Answer
The equation has the solution $\left\{ -1,2,-3 \right\}$.
Work Step by Step
Consider the cubic equation ${{x}^{3}}+2{{x}^{2}}-5x-6=0$.
Apply the hit and trial method and obtain that $\left( x+1 \right)$ is one of the factors of
${{x}^{3}}+2{{x}^{2}}-5x-6=0$
Simplify:
${{x}^{3}}+2{{x}^{2}}-5x-6=\left( x+1 \right)\left( {{x}^{2}}+x-6 \right)$
Factorize $\left( {{x}^{2}}+x-6 \right)$:
$\begin{align}
& \left( {{x}^{2}}+x-6 \right)={{x}^{2}}+3x-2x-6 \\
& =x\left( x+3 \right)-2\left( x+3 \right) \\
& =\left( x-2 \right)\left( x+3 \right)
\end{align}$
Thus,
${{x}^{3}}+2{{x}^{2}}-5x-6=\left( x+1 \right)\left( x-2 \right)\left( x+3 \right)$
Equate the value of the expression to 0.
$\begin{align}
& \left( x+1 \right)\left( x-2 \right)\left( x+3 \right)=0 \\
& x=-1,2,-3
\end{align}$
