Answer
The roots of the polynomial are\[\left\{ -1,2,2i,-2i \right\}\]
Work Step by Step
Consider the given polynomial $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10$.
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 ${{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-4x-8=0$ has three sign changes. Since the function $f\left( x \right)$ has two variations in sign, the function $f\left( x \right)$ has three or one positive real root.
Further, evaluate $f\left( -x \right):$
$\begin{align}
& f\left( x \right)={{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-4x-8 \\
& f\left( -x \right)={{\left( -x \right)}^{4}}-{{\left( -x \right)}^{3}}+2{{\left( -x \right)}^{2}}-4\left( -x \right)-8 \\
& ={{x}^{4}}+{{x}^{3}}+2{{x}^{2}}+4x-8
\end{align}$
There is one variation in sign. Thus, there is only one negative real zero.
Test $x=-1$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)={{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-4x-8 \\
& f\left( -1 \right)={{\left( -1 \right)}^{4}}-{{\left( -1 \right)}^{3}}+2{{\left( -1 \right)}^{2}}-4\left( -1 \right)-8 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by $\left( x+1 \right)$.
$\frac{{{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-4x-8}{\left( x+1 \right)}=\left( x-2 \right)\left( {{x}^{2}}+4 \right)$
Thus, the polynomial can be expressed as $f\left( x \right)=\left( x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+4 \right)$.
Equate $f\left( x \right)=\left( x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+4 \right)$ to zero.
$\begin{align}
& \left( x+1 \right)\left( x-2 \right)\left( {{x}^{2}}+4 \right)=0 \\
& \left( x+1 \right)\left( x-2 \right)\left( x-2i \right)\left( x+2i \right)=0 \\
& x=-1,2,2i,-2i
\end{align}$
The solution of the polynomial is $\left\{ -1,2,2i,-2i \right\}$.
