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

Feb 5, 2016

$6 {x}^{5} - 51 {x}^{3} - 27 x$

$= 3 x \left(2 {x}^{2} + 1\right) \left(x - 3\right) \left(x + 3\right)$

$= 3 x \left(\sqrt{2} x - i\right) \left(\sqrt{2} x + i\right) \left(x - 3\right) \left(x + 3\right)$

#### Explanation:

We will use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = x$ and $b = 3$ later...

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

$6 {x}^{5} - 51 {x}^{3} - 27 x$

$= 3 x \left(2 {x}^{4} - 17 {x}^{2} - 9\right)$

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 $A C = 2 \cdot 9 = 18$ which differ by $B = 17$. The pair $18 , 1$ works. Then use this pair to split the middle term and factor by grouping:

$= 3 x \left(2 {x}^{4} - 18 {x}^{2} + {x}^{2} - 9\right)$

$= 3 x \left(\left(2 {x}^{4} - 18 {x}^{2}\right) + \left({x}^{2} - 9\right)\right)$

$= 3 x \left(2 {x}^{2} \left({x}^{2} - 9\right) + 1 \left({x}^{2} - 9\right)\right)$

$= 3 x \left(2 {x}^{2} + 1\right) \left({x}^{2} - 9\right)$

$= 3 x \left(2 {x}^{2} + 1\right) \left({x}^{2} - {3}^{2}\right)$

$= 3 x \left(2 {x}^{2} + 1\right) \left(x - 3\right) \left(x + 3\right)$

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

$= 3 x \left(\sqrt{2} x - i\right) \left(\sqrt{2} x + i\right) \left(x - 3\right) \left(x + 3\right)$