Proving the existence and number of *real* roots for $x^3 - 3x + 2$ I need to find how many real roots this polynomial has and prove there existence. I was wondering if my logic and thought process was correct.

Determine the number of real roots and prove it for $x^3 - 3x + 2$

First, note that $f'(x) = 3x^2 - 3$ and so 
$f'(x) > 0$ for $x \in (-\infty, -1) \cup (1, \infty)$ and since $f'$ is strictly increasing on those intervals, there can be at most one root in each of them.
$f'(x) < 0$ for $x \in (-1,1)$ and since $f'$ is strictly decreasing on this interval it can have at most one root. 
Now examine $f(-3) = -16$ and $f(-1) = 4$. By the Intermediate Value Theorem (IVT) $f(c) = 0$ for some $c \in (-3, 1)$ and so $f$ has a root on the interval $(-\infty, 1)$.
Again examine $f(-1) = 4$ and $f(1) = 0$. We cannot say anything about $f$ having a root on the interval $(-1, 1)$. 
Likewise examine $f(1) = 0$ and $f(3) = 16$. Again, we cannot say anything about $f$ having a root on $(1, \infty)$.
However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two. 
Hence, $f(x)$ has two real roots. 

Additional Comments
Also, do we say two real roots (because of the multiplicity), or three real roots, or do we say two distinct real roots?
While I realize factoring the polynomial gives me the answer I believe the purpose of the question was to do the former analysis, which when the polynomial isn't easily factorized, can provide a lot of insight. That is why I did it all
 A: 
However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two. 

All the work you did before this becomes unnecessary; after factoring, the roots (and hence the number of roots) are clear - right?

Addition after some comments: when you are asked about the number of roots (real or not), it is usually meant to count the number of distinct (i.e. different) roots. Your equation has two (real) roots, one of which has multiplicity 2 but that doesn't change the fact that there are only two real numbers where the polynomial becomes 0.
A: Since we have
$$
x^3-3x+2=(x-1)^2(x+2),
$$
we have three real roots $1,1,-2$. Here we count with multiplicities (which is standard for many results in geometry and other areas).
A: $x^3 - 3x + 2$
Possible roots are ±1 and ±2:
$f(1) = 1 - 3 + 2 = 0$
$f(2) = 8 - 6 + 2 = 4$
$f(-1) = (-1) - (-3) + 2 = 6$
$f(-2) = (-8) - (-6) + 2 = 0$
$x=1$ and $x=-2$ are the roots so we can factor the expression into $(x-1)(x+2)(x-1)$
We have the equation $(x-1)^2(x+2) = 0$
$(x-1)^2 = 0 \lor x + 2 = 0$
$\sqrt{(x-1)^2} = ±\sqrt{0} \lor x = -2$
$x-1 = ±0 \lor x = -2$
$x = 1±0 \lor x = -2$
$x = 1+0 \lor x = 1-0 \lor x = -2$
$x = 1 \lor x = 1 \lor x = -2$
The fundamental theorem of algebra says every polynomial of degree n has n roots. Since this polynomial is cubic, it has three roots. $x^3 - 3x + 2$ does have three real roots. It's just that two of them happened to be equal.
