Answer
The roots of the polynomial are\[\left\{ -\frac{3}{4},1,i\sqrt{2},-i\sqrt{2} \right\}\]
Work Step by Step
Consider the given polynomial $f\left( x \right)=4{{x}^{4}}-{{x}^{3}}+5{{x}^{2}}-2x-6$.
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.
$p=\pm 1,\pm 2,\pm 3,\pm 6$
$q=\pm 1,\pm 2,\pm 4$
So, the possible rational zeros are:
$\begin{align}
& \text{Possible rational zero}=\frac{\text{Factor of}\left( -6 \right)}{\text{Factor of}\left( 4 \right)} \\
& =\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 2,\pm 4} \\
& =\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{3}{4},\pm \frac{3}{2}
\end{align}$
According to the Descartes’ rule of signs, the function $f\left( x \right)=4{{x}^{4}}-{{x}^{3}}+5{{x}^{2}}-2x-6$ has three sign changes. Since the function $f\left( x \right)$ has three variations in sign, the function $f\left( x \right)$ has three or one positive real root.
Further, evaluate $f\left( -x \right):$
$\begin{align}
& f\left( x \right)=4{{x}^{4}}-{{x}^{3}}+5{{x}^{2}}-2x-6 \\
& f\left( -x \right)=4{{\left( -x \right)}^{4}}-{{\left( -x \right)}^{3}}+5{{\left( -x \right)}^{2}}-2\left( -x \right)-6 \\
& =4{{x}^{4}}+{{x}^{3}}+5{{x}^{2}}+2x-6
\end{align}$
There is only one variation in sign. Thus, there is 1 negative real zero.
Test $x=1$ as a root of the polynomial:
$\begin{align}
& f\left( x \right)=4{{x}^{4}}-{{x}^{3}}+5{{x}^{2}}-2x-6 \\
& f\left( 1 \right)=4{{\left( 1 \right)}^{4}}-{{\left( 1 \right)}^{3}}+5{{\left( 1 \right)}^{2}}-2\left( 1 \right)-6 \\
& =0
\end{align}$
Divide the equation $f\left( x \right)$ by $\left( x-1 \right)$.
$\frac{4{{x}^{4}}-{{x}^{3}}+5{{x}^{2}}-2x-6}{\left( x-1 \right)}=\left( 4x+3 \right)\left( {{x}^{2}}+2 \right)$
Thus, function can be expressed as $f\left( x \right)=\left( x-1 \right)\left( 4x+3 \right)\left( {{x}^{2}}+2 \right)$.
Equate $f\left( x \right)=\left( x-1 \right)\left( 4x+3 \right)\left( {{x}^{2}}+2 \right)$ to zero.
$\begin{align}
& \left( x-1 \right)\left( 4x+3 \right)\left( {{x}^{2}}+2 \right)=0 \\
& x=-\frac{3}{4},1,i\sqrt{2},-i\sqrt{2}
\end{align}$
The solution of the polynomial is $\left\{ -\frac{3}{4},1,i\sqrt{2},-i\sqrt{2} \right\}$.
