$$4x^3-10x^2 = 3000$$

$$ \begin{matrix}x_1 = 10 & x_2 = -\dfrac{ 15 }{ 4 }+\dfrac{ 5 \sqrt{ 39}}{ 4 }i & x_3 = -\dfrac{ 15 }{ 4 }-\dfrac{ 5 \sqrt{ 39}}{ 4 }i \end{matrix} $$

$ \color{blue}{ 4x^{3}-10x^{2}-3000 } $ 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 ( 4 ) are 1 2 4 .The factors of the constant term (-3000) are 1 2 3 4 5 6 8 10 12 15 20 24 25 30 40 50 60 75 100 120 125 150 200 250 300 375 500 600 750 1000 1500 3000 . Then the Rational Roots Tests yields the following possible solutions:

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

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(10) = 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 - 10} $

$$ \frac{ 4x^{3}-10x^{2}-3000 }{ \color{blue}{ x - 10 } } = 4x^{2}+30x+300 $$

Polynomial $ 4x^{2}+30x+300 $ can be used to find the remaining roots.

$ \color{blue}{ 4x^{2}+30x+300 } $ 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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