The assertion of the Intermediate Value Theorem is something which is probably ‘intuitively obvious’, and is also provably true: if a function $f$ is continuous on an interval $[a,b]$ and if $f(a) < 0$ and $f(b) > 0$ (or vice-versa), then there is some third point $c$ with $a < c < b$ so that $f(c)=0$. This result has many relatively ‘theoretical’ uses, but for our purposes can be used to give a crude but simple way to locate the roots of functions. There is a lot of guessing, or trial-and-error, involved here, but that is fair. Again, in this situation, it is to our advantage if we are reasonably proficient in using a calculator to do simple tasks like evaluating polynomials! If this approach to estimating roots is taken to its logical conclusion, it is called the method of interval bisection, for a reason we'll see below. We will not pursue this method very far, because there are better methods to use once we have invoked this just to get going.
For example, we probably don't know a formula to solve the cubic equation $$x^3-x+1=0$$ But the function $f(x)=x^3-x+1$ is certainly continuous, so we can invoke the Intermediate Value Theorem as much as we'd like. For example, $f(2)=7 > 0$ and $f(-2)=-5 < 0$, so we know that there is a root in the interval $[-2,2]$. We'd like to cut down the size of the interval, so we look at what happens at the midpoint, bisecting the interval $[-2,2]$: we have $f(0)=1 > 0$. Therefore, since $f(-2)=-5 < 0$, we can conclude that there is a root in $[-2,0]$. Since both $f(0) > 0$ and $f(2) > 0$, we can't say anything at this point about whether or not there are roots in $[0,2]$. Again bisecting the interval $[-2,0]$ where we know there is a root, we compute $f(-1)=1 > 0$. Thus, since $f(-2) < 0$, we know that there is a root in $[-2,-1]$ (and have no information about $[-1,0]$).
If we continue with this method, we can obtain as good an approximation as we want! But there are faster ways to get a really good approximation, as we'll see.
Unless a person has an amazing intuition for polynomials (or whatever), there is really no way to anticipate what guess is better than any other in getting started.
Invoke the Intermediate Value Theorem to find an interval of length $1$ or less in which there is a root of $x^3+x+3 =0$: Let $f(x)=x^3+x+3$. Just, guessing, we compute $f(0)=3 > 0$. Realizing that the $x^3$ term probably ‘dominates’ $f$ when $x$ is large positive or large negative, and since we want to find a point where $f$ is negative, our next guess will be a ‘large’ negative number: how about $-1$? Well, $f(-1)=1 > 0$, so evidently $-1$ is not negative enough. How about $-2$? Well, $f(-2)=-7 < 0$, so we have succeeded. Further, the failed guess $-1$ actually was worthwhile, since now we know that $f(-2) < 0$ and $f(-1) > 0$. Then, invoking the Intermediate Value Theorem, there is a root in the interval $[-2,-1]$.
Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. We can use the Intermediate Value Theorem to get an idea where all of them are.
Invoke the Intermediate Value Theorem to find three different intervals of length $1$ or less in each of which there is a root of $x^3-4x+1=0$: first, just starting anywhere, $f(0)=1 > 0$. Next, $f(1)=-2 < 0$. So, since $f(0) > 0$ and $f(1) < 0$, there is at least one root in $[0,1]$, by the Intermediate Value Theorem. Next, $f(2)=1 > 0$. So, with some luck here, since $f(1) < 0$ and $f(2) > 0$, by the Intermediate Value Theorem there is a root in $[1,2]$. Now if we somehow imagine that there is a negative root as well, then we try $-1$: $f(-1)=4 > 0$. So we know nothing about roots in $[-1,0]$. But continue: $f(-2)=1 > 0$, and still no new conclusion. Continue: $f(-3)=-14 < 0$. Aha! So since $f(-3) < 0$ and $f(2) > 0$, by the Intermediate Value Theorem there is a third root in the interval $[-3,-2]$.
Notice how even the ‘bad’ guesses were not entirely wasted.