TITLE: Solving for the value of $a,b$ in $f(x)=(ax+b)(x^5+1)-(5x+1)$ s.t. $(x^2+1)|f(x)$
QUESTION [5 upvotes]: $Q.$ 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 .

REPLY [2 votes]: You have 90% of the solution as you states $a-b=-1$ similarly equating imaginary parts to $0$ you should have $a+b=5$ which give $a=2,b=3$.