How do you factor $x^{3} - 8 x^{2} + 16 x$ $x^3-8x^2+16x$ ?

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How do you factor $x^{3} - 8 x^{2} + 16 x$ $x^3-8x^2+16x$ ?

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Answer 1 · 2018-03-14T00:45:44

$x {(x - 4)}^{2}$ $x(x-4)^2$

Explanation:

$x (x^{2} - 8 x + 16) =$ $x(x^2-8x+16)=$
$x (x - 4) (x - 4) =$ $x(x-4)(x-4)=$
$x {(x - 4)}^{2}$ $x(x-4)^2$

It was factored this way because x can be factored out since it is a common factor. $- 4 + (- 4) = - 8$ $-4+(-4)=-8$ and $- 4 \cdot (- 4) = 16$ $-4*(-4)=16$ , so you write $x - 4$ $x-4$ for both. It can be simplified by squaring $x - 4$ $x-4$ .

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How do you factor $x^{3} - 8 x^{2} + 16 x$ $x^3-8x^2+16x$ ?

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How do you factor $x^{3} - 8 x^{2} + 16 x$ $x^3-8x^2+16x$ ?