$$-x^3-28x^2+1175x+3750 = 0$$
Answer
Explanation
$ \color{blue}{ -x^{3}-28x^{2}+1175x+3750 } $ 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 (3750) are 1 2 3 5 6 10 15 25 30 50 75 125 150 250 375 625 750 1250 1875 3750 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 150 }{ 1 } , ~ \pm \frac{ 250 }{ 1 } , ~ \pm \frac{ 375 }{ 1 } , ~ \pm \frac{ 625 }{ 1 } , ~ \pm \frac{ 750 }{ 1 } , ~ \pm \frac{ 1250 }{ 1 } , ~ \pm \frac{ 1875 }{ 1 } , ~ \pm \frac{ 3750 }{ 1 } ~ $$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(-3) = 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 + 3} $
$$ \frac{ -x^{3}-28x^{2}+1175x+3750 }{ \color{blue}{ x + 3 } } = -x^{2}-25x+1250 $$Polynomial $ -x^{2}-25x+1250 $ can be used to find the remaining roots.
$ \color{blue}{ -x^{2}-25x+1250 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.