$$\frac{(x+3)(2x-1)(x-6)}{(3x+5)(4x+1)(x-1)} = 0$$
$$ \begin{matrix}x_1 = 1 & x_2 = \dfrac{ 1 }{ 2 } & x_3 = -\dfrac{ 1 }{ 4 } \\[1 em] x_4 = -3 & x_5 = 6 \\[1 em] \end{matrix} $$
$ \color{blue}{ 32x^{7}-160x^{6}-358x^{5}+1085x^{4}-505x^{3}-187x^{2}+75x+18 } $ is a polynomial of degree 7. 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 ( 32 ) are 1 2 4 8 16 32 .The factors of the constant term (18) are 1 2 3 6 9 18 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 2 } , ~ \pm \frac{ 1 }{ 4 } , ~ \pm \frac{ 1 }{ 8 } , ~ \pm \frac{ 1 }{ 16 } , ~ \pm \frac{ 1 }{ 32 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 2 } , ~ \pm \frac{ 2 }{ 4 } , ~ \pm \frac{ 2 }{ 8 } , ~ \pm \frac{ 2 }{ 16 } , ~ \pm \frac{ 2 }{ 32 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 2 } , ~ \pm \frac{ 3 }{ 4 } , ~ \pm \frac{ 3 }{ 8 } , ~ \pm \frac{ 3 }{ 16 } , ~ \pm \frac{ 3 }{ 32 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 2 } , ~ \pm \frac{ 6 }{ 4 } , ~ \pm \frac{ 6 }{ 8 } , ~ \pm \frac{ 6 }{ 16 } , ~ \pm \frac{ 6 }{ 32 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 9 }{ 2 } , ~ \pm \frac{ 9 }{ 4 } , ~ \pm \frac{ 9 }{ 8 } , ~ \pm \frac{ 9 }{ 16 } , ~ \pm \frac{ 9 }{ 32 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 18 }{ 2 } , ~ \pm \frac{ 18 }{ 4 } , ~ \pm \frac{ 18 }{ 8 } , ~ \pm \frac{ 18 }{ 16 } , ~ \pm \frac{ 18 }{ 32 } ~ $$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} $
$$ \frac{ 32x^{7}-160x^{6}-358x^{5}+1085x^{4}-505x^{3}-187x^{2}+75x+18 }{ \color{blue}{ x - 1 } } = 32x^{6}-128x^{5}-486x^{4}+599x^{3}+94x^{2}-93x-18 $$Polynomial $ 32x^{6}-128x^{5}-486x^{4}+599x^{3}+94x^{2}-93x-18 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ 32x^{6}-128x^{5}-486x^{4}+599x^{3}+94x^{2}-93x-18 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.