Counting the Number of Real Roots of $y^{3}-3y+1$

Here's my question:

How many real roots does the cubic equation $y^3-3y +1$ have?

I graphed the function and it crossed the x-axis $3$ times. But my professor doesn't want a graphical explanation. So in that case, I was looking at the Fundamental Theorem of Algebra and states that a polynomial of degree n can have at most n distinct real roots. so therefore, there must be 3 real roots?

EDIT

It seems that there are numerous ways to approach this problem after all. And we can expand this to other types of polynomials as well, not just cubics.

The given polynomial evaluated at $y\in\{-2,0,1,2\}$ exhibits three sign changes, hence it has at least $3$ real roots, and obviously cannot have more than three roots.

The function has extrema where

$$3y^2-3=0$$ i.e. at $y=\pm1$. The values at these extrema are $3$ and $-1$.

So the variations of this continuous function are $-\infty,3,-1,\infty$, proving that there are three changes of sign.

For a cubic polynomial, the search for the extrema may be more efficient than trial-and-error because it is immediately conclusive.

In the case of a depleted polynomial $$x^3+px+q$$

the extrema are located at $$x=\pm\sqrt{-\frac p3}$$ if $p<0$.

In this case there are three real roots if

$$\sqrt{-\frac p3}^3-p\sqrt{-\frac p3}+q>0\text{ and }-\sqrt{-\frac p3}^3+p\sqrt{-\frac p3}+q<0,$$ or

$$\frac{2p}3\sqrt{-\frac p3}<-|q|.$$

Let $f(y) = y^3 - 3y + 1$. Then you can observe that $f(0) = 1, f(1) = -1$; thus $f$ has at least one root between $0$ and $1$, by the intermediate value theorem. You can find the other two roots similarly.