The remainder when a polynomial, $f(x)$ is divided by $x-1$ is the same as the quotient. Find the value of $f(0)$.
So I've assumed the remainder but be a linear expression and so must the quotient meaning that $f(x)$ must be a quadratic function?
So if $f(x)$ is a quadratic then $f(x)=ax^2+bx+c$
$$\frac{f(x)}{x-1}=(ax+b+a) + \frac{r}{x-1}$$
But I am unsure where to go from here? any help would be appreciated.
The remainder is the same as the quotient, so the quotient is a constant. This implies $f$ is linear – write it as $ax+b$. Then $$ax+b=a(x-1)+a+b$$ $$a+b=a\qquad b=0$$ Thus $f(x)=ax$ for some real $a$. Regardless of the choice, $f(0)=0$.