$f(x) = (x-2)(x-4)(x-6) +2$ 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?
 A: Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$
A: 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.
A: In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment): 
f(1) = -13
f(2) = 2
By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.
f(4) = 2
f(5) = -1
Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5. 
f(6) = 2
You also know there must be a root between x=5 and x=6. 
A: 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?
