The equation x3−6x2+15x+3=0 has

Let f(x)=x3−6x2+15x+3.T Then, f′(x)=3x2−12x+15=3x2−4x+5⇒ f′(x)=3(x−2)2+1>0 for all x∈R⇒ f(x) is strictly increasing on R.  Also f(0)=3>0Thus f(x)>0  for all  x>0 Hence, f (x) has no positive root.