$$\frac{5x+3}{x-2}+\frac{2x-1}{x+3}-\frac{16x+33}{x^2+x-6} = 1-2 \cdot \frac{x}{x+3}$$
$$ \begin{matrix}x_1 = -1 & x_2 = 2 & x_3 = -3 \end{matrix} $$
$ \color{blue}{ 8x^{4}-72x^{2}+32x+96 } $ is a polynomial of degree 4. 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 ( 8 ) are 1 2 4 8 .The factors of the constant term (96) are 1 2 3 4 6 8 12 16 24 32 48 96 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 8 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 8 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 4 }{ 2 } , ~ \pm \frac{ 4 }{ 4 } , ~ \pm \frac{ 4 }{ 8 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 6 }{ 8 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 8 }{ 2 } , ~ \pm \frac{ 8 }{ 4 } , ~ \pm \frac{ 8 }{ 8 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 12 }{ 2 } , ~ \pm \frac{ 12 }{ 4 } , ~ \pm \frac{ 12 }{ 8 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 16 }{ 2 } , ~ \pm \frac{ 16 }{ 4 } , ~ \pm \frac{ 16 }{ 8 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 24 }{ 2 } , ~ \pm \frac{ 24 }{ 4 } , ~ \pm \frac{ 24 }{ 8 } , ~ \pm \frac{ 32 }{ 1 } , ~ \pm \frac{ 32 }{ 2 } , ~ \pm \frac{ 32 }{ 4 } , ~ \pm \frac{ 32 }{ 8 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 48 }{ 2 } , ~ \pm \frac{ 48 }{ 4 } , ~ \pm \frac{ 48 }{ 8 } , ~ \pm \frac{ 96 }{ 1 } , ~ \pm \frac{ 96 }{ 2 } , ~ \pm \frac{ 96 }{ 4 } , ~ \pm \frac{ 96 }{ 8 } ~ $$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} $
$$ \frac{ 8x^{4}-72x^{2}+32x+96 }{ \color{blue}{ x + 1 } } = 8x^{3}-8x^{2}-64x+96 $$Polynomial $ 8x^{3}-8x^{2}-64x+96 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 8x^{3}-8x^{2}-64x+96 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.