Well, I can't say I'm overly familiar with the Upper and Lower Bound Theorems, so I don't know if your reasoning is correct; however, you can see here that your bounds of -4 to 2 are correct.
Here's an exercise I was looking at today.
Use the Upper Bound Theorem to find an integral upper bound and the Lower Bound Theorem to find an integral lower bound of the zeros of
Here was my explanation:
According to the Upper Bound Theorem, c is an upper bound of the zeros of f(x) if
when you divide f(x) by (x - c), there is no sign change in the quotient or remainder.
So I picked a few possible factors of f(x). First I used (x - 1).
The depressed polynomial produced was
There is a sign change so 1 is not an upper bound.
Next, I tried (x-2). The depressed polynomial was
There was no sign change in the quotient or remainder so 2 is an upper bound.
I contend that 2 is the greatest upper bound and there are no zeros of f(x) greater than 2.
Now on to the lower bound.
According to the Lower Bound Theorem, if c is an upper bound of the zeros of f(-x),
then -c is a lower bound of the zeros of f(x).
I went through the same process to find the upper bound for f(-x) as I did for f(x).
Turns out that (x - 2) doesn't yield a quotient and remainder with no sign changes.
Neither does (x - 3) or (x - 4). However, (x - 5) does.
This means that all real zeros of f(x) can be found in the interval
I agree with this, but is -5 the greatest lower bound?
If you solve the thing algebraically, the roots are
This shows the greatest (integral) lower bound is -4.
Shouldn't all the zeros of f(x) be found on this interval:
Since the exercise did not specify the words "greatest" or "least",
maybe they were just looking for any upper and lower bound.
I don't see the significance of that, though.
If that were the case, the interval could be betwen -infinity and +infinity.