## Intermediate Algebra (12th Edition)

$\left\{ -\dfrac{1}{3},0,\dfrac{5}{2} \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $6x^3-13x^2-5x=0 ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ The $GCF$ of the constants of the terms $\{ 6,-13,-5 \}$ is $1$ 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 $\{ x^3,x^2,x \}$ is $x .$ Hence, the entire expression has $GCF= x .$ Factoring the $GCF= x ,$ the expression above is equivalent to \begin{array}{l}\require{cancel} x(6x^2-13x-5)=0 .\end{array} Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $6(-5)=-30$ and the value of $b$ is $-13 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 2,-15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} x(6x^2+2x-15x-5)=0 .\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} x[(6x^2+2x)-(15x+5)]=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x[2x(3x+1)-5(3x+1)]=0 .\end{array} Factoring the $GCF= 3x+1$ of the entire expression above results to \begin{array}{l}\require{cancel} x[(3x+1)(2x-5)]=0 \\\\ x(3x+1)(2x-5)=0 .\end{array} Equating each factor zero (Zero Product Property), then \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 3x+1=0 \\\\\text{OR}\\\\ 2x-5=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x=0 \\\\\text{OR}\\\\ 3x+1=0 \\\\ 3x=-1 \\\\ x=-\dfrac{1}{3} \\\\\text{OR}\\\\ 2x-5=0 \\\\ 2x=5 \\\\ x=\dfrac{5}{2} .\end{array} Hence, the solutions are $\left\{ -\dfrac{1}{3},0,\dfrac{5}{2} \right\} .$