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: 35


$(3-x) \left( 6+2x-x^2 \right)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $ 4(3-x)^2-(3-x)^3+3(3-x) .$ 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 $\{ 4,-1,3 \}$ is $ 1 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{ (3-x)^2,(3-x)^3,(3-x) \}$ is $ (3-x) .$ Hence, the entire expression has $GCF= (3-x) .$ Factoring the $GCF= (3-x) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (3-x) \left( \dfrac{4(3-x)^2}{(3-x)}-\dfrac{(3-x)^3}{(3-x)}+\dfrac{3(3-x)}{(3-x)} \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} (3-x) \left( 4(3-x)^{2-1}-(3-x)^{3-1}+3(3-x)^{1-1} \right) \\\\= (3-x) \left( 4(3-x)^{1}-(3-x)^{2}+3(3-x)^{0} \right) \\\\= (3-x) \left( 4(3-x)-(3-x)^{2}+3(1) \right) \\\\= (3-x) \left( 4(3-x)-(3-x)^{2}+3 \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} (3-x) \left( 4(3-x)-(9-6x+x^2)+3 \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} (3-x) \left( 12-4x-9+6x-x^2+3 \right) \\\\= (3-x) \left( (12-9+3)+(-4x+6x)-x^2 \right) \\\\= (3-x) \left( 6+2x-x^2 \right) .\end{array}
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