Answer
The roots of the polynomial $f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10$ are $\left\{ -1,-10 \right\}$.
Work Step by Step
Determine values of p and q where p are the factors of the constant in the polynomial and q are the factors of the leading coefficient of the polynomial.
$p=\pm 1,\pm 2,\pm 5,\pm 10$
$q=\pm 1$
Calculate $\frac{p}{q}$.
$\frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$
According to Descartes’ s rule of signs, the function $f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10$ has zero sign changes. Since the function $f\left( x \right)$ has no variations in sign, the function $f\left( x \right)$ has zero positive real roots.
Further, evaluate $f\left( -x \right):$
$\begin{align}
& f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10 \\
& f\left( -x \right)={{\left( -x \right)}^{3}}+12{{\left( -x \right)}^{2}}+21\left( -x \right)+10 \\
& =-{{x}^{3}}+12{{x}^{2}}-21x+10
\end{align}$
There are three variations in sign. Thus, there are 3 negative real zeros or $3-2=1$ negative real zero.
Test $x=-1$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)={{x}^{3}}+12{{x}^{2}}+21x+10 \\
& f\left( -1 \right)={{\left( -1 \right)}^{3}}+12{{\left( -1 \right)}^{2}}+21\left( -1 \right)+10 \\
& =-1+12-21+10 \\
& =0
\end{align}$
Therefore $\left( x+1 \right)$ is a factor of the polynomial.
Divide the equation $f\left( x \right)$ by $\left( x+1 \right)$:
$\frac{{{x}^{3}}+12{{x}^{2}}+21x+10}{x+1}={{x}^{2}}+11x+10$
Thus, the polynomial can be expressed as $f\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}+11x+10 \right)$.
Equatethe polynomial equal to zero.
$\begin{align}
& \left( x+1 \right)\left( {{x}^{2}}+11x+10 \right)=0 \\
& \left( x+1 \right)\left( x+1 \right)\left( x+10 \right)=0 \\
& x=-1,-1,-10
\end{align}$
The solution of the polynomial is $\left\{ -1,-10 \right\}$.
