# If p(x) is a polynomial of odd degree, show that the equation p(x)=0 has at least one solution. how would I do this?

Feb 20, 2018

Odd degree functions come in two cases, both depending on their highest degree terms.

When $y = + a {x}^{b}$

We know that $b$ is odd and $a$ is positive. Thus, ${\lim}_{x \to - \infty} = - \infty$ because a negative value of $x$ paired with an odd exponent will give a negative number. Meanwhile, ${\lim}_{x \to \infty} = \infty$, because a positive value of $x$ paired with an odd exponent will still give a positive number. Thus in this case, the function must cross the x-axis at least once, or else the limits as $x \to \pm \infty$ would be on the same side (either positive or negative infinity). We also note that polynomials are continuous for all $x$.

When $y = - a {x}^{b}$

Everything is reversed from example one. The limit as $x \to - \infty = \infty$ because the negative cancels the positive in example $1$. On the same vein, we see that limit as $x \to \infty = - \infty$. Therefore, as above, the function must cross the x-axis at least once otherwise the limits couldn't have $+ \infty$ on one side and $- \infty$ on the other.

Hopefully this helps!

Feb 20, 2018

Doesn't quite fit in a sentence, check below.

#### Explanation:

Since polynomial functions are continuous everywhere, it would be enough to find two points "at" which the polynomial has a different sign (one at which it's positive, one at which it's negative, and let's call them $a$ and $b$ with $a < b$), then we can use Bolzano's theorem to show that there will be a point $c \in \left(a , b\right)$ such that the polynomial at $c$ is $0$, or $c$ is a root.

Our polynomial of odd degree has the general form

$p \left(x\right) = {a}_{k} {x}^{k} + {a}_{k - 1} {x}^{k - 1} + \ldots + {a}_{1} x + {a}_{0}$

where $k$ is an odd natural number and ${a}_{k} \ne 0$.

Let's take the limits at positive and negative infinity of $p \left(x\right)$, and without any loss of generality* we will assume that ${a}_{k} > 0$.

${\lim}_{x \to + \infty} p \left(x\right) = {\lim}_{x \to + \infty} {a}_{k} {x}^{k} = {a}_{k} {\lim}_{x \to + \infty} {x}^{k} = + \infty$

${\lim}_{x \to - \infty} p \left(x\right) = {\lim}_{x \to - \infty} {a}_{k} {x}^{k} = {a}_{k} {\lim}_{x \to - \infty} {x}^{k} = - \infty$

The second line is explained because $k$ is odd, any negative raised to an odd power remains negative.

From the above limits we can see that there will be a value for $x$, let's call it $a$, sufficiently large and negative so that $p \left(a\right) < 0$ and there will be a value for $x$, we'll call it $b$, sufficiently large and positive so that $p \left(b\right) > 0$. Therefore there exist a $c \in \left(a , b\right)$ such that $p \left(c\right) = 0$ and there's our root.

*Now if you were to consider the case where ${a}_{k} < 0$, we'd just see that the limit at $+ \infty$ becomes $- \infty$ and the limit at $- \infty$ becomes $+ \infty$. We'd just have to switch places for our $a$ and $b$ and we'd still be fine. Here when using Bolzano's theorem, if $a < b$, it doesn't matter if $p \left(a\right) < 0 < p \left(b\right)$ or $p \left(b\right) < 0 < p \left(a\right)$.