Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.5 Special Factoring Techniques - 3.5 Exercises - Page 280: 65



Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ -30x^3+58x^2+28x ,$ factor first the negative $GCF.$ Then find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Factoring the negative $GCF= -2x ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -2x(15x^2-29x-14) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 15(-14)=-210 $ and the value of $b$ is $ -29 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -35,6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -2x(15x^2-35x+6x-14) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -2x[(15x^2-35x)+(6x-14)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2x[5x(3x-7)+2(3x-7)] .\end{array} Factoring the $GCF= (3x-7) $ of the entire expression above results to \begin{array}{l}\require{cancel} -2x[(3x-7)(5x+2)] \\\\= -2x(3x-7)(5x+2) .\end{array}
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