The question states: Find $m$ and $n$ for the polynomial $x^2+mx+n$ when the polynomial is divided by $(x-m)$, the remainder is $m$, and when the polynomial is divided by $(x-n)$, the remainder is $n$.
I essentially started with the remainder theorem:
$$\frac{p(x)}{x-m} = f(x)+\frac{m}{(x-m)}$$
$$\frac{p(x)}{x-n} = g(x)+\frac{n}{(x-n)}$$
where $f(x)$ and $g(x)$ are quotients. This expands to:
$$p(x)=(x-m)f(x)+m$$
$$p(x)=(x-n)g(x)+n$$
However, I cannot from this information deduce the values of $m$ and $n$.
Any help would be appreciated.
Best Answer
According to the Remainder Theorem, we have: $$p(m)=m \implies 2m^2+n=m \tag1$$ $$p(n) = n \implies n^2+mn + n =n \implies n(m+n)=0 \tag2$$
From $(2)$, it follows that either $n=0$ or $n=-m$.
If $n=0$, then from $(1)$, it must be that $m=0$ or $m=\frac12$.
If $n=-m$, then from $(1)$, it must be that $m=0$ or $m=1$.
Putting this all together, we have the following solutions: $$(m,n)=(0,0), \,(0,\tfrac12), \, (1,-1)$$