Question

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

Answer 1

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)`