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

Question

1701 views around the world

You can reuse this answer
Creative Commons License

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)#