$$\frac{x}{x+5} = \frac{1}{x+2}-3\frac{x}{x^2+7x+10}$$
Answer
Explanation
$ \color{blue}{ x^{4}+11x^{3}+33x^{2}+5x-50 } $ 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 factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (-50) are 1 2 5 10 25 50 . Then the Rational Roots Tests yields the following possible solutions:
$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 25 }{ 1 } , ~ \pm \frac{ 50 }{ 1 } ~ $$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(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} $
$$ \frac{ x^{4}+11x^{3}+33x^{2}+5x-50 }{ \color{blue}{ x - 1 } } = x^{3}+12x^{2}+45x+50 $$Polynomial $ x^{3}+12x^{2}+45x+50 $ can be used to find the remaining roots.
Use the same procedure to find roots of $ x^{3}+12x^{2}+45x+50 $
When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.