$ \color{blue}{ -x^{3}-50x^{2}+57x+20394 } $ 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 factor of the leading coefficient ( -1 ) is 1 .The factors of the constant term (20394) are 1 2 3 6 9 11 18 22 33 66 99 103 198 206 309 618 927 1133 1854 2266 3399 6798 10197 20394 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 11 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 22 }{ 1 } , ~ \pm \frac{ 33 }{ 1 } , ~ \pm \frac{ 66 }{ 1 } , ~ \pm \frac{ 99 }{ 1 } , ~ \pm \frac{ 103 }{ 1 } , ~ \pm \frac{ 198 }{ 1 } , ~ \pm \frac{ 206 }{ 1 } , ~ \pm \frac{ 309 }{ 1 } , ~ \pm \frac{ 618 }{ 1 } , ~ \pm \frac{ 927 }{ 1 } , ~ \pm \frac{ 1133 }{ 1 } , ~ \pm \frac{ 1854 }{ 1 } , ~ \pm \frac{ 2266 }{ 1 } , ~ \pm \frac{ 3399 }{ 1 } , ~ \pm \frac{ 6798 }{ 1 } , ~ \pm \frac{ 10197 }{ 1 } , ~ \pm \frac{ 20394 }{ 1 } ~ $$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(-33) = 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 + 33} $
$$ \frac{ -x^{3}-50x^{2}+57x+20394 }{ \color{blue}{ x + 33 } } = -x^{2}-17x+618 $$Polynomial $ -x^{2}-17x+618 $ can be used to find the remaining roots.
$ \color{blue}{ -x^{2}-17x+618 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.