$$x^3-30000x-2000000 = 0$$

$$ \begin{matrix}x_1 = -100 & x_2 = 200 \\[1 em] \end{matrix} $$

$ \color{blue}{ x^{3}-30000x-2000000 } $ 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 (-2000000) are 1 2 4 5 8 10 16 20 25 32 40 50 64 80 100 125 128 160 200 250 320 400 500 625 640 800 1000 1250 1600 2000 2500 3125 3200 4000 5000 6250 8000 10000 12500 15625 16000 20000 25000 31250 40000 50000 62500 80000 100000 125000 200000 250000 400000 500000 1000000 2000000 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 64 }{ 1 } , ~ \pm \frac{ 80 }{ 1 } , ~ \pm \frac{ 100 }{ 1 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 128 }{ 1 } , ~ \pm \frac{ 160 }{ 1 } , ~ \pm \frac{ 200 }{ 1 } , ~ \pm \frac{ 250 }{ 1 } , ~ \pm \frac{ 320 }{ 1 } , ~ \pm \frac{ 400 }{ 1 } , ~ \pm \frac{ 500 }{ 1 } , ~ \pm \frac{ 625 }{ 1 } , ~ \pm \frac{ 640 }{ 1 } , ~ \pm \frac{ 800 }{ 1 } , ~ \pm \frac{ 1000 }{ 1 } , ~ \pm \frac{ 1250 }{ 1 } , ~ \pm \frac{ 1600 }{ 1 } , ~ \pm \frac{ 2000 }{ 1 } , ~ \pm \frac{ 2500 }{ 1 } , ~ \pm \frac{ 3125 }{ 1 } , ~ \pm \frac{ 3200 }{ 1 } , ~ \pm \frac{ 4000 }{ 1 } , ~ \pm \frac{ 5000 }{ 1 } , ~ \pm \frac{ 6250 }{ 1 } , ~ \pm \frac{ 8000 }{ 1 } , ~ \pm \frac{ 10000 }{ 1 } , ~ \pm \frac{ 12500 }{ 1 } , ~ \pm \frac{ 15625 }{ 1 } , ~ \pm \frac{ 16000 }{ 1 } , ~ \pm \frac{ 20000 }{ 1 } , ~ \pm \frac{ 25000 }{ 1 } , ~ \pm \frac{ 31250 }{ 1 } , ~ \pm \frac{ 40000 }{ 1 } , ~ \pm \frac{ 50000 }{ 1 } , ~ \pm \frac{ 62500 }{ 1 } , ~ \pm \frac{ 80000 }{ 1 } , ~ \pm \frac{ 100000 }{ 1 } , ~ \pm \frac{ 125000 }{ 1 } , ~ \pm \frac{ 200000 }{ 1 } , ~ \pm \frac{ 250000 }{ 1 } , ~ \pm \frac{ 400000 }{ 1 } , ~ \pm \frac{ 500000 }{ 1 } , ~ \pm \frac{ 1000000 }{ 1 } , ~ \pm \frac{ 2000000 }{ 1 } ~ $$

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(-100) = 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 + 100} $

$$ \frac{ x^{3}-30000x-2000000 }{ \color{blue}{ x + 100 } } = x^{2}-100x-20000 $$

Polynomial $ x^{2}-100x-20000 $ can be used to find the remaining roots.

$ \color{blue}{ x^{2}-100x-20000 } $ 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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