How do you completely factor $x ^ { 6} - 18x ^ { 5} + 81x ^ { 4}$ ?

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Answer 1 · 2017-07-16T03:28:37

$x^4(x-9)^2$

If you take $x^4$ out as the initial factor, you're left with:

$x^4(x^2-18x+81)$

This quadratic can further be factorised to:

$(x-9)(x-9)$ which gives the answer above.

How do you completely factor x ^ { 6} - 18x ^ { 5} + 81x ^ { 4}x ^ { 6} - 18x ^ { 5} + 81x ^ { 4}?