How do you factor $x^{3} - 18 x^{2} + 81 x$ $x^3-18x^2+81x$ ?

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How do you factor $x^{3} - 18 x^{2} + 81 x$ $x^3-18x^2+81x$ ?

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Answer 1 · 2015-06-14T15:54:36

Separate the common $x$ $x$ factor then recognise remaining the perfect square trinomial to find:

$x^{3} - 18 x^{2} + 81 x = x {(x - 9)}^{2}$ $x^3-18x^2+81x = x(x-9)^2$

Explanation:

$x^{3} - 18 x^{2} + 81 x = x (x^{2} - 18 x + 81)$ $x^3-18x^2+81x = x(x^2-18x+81)$

We can recognise $x^{2} - 18 x + 81$ $x^2-18x+81$ as being a perfect square trinomial, as it is of the form $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2 = (a-b)^2$ with $a = x$ $a=x$ and $b = 9$ $b=9$ .

Thus:

$x^{3} - 18 x^{2} + 81 x$ $x^3-18x^2+81x$

$= x (x^{2} - 18 x + 81)$ $= x(x^2-18x+81)$

$= x (x^{2} - (2 \cdot x \cdot 9) + 9^{2})$ $= x(x^2-(2*x*9)+9^2)$

$= x {(x - 9)}^{2}$ $= x(x-9)^2$

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How do you factor $x^{3} - 18 x^{2} + 81 x$ $x^3-18x^2+81x$ ?

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