Below is a solution to a 4th order polynomial equation. 


Solve the equation $2^4-3x^3-x^2-3x+2=0$
Divide through by $x^2$
$2x^2-3x-1-\frac{3}{x}+\frac{2}{x^2}=0$
Collect like terms.
$2x^2+\frac{2}{x^2}-3x-\frac{3}{x}-1=0$
Factor out coefficients
$2(x^2+\frac{1}{x^2})-3(x+\frac{1}{x})-1=0$...............................(i)
Let &$t=x+\frac{1}{x}$ so that $t^2=x^2+2+\frac{1}{x^2}$
Substitute $t$ into (i)
$2(t^2-2)-3t-1=0$
$2t^2-4-3t-1=0$
$2t^2-3t-5=0$
Factorizing the quadratic equation yields:
$2t^2-3t-5=(2t-5)(t+1)=0$
$t=\frac{5}{2}$ or $-1$
Substitute $t=x+\frac{1}{x}$ into the roots of the equation
=$x+\frac{1}{x}=\frac{5}{2}$ or $x+\frac{1}{x}=-1$
Simplify the roots by clearing the denominator, we get
$2x^2+5x+2=0$ or $x^2+x+1=0$
Solving for $x$ in $2x^2+5x+2=0$ we get $x=\frac{1}{2}, 2$
And solving for $x$ in $x^2+x+1=0$ we get $x=\frac{-1\pm{i} \sqrt{3}}{2}$. 

Thetefore: the root of the polynomial equation  $2^4-3x^3-x^2-3x+2=0$ is  $x=\frac{1}{2}, 2$, and  $x=\frac{-1\pm{i} \sqrt{3}}{2}$.