Partial Fraction problem expansion dead end

Question

I am having a problem solving this partial fraction,

$$\frac{18+21x-x^2}{(x-5)(x+2)^2}$$

Which Solves to:

$$\frac{2}{x-5}-\frac3{x+2}+\frac4{(x+2)^2}$$

However no matter how hard I try I always end here,

Answer 1

Consider \begin{align*}\frac{18+21x-x^2}{(x-5)(x+2)^2}&=\frac{A}{x-5}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\\ \iff 18+21x-x^2&=A\cdot(x+2)^2+B\cdot(x-5)(x+2)+C\cdot (x-5)\\ \iff -x^2+21x+18&= x^2\cdot (A+B)+x\cdot(4A-3B+C)+(4A-10B-5C) \end{align*}

Since the coefficients of the polynomials must be identic \begin{cases}-1=A+B\\ 21=4A-3B+C\\ 18=4A-10B-5C\end{cases} Now it's really a matter of solving the system. You might want to use for instance the Gaussian Algorithm. Notwithstanding, I believe that this can be solved faster: Observe that after multiplying the second equation with $5$ and adding it to the third, you obtain $$123=24A-25B$$ Combined with the first equation $-1=A+B$, you have $A=2, B=-3$. Substitute these values in another equation containing $C$ in order to obtain $C=4$.

Answer 2

$$\frac{18+21x-x^2}{(x-5)(x+2)^2}=\frac{-x^2-4x-4+25x+22}{(x-5)(x+2)^2}=$$ $$=-\frac{1}{x-5}+\frac{25x+22}{(x-5)(x+2)^2}=-\frac{1}{x-5}+\frac{25x-125+147}{(x-5)(x+2)^2}=$$ $$=-\frac{1}{x-5}+\frac{25}{(x+2)^2}+\frac{147}{(x-5)(x+2)^2}=$$ $$=-\frac{1}{x-5}+\frac{25}{(x+2)^2}+\frac{21}{x+2}\left(\frac{1}{x-5}-\frac{1}{x+2}\right)=$$

$$=-\frac{1}{x-5}+\frac{4}{(x+2)^2}+3\left(\frac{1}{x-5}-\frac{1}{x+2}\right)=$$

$$=\frac{2}{x-5}+\frac{4}{(x+2)^2}-\frac{3}{x+2}.$$

Answer 3

$$\lim_{x\to5}(x-5)\left(\frac a{x-5}+\frac b{x+2}+\frac c{(x+2)^2}\right)=a=\lim_{x\to5}\frac{18+21x-x^2}{(x+2)^2}=\frac{98}{49}=2.$$

$$\lim_{x\to-2}(x+2)^2\left(\frac a{x-5}+\frac b{x+2}+\frac c{(x+2)^2}\right)=c=\lim_{x\to-2}\frac{18+21x-x^2}{x-5}=\frac{-28}{-7}=4.$$

Then

$$\frac{18+21x-x^2}{(x-5)(x+2)^2}-\frac2{x-5}-\frac4{(x+2)^2}=-\frac3{x+2}.$$