Advanced mathematics exercises F (x) is continuous in [0,1], differentiable in (0,1), ∫ (0,1) f (x) DX = 0 To prove that there is a point x belonging to (0,1) such that 2F (x) + XF (x) = 0

Advanced mathematics exercises F (x) is continuous in [0,1], differentiable in (0,1), ∫ (0,1) f (x) DX = 0 To prove that there is a point x belonging to (0,1) such that 2F (x) + XF (x) = 0

Wrong copy, right
If we want to prove that 2F (x) + XF (x) = 0, because 2F (x) + XF (x) = f (x) [2 + x], we only need to prove that x belongs to (0,1), so that f (x) = 0, which is very easy
Because ∫ (0,1) f (x) DX = 0, according to the integral mean value theorem, there exists x belonging to (0,1), such that f (x) = 0

What is the greatest common factor
The concept is absolutely no problem! Please click ch0425

The answer is like this
The greatest common factor is a term in Higher Algebra (not linear algebra)
Polynomials
Generally speaking, it is a concept in real number field
The greatest common factor of (x-1) (x ^ 2 + 1) and (x ^ 2 + 1) (X-2) is (x ^ 2 + 1)
The greatest common factor of (x-1) (x ^ 2 + 1) and (x-1) is (x-1)
The first common factor is (x ^ 2 + 1)
It is because it belongs to a nonseparable polynomial in the real field
The second one is more obvious. Don't you know this is the answer you want?
Ask me if you still don't understand

What is the first greatest common factor?

It's the greatest common factor with the first coefficient of 1
That is, the greatest common factor of which the coefficient of the highest degree term is 1
If you have any questions, please ask. If you are satisfied, please choose as the satisfactory answer