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

1 Answer
George C.
Dec 22, 2015

Find: #18x^3+9x^5-27x^2=9x^2(x-1)(x^2+x+3)#

as shown below...

Explanation:

Rearrange in standard order (descending powers of #x#) and separate out the common factor #9x^2# which all the terms are divisible by:

#18x^3+9x^5-27x^2#

#=9x^5+18x^3-27x^2#

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

Next note that the sum of the coefficients of #x^3+2x-3# is #0#, so #x=1# is a zero of this cubic and #(x-1)# is a factor:

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

The discriminant of the remaining quadratic factor is #1^2-(4xx1xx3) = -11# which is negative, so there are no simpler factors with Real coefficients.

If you still want to factor it further you can use Complex coefficients:

#(x^2+x+3) = (x+1/2)^2+(sqrt(11)/2)^2#

#= (x+1/2-sqrt(11)/2 i)(x+1/2+sqrt(11)/2 i)#

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