How do you factor $y=18x^3 + 9x^5 - 27x^2$ ?

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Answer 1 · 2016-01-02T00:10:58

$y = 18x^3+9x^5-27x^2$

$=9x^2(x-1)(x^2+x+3)$

$=9x^2(x-1)(x+1/2-sqrt(11)/2 i)(x+1/2+sqrt(11)/2 i)$

First note that all of the terms are divisible by $9x^2$ , so separate out that common factor first.

$y = 18x^3+9x^5-27x^2$

$=9x^2(x^3+2x-3)$

Then notice that the coefficients of the remaining cubic factor add up to $0$ (that is $1+2-3 = 0$ ), so $x=1$ is a zero and $(x-1)$ a factor:

$=9x^2(x-1)(x^2+x+3)$

The remaining quadratic factor has negative discriminant:

$Delta = b^2-4ac = 1^2 - (4xx1xx3) = 1-12 = -11$

So it has no factors with Real coefficients, but it can be factored using Complex coefficients that we can find using the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a) = (-1+-i sqrt(11))/2$

So:

$(x^2+x+3) = (x+1/2-sqrt(11)/2 i)(x+1/2+sqrt(11)/2 i)$

How do you factor y=18x^3 + 9x^5 - 27x^2 y=18x^3 + 9x^5 - 27x^2 ?