Solving for the value of $a,b$ in $f(x)=(ax+b)(x^5+1)-(5x+1)$ s.t. $(x^2+1)|f(x)$ 
$Q.$ If  $f(x)=(ax+b)(x^5+1)-5x-1$ is divisible by $x^2+1$ . Then the value of $2a+3b$ $?$

MY APPROACH : We have , $(x^2+1)|f(x)$ then $(x-i)|f(x)$ and $(x+i)|f(x)$ .
So by Factor Theorem we have $f(-i)=0$ and $f(i)=0$ $$f(i)=-a+bi+(a-5)i+b-1=0$$ $$f(-i)=-a-bi-(a-5)i+b-1=0$$
By this I concluded that $a-b=-1$ . But this is not enough information to solve the problem .

We have
$$\begin{align}&ax^6+bx^5+x(a-5)+(b-1)=P(x)(x^2+1)+mx+n \\\\
\implies &\begin{cases} mi+n=-a+bi+i(a-5)+(b-1)\\-mi+n=-a-bi-i(a-5)+(b-1)\end{cases}\\\\
\implies &\begin{cases} n=\frac{2(b-1)-2a}{2}=b-a-1\\m=\frac{2bi+2i(a-5)}{2i}=a+b-5\\
\end{cases}\\\\
\implies&\begin{cases}n=0,\thinspace b-a=1\\m=0,\thinspace a+b=5
\end{cases} \\\\
\implies &~~a=2, \thinspace b=3\\\\ \implies &~~2a+3b=13. \end{align}$$
