$$20x^3+5x^2-12x-3 = 0$$

$$ \begin{matrix}x_1 = -\dfrac{ 1 }{ 4 } & x_2 = - \dfrac{\sqrt{ 15 }}{ 5 } & x_3 = \dfrac{\sqrt{ 15 }}{ 5 } \end{matrix} $$

$ \color{blue}{ 20x^{3}+5x^{2}-12x-3 } $ is a polynomial of degree 3. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factors of the leading coefficient ( 20 ) are 1 2 4 5 10 20 .The factors of the constant term (-3) are 1 3 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 1 }{ 20 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 10 } , ~ \pm \frac{ 3 }{ 20 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(-\frac{ 1 }{ 4 }) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{ 4 x + 1 } $

$$ \frac{ 20x^{3}+5x^{2}-12x-3 }{ \color{blue}{ 4x + 1 } } = 5x^{2}-3 $$

Polynomial $ 5x^{2}-3 $ can be used to find the remaining roots.

$ \color{blue}{ 5x^{2}-3 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

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