# How do you factor 6x^6-3x^4-9x^2?

Jul 26, 2016

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

#### Explanation:

The Expression$= 6 {x}^{6} - 3 {x}^{4} - 9 {x}^{2}$

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

Now, to factorise $= 2 {x}^{4} - {x}^{2} - 3$, let us put ${x}^{2} = t$ so that the

poly. will become,

$2 {t}^{2} - t - 3$

$= 2 {t}^{2} - 3 t + 2 t - 3$

$= t \left(2 t - 3\right) + 1 \left(2 t - 3\right)$

$= \left(2 t - 3\right) \left(t + 1\right)$

$= \left(2 {x}^{2} - 3\right) \left({x}^{2} + 1\right)$..................[ as, $t = {x}^{2}$]

Therefore, the Exp.$= 3 {x}^{2} \left(2 {x}^{2} - 3\right) \left({x}^{2} + 1\right)$.