<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] $f(x)=ax^3-ax^2+bx+4$
Since $f(4) = 51$,
$51
=a(64-16)+4b+4
=48a+4b+4
$
so
$12a+b
=47/4
$.
Since $f(i) = 0$,
$0
=a(-i+1)+ib+4
=i(b-a)+a+4
$
so
$ a+4 = 0,
a=-4,
b-a = 0,
b=a=-4
$.
Therefore
$f(x) = -4x^3+4x^2-4x+4
$.
But this does not satisfy
$f(4) = 51$.
Therefore the problem is wrong.
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