# How do you solve 6x^3+x^2-6x-1=0?

Feb 5, 2015

The answer is: $x = \pm 1 \mathmr{and} x = - \frac{1}{6}$.

We can recollection as a partial common factor between the first and the third, and we can notice that the second and the fourth are a difference of squares.

So:

$6 x \left({x}^{2} - 1\right) + \left({x}^{2} - 1\right) = 0 \Rightarrow \left({x}^{2} - 1\right) \left(6 x + 1\right) = 0 \Rightarrow$

$\left(x + 1\right) \left(x - 1\right) \left(6 x + 1\right) = 0$, and than:

$x = \pm 1 \mathmr{and} x = - \frac{1}{6}$.