## Intermediate Algebra (12th Edition)

$3b(b+1)(2b-5)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $6b^3-9b^2-15b ,$ 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 $\{ 6,-9,-15 \}$ is $3$ 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 $\{ b^3,b^2,b \}$ is $b .$ Hence, the entire expression has $GCF= 3b .$ Factoring the $GCF= 3b ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} 3b \left( \dfrac{6b^3}{3b}-\dfrac{9b^2}{3b}-\dfrac{15b}{3b} \right) \\\\= 3b(2b^2-3b-5) .\end{array} To factor the trinomial expression above, note that the value of $ac$ is $2(-5)=-10$ and the value of $b$ is $-3 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-10\}, \{2,-5\}, \\ \{-1,10\}, \{-2,5\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 2,-5 \}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3b(2b^2+2b-5b-5) .\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} 3b[(2b^2+2b)-(5b+5)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3b[2b(b+1)-5(b+1)] .\end{array} Factoring the $GCF= (b+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} 3b[(b+1)(2b-5)] \\\\= 3b(b+1)(2b-5) .\end{array}