$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$. True or false? $f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?

Well, 
$$\alpha >0 \to f(-\alpha)=2-(\alpha+2)(\alpha+4)(\alpha+6) < -46$$
and 
$$f(6+\alpha)=2+(4+\alpha)(2+\alpha)(\alpha)> 2$$
So at the very least all its real roots are $\in (0,6)$
You just need to show all of its roots are real.
