## College Algebra 7th Edition

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