# Remainder Theorem Question: Deduce the Polynomial Given the Remainder

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.

• Welcome to Mathematics Stack Exchange. How about $m=n=0$? Dec 23, 2019 at 20:33

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: $$(m,n)=(0,0), \,(0,\tfrac12), \, (1,-1)$$

I used the idea that if f(x) has remainder r when divided by (x-a), then f(a)=r 