How to find the sum of the roots of the following equation: $\frac{(x+3)^2}{(x+1)^2} + 3\frac{(x+3)}{x+2} - 10\frac{(x+1)^2}{(x+2)^2} = 0$
I have tried a lot of substitutions, but I always end up with 4th degree polynomial. Is there a way to simplify it?
The sufficient conditions for the equation to be true are $x\neq -1$, and $ x\neq -2$. Expand, and simplify the equation above, and you'll get something like $$-\frac{6x^4+9x^3-28x^2-71x-44}{(x+1)^2(x+2)^2}=0.$$ Since $x\neq -1$, and $ x\neq -2$, we can get rid of the denominator and obtain $$6x^4+9x^3-28x^2-71x-44=0.$$ Now, the trick here is to apply Vieta's formulas and obtain the sum of all complex roots to be $$r_1+r_2+r_3+r_4=-\frac{a_{n-1}}{a_n}=-\frac{9}{6}=-\frac{3}{2}.$$ If you want the sum of real roots only then $6x^4+9x^3-28x^2-71x-44$ can be factorized as $$(x^2-x-4)(6x^2+15x+11),$$ which gives the sum of all real roots to be $1$.