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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.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. How about $m=n=0$? $\endgroup$ Dec 23 '19 at 20:33
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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: $$(m,n)=(0,0), \,(0,\tfrac12), \, (1,-1)$$

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

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