$$-x^3+136x^2-2559x+11160 = 0$$

$$ \begin{matrix}x_1 = 15 & x_2 = \dfrac{ 121 }{ 2 }-\dfrac{\sqrt{ 11665 }}{ 2 } & x_3 = \dfrac{ 121 }{ 2 }+\dfrac{\sqrt{ 11665 }}{ 2 } \end{matrix} $$

$ \color{blue}{ -x^{3}+136x^{2}-2559x+11160 } $ 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 (11160) are 1 2 3 4 5 6 8 9 10 12 15 18 20 24 30 31 36 40 45 60 62 72 90 93 120 124 155 180 186 248 279 310 360 372 465 558 620 744 930 1116 1240 1395 1860 2232 2790 3720 5580 11160 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 31 }{ 1 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 62 }{ 1 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } , ~ \pm \frac{ 93 }{ 1 } , ~ \pm \frac{ 120 }{ 1 } , ~ \pm \frac{ 124 }{ 1 } , ~ \pm \frac{ 155 }{ 1 } , ~ \pm \frac{ 180 }{ 1 } , ~ \pm \frac{ 186 }{ 1 } , ~ \pm \frac{ 248 }{ 1 } , ~ \pm \frac{ 279 }{ 1 } , ~ \pm \frac{ 310 }{ 1 } , ~ \pm \frac{ 360 }{ 1 } , ~ \pm \frac{ 372 }{ 1 } , ~ \pm \frac{ 465 }{ 1 } , ~ \pm \frac{ 558 }{ 1 } , ~ \pm \frac{ 620 }{ 1 } , ~ \pm \frac{ 744 }{ 1 } , ~ \pm \frac{ 930 }{ 1 } , ~ \pm \frac{ 1116 }{ 1 } , ~ \pm \frac{ 1240 }{ 1 } , ~ \pm \frac{ 1395 }{ 1 } , ~ \pm \frac{ 1860 }{ 1 } , ~ \pm \frac{ 2232 }{ 1 } , ~ \pm \frac{ 2790 }{ 1 } , ~ \pm \frac{ 3720 }{ 1 } , ~ \pm \frac{ 5580 }{ 1 } , ~ \pm \frac{ 11160 }{ 1 } ~ $$

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(15) = 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 - 15} $

$$ \frac{ -x^{3}+136x^{2}-2559x+11160 }{ \color{blue}{ x - 15 } } = -x^{2}+121x-744 $$

Polynomial $ -x^{2}+121x-744 $ can be used to find the remaining roots.

$ \color{blue}{ -x^{2}+121x-744 } $ 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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