\(\frac{3 x^{2}+1}{x^{2}-6 x+8}\) is equal to

Given, \(\frac{3 x^{2}+1}{x^{2}-6 x+8}\)
On dividing, we get
\(\quad \frac{3 x^{2}+1}{x^{2}-6 x+8}=3+\frac{18 x-23}{x^{2}-6 x+8}\) Now, \(\frac{18 x-23}{(x-2)(x-4)}=\frac{A}{x-2}+\frac{B}{x-4}\) \(\Rightarrow \quad 18 x-23=A(x-4)+B(x-2)\) \(\Rightarrow \quad 18 x-23=(A+B) x-4 A-2 B\) Equating the coefficient of \(x\) and constant term, we get \(\quad A+B=18\) \(\quad-4 A-2 B=-23\) get
\(A+B=18\) \(-4 A-2 B=-23\) On solving these equations, we get \[ A=-\frac{13}{2}, B=\frac{49}{2} \] \(\therefore \frac{18 x-23}{(x-2)(x-4)}=-\frac{13}{2(x-2)}+\frac{49}{2(x-4)}\)
Then, from Eq. (i), we get
\[
\frac{3 x^{2}+1}{x^{2}-6 x+8}=3-\frac{13}{2(x-2)}+\frac{49}{2(x-4)}
\]