Intermediate Algebra (12th Edition)

$-12(3x+2)$
$\bf{\text{Solution Outline:}}$ Get the $GCF$ of the given expression, $(3x+2)(x-4)-(3x+2)(x+8) .$ 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 $\{ 1,1 \}$ is $1 .$ The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variables $\{ (3x+2),(3x+2) \}$ is $(3x+2) .$ Hence, the entire expression has $GCF= (3x+2) .$ Factoring the $GCF= (3x+2) ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} (3x+2) \left( \dfrac{(3x+2)(x-4)}{(3x+2)}-\dfrac{(3x+2)(x+8)}{(3x+2)} \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} (3x+2) \left( (3x+2)^{1-1}(x-4)-(3x+2)^{1-1}(x+8) \right) \\\\= (3x+2) \left( (3x+2)^{0}(x-4)-(3x+2)^{0}(x+8) \right) \\\\= (3x+2) \left( (1)(x-4)-(1)(x+8) \right) \\\\= (3x+2) \left( x-4-x-8 \right) \\\\= (3x+2) \left( -12 \right) \\\\= -12(3x+2) .\end{array}