Intermediate Algebra (12th Edition)

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

Chapter 5 - Mixed Review Exercises: 5

Answer

$(5z-3m)^2$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $ 25z^2-30zm+9m^2 ,$ 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. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $ 25(9)=225 $ and the value of $b$ is $ -30 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -15,-15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 25z^2-15zm-15zm+9m^2 .\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} (25z^2-15zm)-(15zm-9m^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5z(5z-3m)-3m(5z-3m) .\end{array} Factoring the $GCF= (5z-3m) $ of the entire expression above results to \begin{array}{l}\require{cancel} (5z-3m)(5z-3m) \\\\= (5z-3m)^2 .\end{array}
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