Answer
The roots of the polynomial are\[\left\{ -1,-1,2-2i,2+2i \right\}\]
Work Step by Step
Consider the given polynomial ${{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+12x+8=0$.
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.
$p=\pm 1,\pm 2,\pm 4,\pm 8$
$q=\pm 1$
Calculate $\frac{p}{q}$.
$\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8$
According to the Descartes’s rule of signs, the function ${{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+12x+8=0$ has two sign changes. Since the function $f\left( x \right)$ has two variations in sign, the function $f\left( x \right)$ has two or zero positive real root.
Further, evaluate $f\left( -x \right):$
$\begin{align}
& f\left( x \right)={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+12x+8 \\
& f\left( -x \right)={{\left( -x \right)}^{4}}-2{{\left( -x \right)}^{3}}+{{\left( -x \right)}^{2}}+12\left( -x \right)+8 \\
& ={{x}^{4}}+2{{x}^{2}}+{{x}^{2}}-12x+8
\end{align}$
There are two variations in sign. Thus, there are 2 negative real zeros or $2-2=0$ negative real zeros.
Test $x=-1$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)={{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+12x+8 \\
& f\left( -1 \right)={{\left( -1 \right)}^{4}}-2{{\left( -1 \right)}^{3}}+{{\left( -1 \right)}^{2}}+12\left( -1 \right)+8 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by ${{\left( x+1 \right)}^{2}}$.
$\frac{\left( {{x}^{4}}-2{{x}^{3}}+{{x}^{2}}+12x+8 \right)}{{{\left( x+1 \right)}^{2}}}={{x}^{2}}-4x+8$
Thus, the function can be expressed as $f\left( x \right)={{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-4x+8 \right)$.
Equate $f\left( x \right)={{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-4x+8 \right)$ to zero.
$\begin{align}
& {{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-4x+8 \right)=0 \\
& {{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-4x+8 \right)=0 \\
& x=-1,-1,2-2i,2+2i
\end{align}$
The solution of the polynomial is $\left\{ -1,-1,2-2i,2+2i \right\}$.
