Precalculus: Mathematics for Calculus, 7th Edition

The solutions are $x=2$, $x=2i$ and $x=-2i$
$x^{3}-2x^{2}+4x-8=0$ Group the first two terms and the last two terms together: $(x^{3}-2x^{2})+(4x-8)=0$ Take out common factor $x^{2}$ from the first parentheses and common factor $4$ from the second parentheses: $x^{2}(x-2)+4(x-2)=0$ Take out common factor $x-2$ from the left side: $(x-2)(x^{2}+4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-2=0$ $x=2$ $x^{2}+4=0$ $x^{2}=-4$ $\sqrt{x^{2}}=\sqrt{-4}$ $x=\pm2i$ The solutions are $x=2$, $x=2i$ and $x=-2i$