## College Algebra 7th Edition

$x=2$ or $x=\pm\sqrt{5}$
We factor by grouping and solve: $x^{3}-2x^{2}-5x+10=0$ $x^{2}(x-2)-5(x-2)=0$ $(x-2)(x^{2}-5)=0$ $x-2=0$ or $x^2-5=0$ $x=2$ or $x=\pm\sqrt{5}$