## Precalculus (6th Edition) Blitzer

$-\dfrac{2}{x+3}+\dfrac{3x-5}{x^2+4}$
Here, we have $\dfrac{x^2+4x-23}{(x+3)(x^2+4)}=\dfrac{A}{x+3}+\dfrac{Bx+C}{x^2+4}$ $x^2+4x-23=(A+B)x^2+(3B+C)x+4A+3C$ Equating the coefficients, we get $A+B=1$ and $3B+C =4$ and $4A+3C=-23$ $\implies -12 =-9B-3C$ gives: $-12-23 =(-9B-3C) +(4A+3C)$ This gives: $4A-9B=-35$ ...(1) and $A+ B=1 \implies 9A+ 9B=9$ ...(2) From equations (1) and (2), we have $A=-2; B= 3; C= -5$ Hence, $\dfrac{x^2+4x-23}{(x+3)(x^2+4)}=-\dfrac{2}{x+3}+\dfrac{3x-5}{x^2+4}$