Answer
$x=\left\{ \left( -3+\sqrt{11} \right),\left( -3-\sqrt{11} \right),-3 \right\}$
Work Step by Step
It is observed that $x=-3$ satisfies the polynomial using the hit-and-trial method
$\begin{align}
& f\left( -3 \right)={{x}^{3}}+9{{x}^{2}}+16x-6 \\
& ={{\left( -3 \right)}^{3}}+9{{\left( -3 \right)}^{2}}+16\left( -3 \right)-6 \\
& =-27+81-48-6 \\
& =0
\end{align}$
Thus, $\left( x+3 \right)$ is a factor of the given polynomial.
Now, using the long division method:
$\left( x+3 \right)\overset{{{x}^{2}}+6x-2}{\overline{\left){\begin{align}
& {{x}^{3}}+9{{x}^{2}}+16x-6 \\
& \pm {{x}^{3}}\pm 3{{x}^{2}} \\
& \overline{\begin{align}
& 6{{x}^{2}}+16x-6 \\
& \pm 6{{x}^{2}}\pm 18x \\
& \overline{\begin{align}
& -2x-6 \\
& \underline{\mp 2x\mp 6} \\
& 0 \\
\end{align}} \\
\end{align}} \\
\end{align}}\right.}}$
Thus, the factors of the polynomial are $\left( x+3 \right)\left( {{x}^{2}}+6x-2 \right)$. Then, we can write the given polynomial function as
$\begin{align}
& f\left( x \right)={{x}^{3}}+9{{x}^{2}}+16x-6 \\
& =\left( x+3 \right)\left( {{x}^{2}}+6x-2 \right)
\end{align}$
To find the solution of the polynomial we equate each factor to 0:
$\begin{align}
& x+3=0 \\
& x=-3
\end{align}$
Or,
${{x}^{2}}+6x-2=0$
Solving the quadratic equation ${{x}^{2}}+6x-2=0$ using the quadratic formula, we get:
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Put:
$\begin{align}
& a=1 \\
& b=6 \\
& c=-2
\end{align}$
$\begin{align}
& x=\frac{-6\pm \sqrt{{{\left( 6 \right)}^{2}}-4\left( 1 \right)\left( -2 \right)}}{2\left( 1 \right)} \\
& x=\frac{-6\pm \sqrt{36+8}}{2} \\
& x=\frac{-6\pm \sqrt{44}}{2} \\
& x=\frac{-6\pm 2\sqrt{11}}{2} \\
\end{align}$
$x=-3\pm \sqrt{11}$
Hence, all the roots of the given polynomial are $x=\left\{ \left( -3+\sqrt{11} \right),\left( -3-\sqrt{11} \right),-3 \right\}$.
