The equation $x^3-6x^2+15x+3=0$ has 

 

Let $f\left(x\right)=x^3-6x^2+15x+3.T$ Then, 

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=3x^2-12x+15=3\left(x^2-4x+5\right)\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=3\left\{(x-2{)}^2+1\right\}>0\text{ for all }x\in R\end{array}$

$\Rightarrow f\left(x\right)$ is strictly increasing on R.  

Also $f\left(0\right)=3>0$

Thus $f\left(x\right)>0$ for all $x>0$ 

Hence, f (x) has no positive root.