$$9x^4-24x^3-16x^2+20x+3 = 0$$

$$ \begin{matrix}x_1 = -1 & x_2 = 3 & x_3 = \dfrac{ 1 }{ 3 }-\dfrac{\sqrt{ 2 }}{ 3 } \\[1 em] x_4 = \dfrac{ 1 }{ 3 }+\dfrac{\sqrt{ 2 }}{ 3 } & \\[1 em] \end{matrix} $$

$ \color{blue}{ 9x^{4}-24x^{3}-16x^{2}+20x+3 } $ is a polynomial of degree 4. 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 ( 9 ) are 1 3 9 .The factors of the constant term (3) are 1 3 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 1 }{ 9 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 3 }{ 9 } ~ $$

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(-1) = 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}{x + 1} $

$$ \frac{ 9x^{4}-24x^{3}-16x^{2}+20x+3 }{ \color{blue}{ x + 1 } } = 9x^{3}-33x^{2}+17x+3 $$

Polynomial $ 9x^{3}-33x^{2}+17x+3 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ 9x^{3}-33x^{2}+17x+3 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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