<s>[INST] 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
 [/INST] $x^3 - 3x + 2$
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.
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