## Intermediate Algebra (12th Edition)

$2x (4+x)(3-x)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $24x-2x^2-2x^3 ,$ factor first the $GCF.$ Then factor the resulting trinomial by finding 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. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 24,-2,-2 \}$ is $2$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{ x,x^2,x^3 \}$ is $x .$ Hence, the entire expression has $GCF= 2x .$ Factoring the $GCF= 2x ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 2x \left( \dfrac{24x}{2x}-\dfrac{2x^2}{2x}-\dfrac{2x^3}{2x} \right) \\\\= 2x (12-x-x^2) .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $-1(12)=-12$ and the value of $b$ is $-1 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-12\}, \{2,-6\}, \{3,-4\}, \\ \{-1,12\}, \{-2,6\}, \{-3,4\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 3,-4 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2x (12+3x-4x-x^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} 2x [(12+3x)-(4x+x^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x [3(4+x)-x(4+x)] .\end{array} Factoring the $GCF= (4+x)$ of the entire expression above results to \begin{array}{l}\require{cancel} 2x [(4+x)(3-x)] \\\\= 2x (4+x)(3-x) .\end{array}