<s>[INST] Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^2+bx+4}{x^2+1}$ is $a+b+x(b-a)$.
Then
$$a+b+x(b-a)=0$$
I can't proceed any further so I'm guessing the other factor of $f(x)$ is $ax+4$.
Then
$$f(x)=(ax+4)(x^2+1)=ax^3+4x^2+ax+4=ax^3-ax^2+bx+4$$
I found that $a=-4$ and $b=a=-4$. Then $f(x)=-4x^3+4x^2-4x+4$. But I doesn't satisfy $f(4)=51$
 [/INST] This question is very interesting......
Only by this statement, the question can be solved
$f(x)= ax^3-ax^2+bx+4$ is a multiple of $x^2+1$;
Because the roots of the equation $x^2+1$ would be the roots of $f(x)$. 
Therefore if $j,k$ are roots of the equation then $j+k=0$ and $j.k =1$.
Let the roots of $f(x)$ be $j,k,l$.
Then,
$j+k+l = -\frac{-a}{a}$
$l=1$
now, $j.k.l= -\frac{4}{a}$
so, $a= -4$
now $j.k + l(k+j) = \frac{b}{-4}$
Therefore  $b=-4$ 
Finally $a+b=-8$ 
But $f(4)= 4(-4^3+4^2-4+1)$ 
Therefore $f(4)= -204$
So the question may be incorrect.
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