College Algebra (10th Edition)

$x\in\{-1,2\}$
We are searching for the real zeros of $f(x)=x^{4}-x^{3}+2x^{2}-4x-8$ Possible rational zeros: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2,\pm 4,\pm 8}{\pm 1}$ Testing with synthetic division, try $x+1$ $\left.\begin{array}{l} -1 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 1 &-1 &2 &-4 &-8 \\\hline &-1 &2 &-4 & 8 \\\hline 1& -2 & 4 &-8 & |\ \ 0 \end{array}$ $f(x)=(x+1)( x^{3}-2x^{2}+4x-8)$ factor in pairs: $x^{3}-2x^{2}+4x-8=x^{2}(x-2)+2(x-2)$ $=(x-2)(x^{2}+2)$ $x^{2}+2$ has no real zeros and can not be factored any further. $f(x)=(x+1)(x-2)(x^{2}+2)$ $f(x) =0$ when $x\in\{-1,2\}$