Answer
The roots of the polynomial are\[\left\{ -\frac{5}{2},1,\sqrt{3},-\sqrt{3} \right\}\]
Work Step by Step
Consider the given polynomial $f\left( x \right)=2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15$.
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.
$p=\pm 1,\pm 3,\pm 5,\pm 15$
$q=\pm 1,\pm 2$
Calculate $\frac{p}{q}$.
$\frac{p}{q}=\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}$
According to the Descartes’s rule of signs, the function $f\left( x \right)=2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15$ 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 $f\left( -x \right):$
$\begin{align}
& f\left( x \right)=2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15 \\
& f\left( -x \right)=2{{\left( -x \right)}^{4}}+3{{\left( -x \right)}^{3}}-11{{\left( -x \right)}^{2}}-9\left( -x \right)+15 \\
& =2{{x}^{4}}-3{{x}^{3}}-11{{x}^{2}}+9x+15
\end{align}$
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:
$\begin{align}
& f\left( x \right)=2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15 \\
& f\left( 1 \right)=2{{\left( 1 \right)}^{4}}+3{{\left( 1 \right)}^{3}}-11{{\left( 1 \right)}^{2}}-9\left( 1 \right)+15 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by $\left( x-1 \right)$.
$\frac{2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15}{\left( x-1 \right)}=\left( 2x+5 \right)\left( {{x}^{2}}-3 \right)$
Thus, the polynomial can be expressed as $f\left( x \right)=\left( x-1 \right)\left( 2x+5 \right)\left( {{x}^{2}}-3 \right)$.
Equate $f\left( x \right)=\left( x-1 \right)\left( 2x+5 \right)\left( {{x}^{2}}-3 \right)$ to zero.
$\begin{align}
& \left( x-1 \right)\left( 2x+5 \right)\left( {{x}^{2}}-3 \right)=0 \\
& x=-\frac{5}{2},1,\sqrt{3},-\sqrt{3}
\end{align}$
The solution of the polynomial is $\left\{ -\frac{5}{2},1,\sqrt{3},-\sqrt{3} \right\}$.
