If f(11) = 0 then one of the two terms inside (x-p) or (x-q) must be 0. So let's arbitrarily assign p=11.
Similarly if f(20) = 0 then we can assign q=20.

So what's f(10)? It will be (10-11) (10-20) = (-1) (-10) = 10

Let f(x)= (x-p)(x-q). If f(11)=f(20)=0, then f(10)=?

A. 1
B. -1
C. 5
D. -5
E. 10

Hi Max@Math Revolution,
Let's take a look at your question.

$$f\left(x\right)=\left(x-p\right)\left(x-q\right)$$
$$f\left(11\right)=\left(11-p\right)\left(11-q\right)$$
$$\text{Since,} f\left(11\right)=0$$
$$\left(11-p\right)\left(11-q\right)=0$$
$$\text{Either }\left(11-p\right)=0 \text{or }\left(11-q\right)=0$$
$$\text{Either }p=11 \text{or }q=11$$

Also ,
$$f\left(20\right)=\left(20-p\right)\left(20-q\right)$$
$$\text{Since,} f\left(20\right)=0$$
$$\left(20-p\right)\left(20-q\right)=0$$
$$\text{Either }\left(20-p\right)=0 \text{or }\left(20-q\right)=0$$
$$\text{Either }p=20 \text{or }q=20$$

Now how we are going to decide which what values of p and q will be selected to find f(10).
We know that f(11)=f(20)=0, therefore,
$$\left(11-p\right)\left(11-q\right)=\left(20-p\right)\left(20-q\right)$$
$$121-11p-11q+pq=400-20p-20q+pq$$
$$121-11p-11q=400-20p-20q$$
$$121-11p-11q-400+20p+20q=0$$
$$-279+9p+9q=0$$
$$-31+p+q=0$$
$$p+q=31$$

So we will be selecting the values of p and q such that there sum is 31.
So we can take p=11, q = 20 or p=20, q=11, either case we will get the same result.

Now we will find f(10).
$$f\left(10\right)=\left(10-p\right)\left(10-q\right)$$
Plugin the values of p and q:
$$f\left(10\right)=\left(10-11\right)\left(10-20\right)=\left(-1\right)\left(-10\right)=10$$

Therefore, Option E is correct.

Hope it helps.
I am available if you'd like any follow up.

=>
Since 11 and 20 are roots of f(x) = 0, we must have f(x) = (x-11)(x-20).
Thus, f(10) = (10-11)(10-20) = (-1)(-10) = 10.

Therefore, the answer is E.
Answer : E

For (x-p)(x-q) = 0, p = 11 and q = 20 (or vice versa).
Using f(x)= (x-p)(x-q), f(10)= (10-20)(10-11) = (-10)(-1) = 10
Answer E