I$f\left(x\right)=(x-1)(x-3)(x-4)(x-6)+19$ have : 

We know ,  Discarte's Rule of Signs No equation can have more $+ve$ real roots than it has changes of sign from $+ve$ to $-ve,$ and from $-\mathrm{ve}$ to $+ve.$

And similarly, number of -ve roots off(x) can not more than the number of changes of sign in f(-x) . 

$\begin{array}{r}f\left(x\right)=(x-1)(x-3)(x-4)(x-6)+19\\ \Rightarrow f\left(x\right)=x^4-14x^3+67x^2-126x+91\end{array}$

Here, $f\left(x\right)$ has four changes of sign. So the equation can not have more than four positive roots.

$f(-x)=x^4+14x^3+67x^2+126x+91$

$f(-x)$ has no changes of sign, so by Discarte's rule, the equation have no negative roots. As 

Thus the equation have four positive roots.