Let p(x) = ax^2 + bx + c
Then p(1-x) = a(1-x)^2 + b(1-x) + c
Since p(x) = p(1-x), then
ax^2 + bx + c = a(1-x)^2 + b(1-x) + c
ax^2 + bx = a(1 - 2x + x^2) + b(1-x)
ax^2 + bx = a - 2ax + ax^2 + b - bx
2bx = (a + b) - 2ax
0 = (a+b) - 2(a+b)x
i.e. a+b = 0
or, a = -b
The polynomial then is p(x) = ax^2 - ax + c
So, all polynomials of the form p(x) = ax^2 - ax + c have the property that p(x) = p(1-x)