Solution to problem of polynomial equation using factorization.

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  $x=\frac{-1\pm{i} \sqrt{3}}{2}$.