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

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Answer 1 · 2016-03-13T03:48:26

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

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

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

Now we can find the zeros:

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

If there are any further questions, let me know

How do you find all the zeros of g(x)=2x3+x2−6x−3g(x) = 2x^3 + x^2 - 6x - 3?