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Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-17$, the remainder is $14$, and when $P(x)$ is divided by $x-13$, the remainder is $6$. What is the remainder when $P(x)$ is divided by $(x-13)(x-17)$?

Here was my process, that I'm not sure if it's right or not:

We can write $P(x)$ in the form of $$P(x)=Q(x)(x-17)(x-13)+cx+d$$

Thus, by the remainder theorem, we have a system of equations:

\begin{align*} 14c+d &=6,\\ 6c+d &=14. \end{align*}

Solving gets $c=-1, d=20.$

Thus, our remainder is $\boxed{-x+20}.$

Did I make any flaws during my process. Thanks in advance for helping. :)

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    $\begingroup$ Not following. We know that $P(17)=14$, say, from which we deduce that $17c+d=14$. Similarly, $13c+d=6$. Not sure where your equations are coming from. $\endgroup$
    – lulu
    May 26 '20 at 16:05
  • $\begingroup$ Wait, so we just solve that system of equations? $\endgroup$
    – Frost Bite
    May 26 '20 at 16:07
  • $\begingroup$ Note, by the way, that you can check your tentative answer (or indeed any linear polynomial): divide $-x+20$ itself by each of $x-17$ and $x-13$—do you get remainders of $14$ and $6$ respectively? $\endgroup$ May 26 '20 at 16:21
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    $\begingroup$ It also might be worth commenting: don't be fooled into believing, from the numbers chosen in the problem, that the polynomial remainder when dividing by $x-17$ must always be between $0$ and $17$ (and similarly for $13$); check the case $P(x) = x^2$ for example. Polynomial remainders have smaller degree, but the size and sign of their coefficients can be arbitrary (literally anything, as linear algebra tells us). $\endgroup$ May 26 '20 at 16:23
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From $$P(x)=Q(x)(x-17)(x-13)+cx+d$$

Now, let $x=17$, then we have $$17c+d=14$$

If we let $x=13$, then we have

$$13c+d=6$$

Now solve for $c$ and $d$.

Subtract the two equations, we ahve $4c=8 \iff c=2$. Proceed on to solve for $d$ to get the remainder.

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    $\begingroup$ $c=2, d=-20.$ Our remainder is $2x-20$ $\endgroup$
    – Frost Bite
    May 26 '20 at 16:10
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    $\begingroup$ yes, that is right. $\endgroup$ May 26 '20 at 16:11
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    $\begingroup$ as a check, $2x-20=2(x-17)+14=2(x-13)+6$ $\endgroup$ May 26 '20 at 16:11
  • $\begingroup$ Thank you all. $+1$ $\checkmark$ $\endgroup$
    – Frost Bite
    May 26 '20 at 16:12

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