$$2x^3+45x^2-2052 = 0$$

$$ \begin{matrix}x_1 = 6 & x_2 = -\dfrac{ 57 }{ 4 }-\dfrac{ 3 \sqrt{ 57}}{ 4 } & x_3 = -\dfrac{ 57 }{ 4 }+\dfrac{ 3 \sqrt{ 57}}{ 4 } \end{matrix} $$

$ \color{blue}{ 2x^{3}+45x^{2}-2052 } $ 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 factors of the leading coefficient ( 2 ) are 1 2 .The factors of the constant term (-2052) are 1 2 3 4 6 9 12 18 19 27 36 38 54 57 76 108 114 171 228 342 513 684 1026 2052 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 19 }{ 1 } , ~ \pm \frac{ 19 }{ 2 } , ~ \pm \frac{ 27 }{ 1 } , ~ \pm \frac{ 27 }{ 2 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 2 } , ~ \pm \frac{ 38 }{ 1 } , ~ \pm \frac{ 38 }{ 2 } , ~ \pm \frac{ 54 }{ 1 } , ~ \pm \frac{ 54 }{ 2 } , ~ \pm \frac{ 57 }{ 1 } , ~ \pm \frac{ 57 }{ 2 } , ~ \pm \frac{ 76 }{ 1 } , ~ \pm \frac{ 76 }{ 2 } , ~ \pm \frac{ 108 }{ 1 } , ~ \pm \frac{ 108 }{ 2 } , ~ \pm \frac{ 114 }{ 1 } , ~ \pm \frac{ 114 }{ 2 } , ~ \pm \frac{ 171 }{ 1 } , ~ \pm \frac{ 171 }{ 2 } , ~ \pm \frac{ 228 }{ 1 } , ~ \pm \frac{ 228 }{ 2 } , ~ \pm \frac{ 342 }{ 1 } , ~ \pm \frac{ 342 }{ 2 } , ~ \pm \frac{ 513 }{ 1 } , ~ \pm \frac{ 513 }{ 2 } , ~ \pm \frac{ 684 }{ 1 } , ~ \pm \frac{ 684 }{ 2 } , ~ \pm \frac{ 1026 }{ 1 } , ~ \pm \frac{ 1026 }{ 2 } , ~ \pm \frac{ 2052 }{ 1 } , ~ \pm \frac{ 2052 }{ 2 } ~ $$

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(6) = 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 - 6} $

$$ \frac{ 2x^{3}+45x^{2}-2052 }{ \color{blue}{ x - 6 } } = 2x^{2}+57x+342 $$

Polynomial $ 2x^{2}+57x+342 $ can be used to find the remaining roots.

$ \color{blue}{ 2x^{2}+57x+342 } $ 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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