Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 6


$(2y+1)(3y-1) $

Work Step by Step

Using factoring of trinomials in the form $ax^2+bx+c,$ the given expression, $ 6y^2+y-1 ,$ has $ac$ equal to $ 6(-1)=-6 $ and $b$ equal to $ 1 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 3,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 6y^2+3y-2y-1 .\end{array} Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (6y^2+3y)-(2y+1) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3y(2y+1)-(2y+1) .\end{array} Factoring the $GCF= (2y+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} (2y+1)(3y-1) .\end{array}
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