# Polynomial Division Under Certain Remainders

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. :)

• 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.
– lulu
May 26 '20 at 16:05
• Wait, so we just solve that system of equations? May 26 '20 at 16:07
• 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? May 26 '20 at 16:21
• 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). May 26 '20 at 16:23

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.

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