# How do you find all the zeros of g(x) = 2x^3 + x^2 - 6x - 3?

Mar 13, 2016

$x = \sqrt{3}$ and $x = - \frac{1}{2}$

#### Explanation:

First of all, let us factorize your equation 2x^3+x^2−6x−3. We can display the equation in the following format:

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

Now we can find the zeros:

$x = \sqrt{3}$ and $x = - \frac{1}{2}$

If there are any further questions, let me know