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: 75



Work Step by Step

Factoring the $GCF=y$, then the given expression, $ 3x^2y-11xy+8y $, is equivalent to \begin{array}{l} y(3x^2-11x+8) .\end{array} The two numbers whose product is $ac= 3(8)=24 $ and whose sum is $b= -11 $ are $\{ -3,-8 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ y(3x^2-11x+8) $, is \begin{array}{l}\require{cancel} y(3x^2-3x-8x+8) \\\\= y[(3x^2-3x)-(8x-8)] \\\\= y[3x(x-1)-8(x-1)] \\\\= y[(x-1)(3x-8)] \\\\= y(x-1)(3x-8) .\end{array}
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