## Intermediate Algebra (12th Edition)

$\left\{ -\dfrac{5}{2},-1,1 \right\}$
$\bf{\text{Solution Outline:}}$ Express the given equation, $2x^3+5x^2-2x-5=0 ,$ in factored form. Then use the Zero Product Property by equating each factor to zero. Finally, solve each of the resulting equations. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (2x^3+5x^2)-(2x+5)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^2(2x+5)-(2x+5)=0 .\end{array} Factoring the $GCF= (2x+5)$ of the entire expression above results to \begin{array}{l}\require{cancel} (2x+5)(x^2-1)=0 .\end{array} The expressions $x^2$ and $1$ are both perfect squares and are separated by a minus sign. Hence, $x^2-1 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (2x+5)(x+1)(x-1)=0 .\end{array} Equating each factor to zero (Zero Product Property), then \begin{array}{l}\require{cancel} 2x+5=0 \\\\\text{OR}\\\\ x+1=0 \\\\\text{OR}\\\\ x-1=0 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 2x+5=0 \\\\ 2x=-5 \\\\ x=-\dfrac{5}{2} \\\\\text{OR}\\\\ x+1=0 \\\\ x=-1 \\\\\text{OR}\\\\ x-1=0 \\\\ x=1 .\end{array} Hence, the solutions are $\left\{ -\dfrac{5}{2},-1,1 \right\} .$