[jacks]

If \( f \) be a non zero polynomial such that \( f(1-x)=f(1+x) \) for all real \( x \)

And \( f(1)=0 \). Then largest positive integer \( m \) such that

\( (x-1)^{m} \) divides polynomial \( f(x) \) for all polynomial \( f(x), \) is

[geómetracat]

Maybe I am missing something obvious, but what about the polynomials \( (x-1)^{2k} \) for an arbitrary \( k>0 \)? They satisfy all the conditions, so there is no such largest \( m \).

[martiniano]

Hello.

The polinomials you have found are not divisibles all them by \( (x-1)^m \) if \( m>2 \). Then \( m\leq{2} \)...

Health.

[geómetracat]

Obviously I have misunderstood the statement of the problem. The question is to find the largest \( m \) dividing all polynomials satisfying those conditions. Thanks, martiniano.

But now it's easy to see that in fact \( m=2 \). Indeed, as martiniano has remarked, my previous example show that \( m \leq 2 \).
Now, if \( f(x) \) is a polynomial satisfying the conditions but not divisible by \( (1-x)^2 \), we can write:
\( f(x) = (x-1)g(x) \), where \( g(x) \) is a polynomial with \( g(1) \neq 0 \).
From \( f(1+x)=f(1-x) \) we obtain:
\( -xg(1+x)=xg(1-x) \), hence:
\( -g(1+x)=g(1-x) \).
Evaluating at \( x=0 \), we obtain \( -g(1) = g(1) \), hence \( 2g(1)=0 \) and \( g(1)=0 \), contradiction.

[jacks]

Thanks Moderator got it