How do you factor 6x ^ { 3} + 2x ^ { 2} - 3x - 1?

Jul 30, 2017

In terms of a factorization involving integers, $6 {x}^{3} + 2 {x}^{2} - 3 x - 1 = \left(3 x + 1\right) \left(2 {x}^{2} - 1\right)$. It can be factored in terms of linear factors if you allow irrational numbers: $6 {x}^{3} + 2 {x}^{2} - 3 x - 1 = 2 \left(3 x + 1\right) \left(x - \frac{1}{\sqrt{2}}\right) \left(x + \frac{1}{\sqrt{2}}\right)$.

Explanation:

Through graphing the function $f \left(x\right) = 6 {x}^{3} + 2 {x}^{2} - 3 x - 1$ and the Rational Root Theorem , you'll see that it looks like $x = - \frac{1}{3}$ is a rational root of $f \left(x\right)$, which you can confirm like this:

$f \left(- \frac{1}{3}\right) = 6 \cdot {\left(- \frac{1}{3}\right)}^{3} + 2 {\left(- \frac{1}{3}\right)}^{2} - 3 \left(- \frac{1}{3}\right) - 1$

$= - \frac{6}{27} + \frac{2}{9} + 1 - 1 = - \frac{2}{9} + \frac{2}{9} = 0$.

Then, after using either long division or synthetic division , you will find that $6 {x}^{3} + 2 {x}^{2} - 3 x - 1 = \left(3 x + 1\right) \left(2 {x}^{2} - 1\right)$.

If you want to factor in terms of irrational numbers as done above, note that $\setminus \pm \frac{1}{\sqrt{2}}$ are the roots of $2 {x}^{2} - 1$.