<s>[INST] Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$ Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder should be $$\operatorname {f}(1) = 1^3+1 = 2$$
so the remainder should be 2 for any value of $x$  
But for $x=3$
$$f(3) = 3^3+1$$
when this is divided by $x-1$ which is $3-1$ which equals to $2$, $$\frac {3^3+1}{3-1}$$ $$= \frac {27+1}{2}$$ $$= \frac {28}{2}$$ $$= 14, remainder = 0 ≠ 2$$
Please explain me why this happened
 [/INST] $$\frac{x^3+1}{x-1}=x^2+x+1+\frac{2}{x-1}$$
Here $f(x)=x^3+1$, so the remainder when $f(x)$ is divided by the linear factor $x-1$ is a constant 2 as seen above and which is nothing but $f(1)=2$.
If the divisor is quadratic, the remainder is linear function of $x$ or a constant.
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