yee, partial fractions
remember some things to solve:
[tex]\frac{px+q}{(x+a)(x+b)}=\frac{A}{x+a}+\frac{B}{x+b}[/tex]
[tex]\frac{px+q}{(x+a)^2}=\frac{A}{x+a}+\frac{B}{(x+a)^2}[/tex]
[tex]\frac{px^2-qx+r}{(x+a)(x^2+bx+c)}=\frac{A}{x+a}+\frac{Bx+C}{x^2+bx+c}[/tex]
for your problem
[tex]\frac{x^2+8}{x^2-5x+6}[/tex]
by long division, we get [tex]1+\frac{5x+2}{x^2-5x+6}[/tex]
let's consider the 2nd part
[tex]\frac{5x+2}{x^2-5x+6}[/tex]
factor
[tex]\frac{5x+2}{(x-3)(x-2)}[/tex]
apply thing
[tex]\frac{5x+2}{(x-3)(x-2)}=\frac{A}{x-3}+\frac{B}{x-2}[/tex]
solve for A and B
cross multiply
5x+2=A(x-2)+B(x-3)
5x+2=Ax-2A+Bx-3B
match powers
5x=Ax+Bx
5=A+B
2=-2A-3B
adding twice of previous equation, we get
12=-B
B=12
5=A+B
5=A+12
-7=A
so expressed as partial fractions, it is[tex]\frac{x^2+8}{x^2-5x+6}=1-\frac{7}{x-3}+\frac{12}{x-2}[/tex]
