3
$\begingroup$

The coefficients of the polynomial are in the ring of real numbers. Prove that a polynomial of odd degree in $ \Bbb R[x]\\$ with no multiple roots must have an off number of real roots. I hate to ask people to solve things, so I was wondering if you could glance at my attempt and give a suggestion. There is a theorem in my textbook that states that every polynomial of odd degree has a root.

Let f(x) be a polynomial of odd degree. Since, I know that since f(x) is odd it has a root, and thus can be factored as follows: Let a be an arbitrary root of f(x). With a $\in \Bbb R$, then f (x) = (x-a)(h(x)) where h(x) $\in \Bbb R[x]$. This implies that h(x) must have even degree, because deg(f(x)) is odd and deg (x-a) is odd. At this point can I assume that, since h(x) must have even degree, it must have an even number of roots and therefore factors? Because, to the contrary, if it had an odd number of factors or roots, our initial assumption that f(x) has odd degree would violated?

$\endgroup$
3
  • 1
    $\begingroup$ Non-real roots come in complex conjugate pairs. Here I am implicitly using the Fundamental Theorem of Algebra. $\endgroup$ Commented Feb 22, 2016 at 7:49
  • $\begingroup$ @AndréNicolas No, you're not using it. $\endgroup$
    – egreg
    Commented Feb 22, 2016 at 9:59
  • $\begingroup$ @egreg: Well, I was, but you showed there is no need. $\endgroup$ Commented Feb 22, 2016 at 12:52

2 Answers 2

3
$\begingroup$

The only deep result you need is

An odd-degree polynomial $f(x)\in\mathbb{R}[x]$ has at least a root.

Also you can check algebraically whether $f$ has multiple roots (in some extension field), by computing the greatest common divisor between $f$ and its derivative.

Another easy observation is that if $u\in\mathbb{C}$ is a root of $f$, then also $\bar{u}$ is a root, which is distinct from $u$ if $u\notin\mathbb{R}$.

Thus you can factor your polynomial as $$ (x-a_1)\dots(x-a_r) (x-u_1)(x-\bar{u}_1)\dots(x-u_s)(x-\bar{u}_s) g(x) $$ where $a_1,\dots a_r$ are the real roots and $u_1,\bar{u}_1,\dots,u_s,\bar{u}_s$ are the complex (nonreal) roots, and $g$ has no root (either real or complex). Note that $g$ must have even degree, otherwise it would have a real root.

Now count degrees.

Note. The FTA says that $g$ has degree zero, but this is not needed.

$\endgroup$
3
  • $\begingroup$ The OP's question title is like saying "Every polynomial with odd degree has odd number of roots" or so...But this is not the case in general of course. For example, $(x+i)(x-i)(x+1)$ does not have $3$ real roots. $\endgroup$
    – NoChance
    Commented Jul 18, 2023 at 0:27
  • $\begingroup$ @NoChance Your polynomial has an odd number of real roots. It is not stated in the question that the number of roots is the same as the degree. $\endgroup$
    – egreg
    Commented Jul 18, 2023 at 6:12
  • $\begingroup$ In fact, the title says so. It is OK. Thanks. $\endgroup$
    – NoChance
    Commented Jul 18, 2023 at 9:59
2
$\begingroup$

Hint: Any roots that aren't real will occur in pairs (why?)

$\endgroup$
4
  • $\begingroup$ It is becaue if a+bi is a root, then a-bi is also a root. But what do I do now, I don't know how to use this to show it has an odd number of roots. $\endgroup$
    – wesssg
    Commented Feb 22, 2016 at 7:58
  • $\begingroup$ Well, then you just count - $\text{number of roots} = \text{number of real roots}+\text{number of complex roots}$. Then consider each of these numbers $\text{mod} 2$, i.e. which are even and odd. $\endgroup$
    – πr8
    Commented Feb 22, 2016 at 8:00
  • $\begingroup$ So total number of real roots must be odd because h(x) has an even number of complex roots because of FTA? $\endgroup$
    – wesssg
    Commented Feb 22, 2016 at 8:03
  • $\begingroup$ FTA guarantees this, but as egreg says, you can actually establish the result by simply noting that any real polynomial of odd degree has a real root. $\endgroup$
    – πr8
    Commented Feb 22, 2016 at 9:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .