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

1 Answer
Feb 5, 2016

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 #a=x# and #b=3# later...

First separate out the common factor #3x# of the terms to find:

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

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

The remaining quartic factor is a quadratic in #x^2#, which we can factor using the AC method with grouping. First find a pair of factors of #AC=2*9=18# which differ by #B=17#. The pair #18, 1# works. Then use this pair to split the middle term and factor by grouping:

#= 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)#