# How do I use the intermediate value theorem to prove every polynomial of odd degree has at least one real root?

Oct 27, 2015

Given any polynomial $f \left(x\right)$ of odd degree and positive leading coefficient find ${x}_{1}$ such that $f \left(- {x}_{1}\right) < 0$ and $f \left({x}_{1}\right) > 0$, so $\exists x \in \left(- {x}_{1} , {x}_{1}\right)$ with $f \left(x\right) = 0$.

#### Explanation:

Let $f \left(x\right) = {a}_{0} {x}^{n} + {a}_{1} {x}^{n - 1} + \ldots + {a}_{n}$ with ${a}_{0} \ne 0$

Note that $f \left(x\right)$ is a continuous function.

If $x$ is sufficiently large and positive, $f \left(x\right) > 0$.

To prove that:

Let ${x}_{1} = \frac{1 + \left\mid {a}_{0} \right\mid + \left\mid {a}_{1} \right\mid + \left\mid {a}_{2} \right\mid + \ldots + \left\mid {a}_{n} \right\mid}{\left\mid {a}_{0} \right\mid}$

Note that ${x}_{1} > 1$, so if $m \le n - 1$ then ${x}^{m} \le {x}^{n - 1}$

So we find:

$f \left({x}_{1}\right) \ge {a}_{0} {x}_{1}^{n} - \left(\left\mid {a}_{1} \right\mid {x}^{n - 1} + \left\mid {a}_{2} \right\mid {x}^{n - 2} + \ldots + \left\mid {a}_{n} \right\mid\right)$

$\ge {a}_{0} {x}_{1}^{n} - \left(\left\mid {a}_{1} \right\mid {x}^{n - 1} + \left\mid {a}_{2} \right\mid {x}^{n - 1} + \ldots + \left\mid {a}_{n} \right\mid {x}^{x - 1}\right)$

$= {x}^{n - 1} \left({a}_{0} x - \left(\left\mid {a}_{1} \right\mid + \left\mid {a}_{2} \right\mid + \ldots + \left\mid {a}_{n} \right\mid\right)\right)$

$= {x}^{n - 1} \left(\left(1 + \left\mid {a}_{0} \right\mid + \left\mid {a}_{1} \right\mid + \left\mid {a}_{2} \right\mid + \ldots + \left\mid {a}_{n} \right\mid\right) - \left(\left\mid {a}_{1} \right\mid + \left\mid {a}_{2} \right\mid + \ldots + \left\mid {a}_{n} \right\mid\right)\right)$

$= {x}^{n - 1} \left(1 + \left\mid {a}_{0} \right\mid\right) > 0$

If $n$ is odd and $x$ sufficiently large and negative then $f \left(x\right) < 0$.

To prove that, note that if $n$ is odd then ${a}_{0} {\left(- x\right)}^{n} = - {a}_{0} {x}^{n}$, so $- f \left(- x\right) = {a}_{0} {x}^{n} - {a}_{1} {x}^{n - 1} + \ldots - {a}_{n}$, hence $f \left(- {x}_{1}\right) < 0$

We have:

• $f \left(x\right)$ continuous over $\left[- {x}_{1} , {x}_{1}\right]$
• $f \left(- {x}_{1}\right) < 0 < f \left({x}_{1}\right)$

So by the intermediate value theorem:

$\exists x \in \left(- {x}_{1} , {x}_{1}\right) : f \left(x\right) = 0$

If the leading coefficient (${a}_{0}$) of $f \left(x\right)$ is negative then the same formula for ${x}_{1}$ yields $f \left({x}_{1}\right) < 0 < f \left(- {x}_{1}\right)$, hence there is a root in $\left(- {x}_{1} , {x}_{1}\right)$ again.