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 - Chapter Review Exercises: 52

Answer

$7(2k-1)(k-1)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 14k^2-21k+7 ,$ factor first the $GCF.$ 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 $GCF= 7 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 7(2k^2-3k+1) .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 2(1)=2 $ and the value of $b$ is $ -3 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -1,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 7(2k^2-1k-2k+1) .\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} 7[(2k^2-1k)-(2k-1)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 7[k(2k-1)-(2k-1)] .\end{array} Factoring the $GCF= (2k-1) $ of the entire expression above results to \begin{array}{l}\require{cancel} 7[(2k-1)(k-1)] \\\\= 7(2k-1)(k-1) .\end{array}
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