$$\frac{1}{x+10}+\frac{20}{x^2-100} = 1$$
Answer
Explanation
$ \color{blue}{ -x^{3}-9x^{2}+120x+1100 } $ 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 (1100) are 1 2 4 5 10 11 20 22 25 44 50 55 100 110 220 275 550 1100 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 11 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 22 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 44 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } , ~ \pm \frac{ 55 }{ 1 } , ~ \pm \frac{ 100 }{ 1 } , ~ \pm \frac{ 110 }{ 1 } , ~ \pm \frac{ 220 }{ 1 } , ~ \pm \frac{ 275 }{ 1 } , ~ \pm \frac{ 550 }{ 1 } , ~ \pm \frac{ 1100 }{ 1 } ~ $$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(-10) = 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 + 10} $
$$ \frac{ -x^{3}-9x^{2}+120x+1100 }{ \color{blue}{ x + 10 } } = -x^{2}+x+110 $$Polynomial $ -x^{2}+x+110 $ can be used to find the remaining roots.
$ \color{blue}{ -x^{2}+x+110 } $ is a second degree polynomial. For a detailed answer how to find its roots you can use step-by-step quadratic equation solver.