Answer
The roots of the polynomial are\[\left\{ -1,-1,3-i,3+i \right\}\]
Work Step by Step
Consider the given polynomial ${{x}^{4}}-4{{x}^{3}}-{{x}^{2}}+14x+10=0$.
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.
$p=\pm 1,\pm 2,\pm 5,\pm 10$
$q=\pm 1$
Calculate $\frac{p}{q}$ .
$\frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$
According to the Descartes’s rule of signs, the function ${{x}^{4}}-4{{x}^{3}}-{{x}^{2}}+14x+10=0$ 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 $f\left( -x \right):$
$\begin{align}
& f\left( x \right)={{x}^{4}}-4{{x}^{3}}-{{x}^{2}}+14x+10 \\
& f\left( -x \right)={{\left( -x \right)}^{4}}-4{{\left( -x \right)}^{3}}-{{\left( -x \right)}^{2}}+14\left( -x \right)+10 \\
& ={{x}^{4}}+4{{x}^{3}}-{{x}^{2}}-14x+10
\end{align}$
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:
$\begin{align}
& f\left( x \right)={{x}^{4}}-4{{x}^{3}}-{{x}^{2}}+14x+10 \\
& f\left( -1 \right)={{\left( -1 \right)}^{4}}-4{{\left( -1 \right)}^{3}}-{{\left( -1 \right)}^{2}}+14\left( -1 \right)+10 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by ${{\left( x+1 \right)}^{2}}$.
$\frac{\left( {{x}^{4}}-4{{x}^{3}}-{{x}^{2}}+14x+10 \right)}{{{\left( x+1 \right)}^{2}}}={{x}^{2}}-6x+10$
Thus, the function can be expressed as $f\left( x \right)={{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-6x+10 \right)$.
Equate $f\left( x \right)={{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-6x+10 \right)$ to zero.
$\begin{align}
& {{\left( x+1 \right)}^{2}}\left( {{x}^{2}}-6x+10 \right)=0 \\
& {{\left( x+1 \right)}^{2}}\left( x-\left( 3+i \right) \right)\left( x-\left( 3-i \right) \right)=0 \\
& x=-1,-1,3-i,3+i
\end{align}$
The solution of the polynomial is $\left\{ -1,-1,3-i,3+i \right\}$.
