Answer
$x=5$ or $x=\pm\sqrt{2}$
Work Step by Step
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 $x=\pm\sqrt{2}$
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