Question

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

Answer 1

`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)`