## Precalculus (6th Edition) Blitzer

$x=-2,-1,1$
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: $x=-2,-1,1$