# Is it possible to factor y=4x^3-13x-6 ? If so, what are the factors?

Mar 14, 2016

$4 {x}^{3} - 13 x - 6 = \left(x - 2\right) \left(2 x + 3\right) \left(2 x + 1\right)$

#### Explanation:

Since there is nothing obvious to factor out at the beginning, start by plugging in small numbers for $x$ to see if you can get the equation to equal $0$.

Plugging in $x = 2$ gives $0$, so $x - 2$ is a factor.

Now divide $4 {x}^{3} - 13 x - 6$ by $x - 2$ using polynomial long division or synthetic division. If you need help with this step, say something.

$\frac{4 {x}^{3} - 13 x - 6}{x - 2} = 4 {x}^{2} + 8 x + 3$

So, from here we know that

$4 {x}^{3} - 13 x - 6 = \left(x - 2\right) \left(4 {x}^{2} + 8 x + 3\right)$

So now all we need to do is factor $4 {x}^{2} + 8 x + 3$. Looking for a pair of factors of $12$ whose sum is $8$, we see that $2 , 6$ works:

$4 {x}^{2} + 8 x + 3$

$= 4 {x}^{2} + 2 x + 6 x + 3$

$= 2 x \left(2 x + 1\right) + 3 \left(2 x + 1\right)$

$= \left(2 x + 3\right) \left(2 x + 1\right)$

Thus,

$4 {x}^{3} - 13 x - 6 = \left(x - 2\right) \left(2 x + 3\right) \left(2 x + 1\right)$