$$4x^3-10x^2 = 1125$$

$$ \begin{matrix}x_1 = \dfrac{ 15 }{ 2 } & x_2 = -\dfrac{ 5 }{ 2 }+\dfrac{ 5 \sqrt{ 5}}{ 2 }i & x_3 = -\dfrac{ 5 }{ 2 }-\dfrac{ 5 \sqrt{ 5}}{ 2 }i \end{matrix} $$

$ \color{blue}{ 4x^{3}-10x^{2}-1125 } $ 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 (-1125) are 1 3 5 9 15 25 45 75 125 225 375 1125 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 5 }{ 2 } , ~ \pm \frac{ 5 }{ 4 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 4 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 15 }{ 2 } , ~ \pm \frac{ 15 }{ 4 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 25 }{ 2 } , ~ \pm \frac{ 25 }{ 4 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 45 }{ 2 } , ~ \pm \frac{ 45 }{ 4 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 75 }{ 2 } , ~ \pm \frac{ 75 }{ 4 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 125 }{ 2 } , ~ \pm \frac{ 125 }{ 4 } , ~ \pm \frac{ 225 }{ 1 } , ~ \pm \frac{ 225 }{ 2 } , ~ \pm \frac{ 225 }{ 4 } , ~ \pm \frac{ 375 }{ 1 } , ~ \pm \frac{ 375 }{ 2 } , ~ \pm \frac{ 375 }{ 4 } , ~ \pm \frac{ 1125 }{ 1 } , ~ \pm \frac{ 1125 }{ 2 } , ~ \pm \frac{ 1125 }{ 4 } ~ $$

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(\frac{ 15 }{ 2 }) = 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}{ 2 x - 15 } $

$$ \frac{ 4x^{3}-10x^{2}-1125 }{ \color{blue}{ 2x - 15 } } = 2x^{2}+10x+75 $$

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

$ \color{blue}{ 2x^{2}+10x+75 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.

Polynomial Equations Solver