# 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

• I assume you mean real roots, because it has 3 complex roots. – ÍgjøgnumMeg Apr 4 '17 at 13:51
• "First, note that $f′(x)=3x-3$ and so", you mean $f′(x)=3x^{\color{red}{2}}-3$...? – StackTD Apr 4 '17 at 13:52
• Haha yes! Let me make these adjustments. – student_t Apr 4 '17 at 13:52

## 2 Answers

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.

• Yes, but I think the purpose of the exercise was to do the former analysis. I only did that to show the multiplicity of the root. But yes in general all my work before would have been a waste haha! – student_t Apr 4 '17 at 13:55

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).

• I would say there are two real roots, one of which has multiplicity two. – gandalf61 Apr 4 '17 at 13:56
• Yes, but I am suppose to do the little analysis before for the question I believe. Of course factoring would be much faster. – student_t Apr 4 '17 at 13:57
• @danny Yes, this may be the case. But I think, it does not matter so much what you are supposed to do or think, but what you yourself think is the best way. – Dietrich Burde Apr 4 '17 at 13:59
• I agree with gandalf61. In the context of the fundamental theorem of algebra, we often say an $n$th-order polynomial "has $n$ complex roots" but this is an abbreviation where we mean to count the multiplicities. That doesn't change the fact that $x^3$ only has one (distinct) root. – StackTD Apr 4 '17 at 13:59
• @danny see comment above; I would say yours has two (distinct) roots, we usually omit 'distinct' and mean different roots when we're talking about the number of roots. – StackTD Apr 4 '17 at 14:00