$ \color{blue}{ x^{3}-x-5814 } $ 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 (-5814) are 1 2 3 6 9 17 18 19 34 38 51 57 102 114 153 171 306 323 342 646 969 1938 2907 5814 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 17 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 19 }{ 1 } , ~ \pm \frac{ 34 }{ 1 } , ~ \pm \frac{ 38 }{ 1 } , ~ \pm \frac{ 51 }{ 1 } , ~ \pm \frac{ 57 }{ 1 } , ~ \pm \frac{ 102 }{ 1 } , ~ \pm \frac{ 114 }{ 1 } , ~ \pm \frac{ 153 }{ 1 } , ~ \pm \frac{ 171 }{ 1 } , ~ \pm \frac{ 306 }{ 1 } , ~ \pm \frac{ 323 }{ 1 } , ~ \pm \frac{ 342 }{ 1 } , ~ \pm \frac{ 646 }{ 1 } , ~ \pm \frac{ 969 }{ 1 } , ~ \pm \frac{ 1938 }{ 1 } , ~ \pm \frac{ 2907 }{ 1 } , ~ \pm \frac{ 5814 }{ 1 } ~ $$

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(18) = 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 - 18} $

$$ \frac{ x^{3}-x-5814 }{ \color{blue}{ x - 18 } } = x^{2}+18x+323 $$

Polynomial $ x^{2}+18x+323 $ can be used to find the remaining roots.

$ \color{blue}{ x^{2}+18x+323 } $ 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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