The maximum possible number of real roots of equation x5-6x2-4x+5=0 is

Let f(x)=x5-6x2-4x+5=0Then the number of change of sign in f(x) is 2 therefore f(x) can have at most two positive real roots.Now f(-x)=-x5-6x4+4x+5=0Then the number of change of sign is 1.Hence f(x) can have at most one negative real root. So that the total possible number of real roots is 3.