$f(x) = (x-2)(x-4)(x-6) +2$ 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?

Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x \rightarrow \infty$ $F(x), f(x) \rightarrow \infty.$
3) $x \rightarrow  -\infty$ $F(x), f(x) \rightarrow -\infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise  for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
