Intermediate Algebra (12th Edition)

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

Chapter 5 - Section 5.1 - Greatest Common Factors and Factoring by Grouping - 5.1 Exercises: 36


$(t-s) \left( 2+4t-4s-t^2+2st-s^2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ 2(t-s)+4(t-s)^2-(t-s)^3 .$ Divide the given expression and the $GCF.$ Express the answer as the product of the $GCF$ and the resulting quotient. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 2,4,-1 \}$ is $ 1 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{ (t-s),(t-s)^2,(t-s)^3 \}$ is $ (t-s) .$ Hence, the entire expression has $GCF= (t-s) .$ Factoring the $GCF= (t-s) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (t-s) \left( \dfrac{2(t-s)}{(t-s)}+\dfrac{4(t-s)^2}{(t-s)}-\dfrac{(t-s)^3}{(t-s)} \right) .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} (t-s) \left( 2(t-s)^{1-1}+4(t-s)^{2-1}-(t-s)^{3-1} \right) \\\\= (t-s) \left( 2(t-s)^{0}+4(t-s)^{1}-(t-s)^{2} \right) \\\\= (t-s) \left( 2(1)+4(t-s)-(t-s)^{2} \right) \\\\= (t-s) \left( 2+4(t-s)-(t-s)^{2} \right) .\end{array} Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the expression above is \begin{array}{l}\require{cancel} (t-s) \left( 2+4(t-s)-(t^2-2st+s^2) \right) .\end{array} Using $a(b+c)=ab+ac$ or the Distributive Property and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} (t-s) \left( 2+4t-4s-t^2+2st-s^2 \right) \\\\= (t-s) \left( 2+4t-4s-t^2+2st-s^2 \right) .\end{array}
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