Polynomial equation solver

Solve equation$$-x^3+110x^2+1725x-236250 = 0$$

solution

$$ \begin{matrix}x_1 = -45 & x_2 = 50 & x_3 = 105 \end{matrix} $$

explanation

$ \color{blue}{ -x^{3}+110x^{2}+1725x-236250 } $ 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 (-236250) are 1 2 3 5 6 7 9 10 14 15 18 21 25 27 30 35 42 45 50 54 63 70 75 90 105 125 126 135 150 175 189 210 225 250 270 315 350 375 378 450 525 625 630 675 750 875 945 1050 1125 1250 1350 1575 1750 1875 1890 2250 2625 3150 3375 3750 4375 4725 5250 5625 6750 7875 8750 9450 11250 13125 15750 16875 23625 26250 33750 39375 47250 78750 118125 236250 . 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{ 7 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 35 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 70 }{ 1 } , ~ \pm \frac{ 75 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } , ~ \pm \frac{ 105 }{ 1 } , ~ \pm \frac{ 125 }{ 1 } , ~ \pm \frac{ 126 }{ 1 } , ~ \pm \frac{ 135 }{ 1 } , ~ \pm \frac{ 150 }{ 1 } , ~ \pm \frac{ 175 }{ 1 } , ~ \pm \frac{ 189 }{ 1 } , ~ \pm \frac{ 210 }{ 1 } , ~ \pm \frac{ 225 }{ 1 } , ~ \pm \frac{ 250 }{ 1 } , ~ \pm \frac{ 270 }{ 1 } , ~ \pm \frac{ 315 }{ 1 } , ~ \pm \frac{ 350 }{ 1 } , ~ \pm \frac{ 375 }{ 1 } , ~ \pm \frac{ 378 }{ 1 } , ~ \pm \frac{ 450 }{ 1 } , ~ \pm \frac{ 525 }{ 1 } , ~ \pm \frac{ 625 }{ 1 } , ~ \pm \frac{ 630 }{ 1 } , ~ \pm \frac{ 675 }{ 1 } , ~ \pm \frac{ 750 }{ 1 } , ~ \pm \frac{ 875 }{ 1 } , ~ \pm \frac{ 945 }{ 1 } , ~ \pm \frac{ 1050 }{ 1 } , ~ \pm \frac{ 1125 }{ 1 } , ~ \pm \frac{ 1250 }{ 1 } , ~ \pm \frac{ 1350 }{ 1 } , ~ \pm \frac{ 1575 }{ 1 } , ~ \pm \frac{ 1750 }{ 1 } , ~ \pm \frac{ 1875 }{ 1 } , ~ \pm \frac{ 1890 }{ 1 } , ~ \pm \frac{ 2250 }{ 1 } , ~ \pm \frac{ 2625 }{ 1 } , ~ \pm \frac{ 3150 }{ 1 } , ~ \pm \frac{ 3375 }{ 1 } , ~ \pm \frac{ 3750 }{ 1 } , ~ \pm \frac{ 4375 }{ 1 } , ~ \pm \frac{ 4725 }{ 1 } , ~ \pm \frac{ 5250 }{ 1 } , ~ \pm \frac{ 5625 }{ 1 } , ~ \pm \frac{ 6750 }{ 1 } , ~ \pm \frac{ 7875 }{ 1 } , ~ \pm \frac{ 8750 }{ 1 } , ~ \pm \frac{ 9450 }{ 1 } , ~ \pm \frac{ 11250 }{ 1 } , ~ \pm \frac{ 13125 }{ 1 } , ~ \pm \frac{ 15750 }{ 1 } , ~ \pm \frac{ 16875 }{ 1 } , ~ \pm \frac{ 23625 }{ 1 } , ~ \pm \frac{ 26250 }{ 1 } , ~ \pm \frac{ 33750 }{ 1 } , ~ \pm \frac{ 39375 }{ 1 } , ~ \pm \frac{ 47250 }{ 1 } , ~ \pm \frac{ 78750 }{ 1 } , ~ \pm \frac{ 118125 }{ 1 } , ~ \pm \frac{ 236250 }{ 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(-45) = 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 + 45} $

$$ \frac{ -x^{3}+110x^{2}+1725x-236250 }{ \color{blue}{ x + 45 } } = -x^{2}+155x-5250 $$

Polynomial $ -x^{2}+155x-5250 $ can be used to find the remaining roots.

$ \color{blue}{ -x^{2}+155x-5250 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.