$$\frac{2x^2-18}{3(x+1)\cdot(2-x)} = 0$$
$$ \begin{matrix}x_1 = -1 & x_2 = 2 & x_3 = 3 \\[1 em] x_4 = -3 & \\[1 em] \end{matrix} $$
$ \color{blue}{ 2x^{6}-4x^{5}-24x^{4}+44x^{3}+62x^{2}-72x-72 } $ is a polynomial of degree 6. 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 (-72) are 1 2 3 4 6 8 9 12 18 24 36 72 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 36 }{ 2 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 72 }{ 2 } ~ $$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(-1) = 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 + 1} $
$$ \frac{ 2x^{6}-4x^{5}-24x^{4}+44x^{3}+62x^{2}-72x-72 }{ \color{blue}{ x + 1 } } = 2x^{5}-6x^{4}-18x^{3}+62x^{2}-72 $$Polynomial $ 2x^{5}-6x^{4}-18x^{3}+62x^{2}-72 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 2x^{5}-6x^{4}-18x^{3}+62x^{2}-72 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.