Question: If you divide the polynomial $p(x)$ by $x-1$ the remainder is $2$.
If you divide $p(x)$ by $x-2$ the remainder is $1$.

Which is the remainder of $\dfrac{p(x)}{(x-1)(x-2)}$?

Answer: 
**Hint:** Let $P(x)=(x-1)(x-2)Q(x)+ax+b$. You know $P(1)$ and $P(2)$.
Given information:
$$\frac{P(x)}{x-1} \ \text{has remainder $2$}$$
$$\frac{P(x)}{x-2} \ \text{has remainder $1$}$$
We need to find the remainder of:
$$\frac{P(x)}{(x-1)(x-2)}$$

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Remainder theorem:
$$\dfrac{P(x)}{D(x)}=Q(x)D(x)+R(x)$$
Where $P(x)$ is the dividend, $D(x)$ is the divisor, $Q(x)$ is the quotient and $R(x)$ is the remainder. We also know that $R(x)$ has to be of a degree less than the divisor. In this case we have our equations:
$$\frac{P(x)}{x-1}=Q\_{1}(x)+2$$
$$\frac{P(x)}{x-2}=Q\_{2}(x)+1$$
$$\frac{P(x)}{(x-1)(x-2)}=Q\_{3}(x)+ax+b$$
We can rewrite these equations:
$$P(x)=Q\_{1}(x)(x-1)+2$$
$$P(x)=Q\_{2}(x)(x-2)+1$$
$$P(x)=Q\_{3}(x)(x-1)(x-2)+ax+b$$
Now what do we do? We can find out two values of $P(x)$. When you plug in a value of $x=1$ to the first equation, you get $P(1)=2$. When you plug in a value of $x=2$ to the second equation, you get a value of $P(2)=1$. Plugging in these values of $x$ into our last equation gives us:
$$2=a+b$$
$$1=2a+b$$
This is a system of equations. We can solve for $a$ and $b$ using elimination. Subtracting the second equation from the first one gives us:
$$1=-a$$
$$a=-1$$
Plugging in the value of $a$ to either equation...
$$2=-1+b$$
$$b=3$$
Therefore our remainder is:
$$-x+3$$
Hope I helped.