#### Answer

$\left\{ -\dfrac{5}{2},-1,1 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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\}
.$