$$(s+30)(s^2-20s+200) = 0$$

$$ \begin{matrix}s_1 = -30 & s_2 = 10+10i & s_3 = 10-10i \end{matrix} $$

$ \color{blue}{ x^{3}+10x^{2}-400x+6000 } $ 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 (6000) are 1 2 3 4 5 6 8 10 12 15 16 20 24 25 30 40 48 50 60 75 80 100 120 125 150 200 240 250 300 375 400 500 600 750 1000 1200 1500 2000 3000 6000 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 80 }{ 1 } , ~ \pm \frac{ 100 }{ 1 } , ~ \pm \frac{ 120 }{ 1 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 150 }{ 1 } , ~ \pm \frac{ 200 }{ 1 } , ~ \pm \frac{ 240 }{ 1 } , ~ \pm \frac{ 250 }{ 1 } , ~ \pm \frac{ 300 }{ 1 } , ~ \pm \frac{ 375 }{ 1 } , ~ \pm \frac{ 400 }{ 1 } , ~ \pm \frac{ 500 }{ 1 } , ~ \pm \frac{ 600 }{ 1 } , ~ \pm \frac{ 750 }{ 1 } , ~ \pm \frac{ 1000 }{ 1 } , ~ \pm \frac{ 1200 }{ 1 } , ~ \pm \frac{ 1500 }{ 1 } , ~ \pm \frac{ 2000 }{ 1 } , ~ \pm \frac{ 3000 }{ 1 } , ~ \pm \frac{ 6000 }{ 1 } ~ $$

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(-30) = 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 + 30} $

$$ \frac{ x^{3}+10x^{2}-400x+6000 }{ \color{blue}{ x + 30 } } = x^{2}-20x+200 $$

Polynomial $ x^{2}-20x+200 $ can be used to find the remaining roots.

$ \color{blue}{ x^{2}-20x+200 } $ 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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