Polynomial equation solver

Solve equation$$-x^3+25x^2-200x+500 = 0$$

solution

$$ \begin{matrix}x_1 = 5 & x_2 = 10 \\[1 em] \end{matrix} $$

explanation

$ \color{blue}{ -x^{3}+25x^{2}-200x+500 } $ 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 (500) are 1 2 4 5 10 20 25 50 100 125 250 500 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 100 }{ 1 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 250 }{ 1 } , ~ \pm \frac{ 500 }{ 1 } ~ $$

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(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} $

$$ \frac{ -x^{3}+25x^{2}-200x+500 }{ \color{blue}{ x - 5 } } = -x^{2}+20x-100 $$

Polynomial $ -x^{2}+20x-100 $ can be used to find the remaining roots.

$ \color{blue}{ -x^{2}+20x-100 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.