Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 32

Answer

$(3y+2)(-5y+9)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression, $ -15y^2+17y+18 ,$ the value of $ac$ is $ -15(18)=-270 $ and the value of $b$ is $ 17 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-270\}, \{2,-135\}, \{3,-90\}, \{5,-54\}, \{6,-45\}, \{9,-30\}, \{10,-27\}, \{15,-18\}, \{-1,270\}, \{-2,135\}, \{-3,90\}, \{-5,54\}, \{-6,45\}, \{-9,30\}, \{-10,27\}, \{-15,18\}, .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -10,27 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} -15y^2-10y+27y+18 .\end{array} Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (-15y^2-10y)+(27y+18) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -5y(3y+2)+9(3y+2) .\end{array} Factoring the $GCF= (3y+2) $ of the entire expression above results to \begin{array}{l}\require{cancel} (3y+2)(-5y+9) .\end{array}
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