Answer
$ x=-2,-1,1$
Work Step by Step
We have:
$ f(x)= x^4+x^3-3x^2-x+2=0$
Apply the rational zeroes theorem.
Here, we have:
$p=\pm1, \pm2$
$q=\pm 1$
$\dfrac{p}{q}=\pm1, \pm 2$
Now, plug in the possible zeros: $ 1,-1, 2, -2$
$ f(-1)=(-1)^4+(-1)^3-3(-1)^2-(-1)+2=0$
$ f(1)=(1)^4+(1)^3-3(1)^2-(1)+2=1+1-3-1+2=0$
$ f(-2)=(-2)^4+(-2)^3-3(-2)^2-(-2)+2=0$
$ f(2)=(2)^4+(2)^3-3(2)^2-(2)+2=16+8-12=12$
So, there are three zeroes: $x=-2,-1,1$
