$$-10x^3+25x^2 = 1875$$

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

$ \color{blue}{ -10x^{3}+25x^{2}-1875 } $ 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 ( -10 ) are 1 2 5 10 .The factors of the constant term (-1875) are 1 3 5 15 25 75 125 375 625 1875 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 5 } , ~ \pm \frac{ 1 }{ 10 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 5 } , ~ \pm \frac{ 3 }{ 10 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 5 } , ~ \pm \frac{ 5 }{ 10 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 5 } , ~ \pm \frac{ 15 }{ 10 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 25 }{ 2 } , ~ \pm \frac{ 25 }{ 5 } , ~ \pm \frac{ 25 }{ 10 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 75 }{ 2 } , ~ \pm \frac{ 75 }{ 5 } , ~ \pm \frac{ 75 }{ 10 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 125 }{ 2 } , ~ \pm \frac{ 125 }{ 5 } , ~ \pm \frac{ 125 }{ 10 } , ~ \pm \frac{ 375 }{ 1 } , ~ \pm \frac{ 375 }{ 2 } , ~ \pm \frac{ 375 }{ 5 } , ~ \pm \frac{ 375 }{ 10 } , ~ \pm \frac{ 625 }{ 1 } , ~ \pm \frac{ 625 }{ 2 } , ~ \pm \frac{ 625 }{ 5 } , ~ \pm \frac{ 625 }{ 10 } , ~ \pm \frac{ 1875 }{ 1 } , ~ \pm \frac{ 1875 }{ 2 } , ~ \pm \frac{ 1875 }{ 5 } , ~ \pm \frac{ 1875 }{ 10 } ~ $$

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(-5) = 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 + 5} $

$$ \frac{ -10x^{3}+25x^{2}-1875 }{ \color{blue}{ x + 5 } } = -10x^{2}+75x-375 $$

Polynomial $ -10x^{2}+75x-375 $ can be used to find the remaining roots.

$ \color{blue}{ -10x^{2}+75x-375 } $ 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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