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 $f(4)=51$

$$x=\pm i$$
so
$$a(\pm i)^3-a(\pm i)^2+b(\pm i)+4=0$$
so
$$ai+a\pm bi+4=0$$
$$ai+a+bi+4=0\tag 1$$
or
$$ai+a-bi+4=0\tag 2$$ 
now we will solve the first equation
$$a+4=0\rightarrow a=-4$$
$$a-b=0\rightarrow b=-4$$
hence $$f(x)=-4x^3+4x^2-4x+4$$
at $x=4$
$$f(4)=-204=-4(51)$$
the second equation gives
$$a=-4$$
$$b=4$$
hence $$f(x)=-4x^3+4x^2+4x+4$$
at $x=4$
$$f(4)=-176=-4(43)$$
as @marty said the problem is wrong
