(6a-7)(a+1)$

Work Step by Step

Let $z= 2a-3 .$ The given expression, $ 3(2a-3)^2+17(2a-3)+10 ,$ is equivalent to \begin{align*} 3z^2+17z+10 .\end{align*} Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression above has $ac= 3(10)=30 $ and $b= 17 .$ The two numbers with a product of $ac$ and a sum of $b$ are $\left\{ 2,15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{align*} 3z^2+2z+15z+10 .\end{align*} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{align*} (3z^2+2z)+(15z+10) .\end{align*} Factoring the $GCF$ in each group results to \begin{align*} z(3z+2)+5(3z+2) .\end{align*} Factoring the $GCF= (3z+2) $ of the entire expression above results to \begin{align*} (3z+2)(z+5) .\end{align*} Substituting back $z= 2a-3 ,$ the expression above is equivalent to \begin{align*} & (3(2a-3)+2)(2a-3+5) \\&= (6a-9+2)(2a-3+5) \\&= (6a-7)(2a+2) \\&= 2(6a-7)(a+1) &\text{ (factor $GCF=2$)} .\end{align*}
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