Example 
Find the zeroes of the polynomial $p(x) = x^4 +x^3-2x^2-x+1$.


Solution 
Since the polynomial is of degree 4, so it has at least 4 zeroes.

Divide the polynomial by $x^2$
therefore the polynomial becomes,

$p(x)=x^4/x^2 +x^3/x^2-{2x^2}/x^2-x/x^2+1/x^2$
$p(x)=x^2 +x-{2}-1/x+1/x^2$

Rearranging the polynomial and equating it ot zero.

Therefore the equation become,

$x^2+1/x^2 +x-1/x -2 = 0$

Let $x-1/x = y$
squaring both the sides

$(x-1/x)^2=y^2$
$ x^2+1/x^2 -2 = y^2$ => $ x^2+1/x^2 =y^2+2$
Substuting in given equation, we have
$y^2+2+y-2=0$			
$y^2+y=0$
$y(y+1)=0$

So y= 0 or -1

Since $x-1/x=y$
So we have,
$x-1/x=0$

$x^2-1=0$
$(x-1)(x+1)=0$
So x= -1 and 1

Also $y=-1$
$x-1/x=-1$	
$x^2-1=-x$
$x^2 +x-1=0$

Solving using method of completing the square,

$x^2+x$ = $x^2+x/2+x/2 +1/4 -1/4$	
        =$(x+1/2)^2 -1/4$
		
Therefore $x^2 +x-1=0$ => $(x+1/2)^2 -1/4 -1=0$
$(x+1/2)^2 -5/4=0$
$(x+1/2)^2 -(√5/2)^2=0$

$(x+1/2+√5/2)=0$  or $(x+1/2-√5/2)=0$

$x+{1+√5}/2=0$  or $(x+{1-√5}/2)=0$

i.e $x= -{1+√5}/2 $ and ${√5-1}/2 $

Answer: Zeroes of the polynomial are ,$ -1$, $1$, $-{1+√5}/2$ and ${√5-1}/2$