(1) Let there be real polynomials f(x)

if it has exactly 3 real zeros and its first derative has exactly 8 real zeros , asuming all there 0 are to be dictinct

then write the least no. of zeros of

g(x) = (f '(x))² + f(x)f ^''(x)

Define H(x) = f'(x)f(x)

This equation has 11 zeroes acc to condition.

Now H'(x) = g(x)

Hence g(x) has 10 zeroes.

PS: I dont know what is the meaning of least no. of zeroes , I think there can be only 10 zeroes.