$$\frac{1}{x+6}+\frac{12}{x^2-36} = 1$$
Answer
Explanation
$ \color{blue}{ -x^{3}-5x^{2}+48x+252 } $ 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 (252) are 1 2 3 4 6 7 9 12 14 18 21 28 36 42 63 84 126 252 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 7 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 14 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 21 }{ 1 } , ~ \pm \frac{ 28 }{ 1 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 42 }{ 1 } , ~ \pm \frac{ 63 }{ 1 } , ~ \pm \frac{ 84 }{ 1 } , ~ \pm \frac{ 126 }{ 1 } , ~ \pm \frac{ 252 }{ 1 } ~ $$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(-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} $
$$ \frac{ -x^{3}-5x^{2}+48x+252 }{ \color{blue}{ x + 6 } } = -x^{2}+x+42 $$Polynomial $ -x^{2}+x+42 $ can be used to find the remaining roots.
$ \color{blue}{ -x^{2}+x+42 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.