How do you factor #x^3-18x^2+81x#?

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Answer 1 · 2015-06-14T15:54:36

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

#x^3-18x^2+81x = x(x-9)^2#

#x^3-18x^2+81x = x(x^2-18x+81)#

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

Thus:

#x^3-18x^2+81x#

#= x(x^2-18x+81)#

#= x(x^2-(2*x*9)+9^2)#

#= x(x-9)^2#