Answer
The roots of the polynomial function are:\[\left\{ 1,-2,5 \right\}\]
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 pare 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 $\frac{p}{q}$ .
$\frac{p}{q}=\pm 1,\pm 2,\pm 5,\pm 10$
By using the Descartes’ rule of sign.
The function $f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10$ 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.
Evaluate $f\left( -x \right):$
$\begin{align}
& f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10 \\
& f\left( -x \right)={{\left( -x \right)}^{3}}-4{{\left( -x \right)}^{2}}-7\left( -x \right)+10 \\
& =-{{x}^{3}}-4{{x}^{2}}+7x+10
\end{align}$
The function has only one sign change, thus the function has one negative real zero.
Test $x=1$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)={{x}^{3}}-4{{x}^{2}}-7x+10 \\
& f\left( 1 \right)={{\left( 1 \right)}^{3}}-4{{\left( 1 \right)}^{2}}-7\left( 1 \right)+10 \\
& =1-4-7+10 \\
& =0
\end{align}$
Therefore, the factor of the polynomial is $\left( x-1 \right)$.
Divide the equation $f\left( x \right)$ by $\left( x-1 \right)$:
$\frac{{{x}^{3}}-4{{x}^{2}}-7x+10}{x-1}={{x}^{2}}-3x-10$
Thus, the polynomial can be expressed as $f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-3x-10 \right)$.
Equate $f\left( x \right)=\left( x-1 \right)\left( {{x}^{2}}-3x-10 \right)$ to zero.
$\begin{align}
& \left( x-1 \right)\left( {{x}^{2}}-3x-10 \right)=0 \\
& \left( x-1 \right)\left( x+2 \right)\left( x-5 \right)=0 \\
& x=1,-2,5
\end{align}$
The solution of the set is $\left\{ 1,-2,5 \right\}$.
