A(x) = x² - 3x + 2
B(x) = x² - 4x + 3
C(x) = x² - 5x + 4

For x = 1, the value of all 3 of these polynomials is 0:

A(1) = B(1) = C(1) = 0

You can also calculate values of their ratios for x ¹ 1, say:

Q(x) = B(x)/A(x) = (x² - 4x + 3)/(x² - 3x + 2)
R(x) = C(x)/A(x) = (x² - 5x + 4)/(x² - 3x + 2)

If 0/0 was equal to anything, it would be reasonable to expect that

Q(1) = B(1)/A(1) = 0/0
R(1) = C(1)/A(1) = 0/0

It would be also reasonable to expect that if x ¹ 1 but it is very very close to 1 (we say x is approaching 1, x ® 1), values of the above ratios would be approaching 0/0, Q(x) ® 0/0 and also R(x) ® 0/0. But the approached values (we call them limits) are easy to calculate:

A(x) = (x - 1)·(x - 2)
B(x) = (x - 1)·(x - 3)
C(x) = (x - 1)·(x - 4)

Q(x) = B(x)/A(x) = (x - 1)·(x - 3)/(x - 1)/(x - 2)
R(x) = C(x)/A(x) = (x - 1)·(x - 4)/(x - 1)/(x - 2)

If x ¹ 1 but it is very very close to 1, then x - 1 ¹ 0 and we can reduce the fractions to

Q(x) = (x - 3)/(x - 2)
R(x) = (x - 4)/(x - 2)

Now we have no problems with x approaching 1:

when x ® 1, Q(x) ® (1 - 3)/(1 - 2) = (-2)/(-1) = 2
when x ® 1, R(x) ® (1 - 4)/(1 - 2) = (-3)/(-1) = 3

Therefore, if 0/0 was equal to anything, it would be reasonable to expect that it equals 0/0 = Q(1) = 2 and also 0/0 = R(1) = 3. By selecting appropriate quadratic polynomials to start with, it would be equally reasonable to expect that 0/0 equals to any number whatsoever. Moreover, all these numbers would have to be equal not only to 0/0, but also to each other. That would be the end of arithmetic. For such reasons, we say that 0/0 is undefined, i.e., it does not equal to any number.

0/0 must be undefined - if we allow it to be any constant, a contradiction arises. This is different from 0^0, which can be defined to be 1 for convenience without giving a contradiction.

Let's say we allowed 0/0 to be some constant "a".

0/0 = a

so

0 = a * 0

but for any b,

0 = b * 0

then

b = 0/0

hence for any b, a = b

If we defined 0/0 = 1, we would be able to work out that 1 = 2.