# How do you factor #6x^5-51x^3-27x#?

##### 1 Answer

#6x^5-51x^3-27x#

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

#= 3x(sqrt(2)x-i)(sqrt(2)x+i)(x-3)(x+3)#

#### Explanation:

We will use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with

First separate out the common factor

#6x^5-51x^3-27x#

#= 3x(2x^4-17x^2-9)#

The remaining quartic factor is a quadratic in

#= 3x(2x^4-18x^2+x^2-9)#

#= 3x((2x^4-18x^2)+(x^2-9))#

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

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

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

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

This is as far as we can go with Real coefficients, but if we allow Complex coefficients then this can be factored further as:

#= 3x(sqrt(2)x-i)(sqrt(2)x+i)(x-3)(x+3)#