Question

Show that a polynomial of odd degree in $\mathbf{\square }\left[x\right]$with no multiple roots must have an odd number of real roots.

Short answer

It is shown that a polynomial of odd degree in $\square \left[x\right]$ with no multiple roots have an odd number of real roots.

Step-by-step solution

Lemma 4.29

If $f\left(x\right)$is a polynomial in $\square \left[x\right]$ and $a+bi$ is a root of $f\left(x\right)$ in $\square$, then $a-bi$ is also a root of $f\left(x\right)$ .

Explanation for an odd number of real roots

Let $f\left(x\right)$ be a polynomial in $\square \left[x\right]$with odd degree n.

Given that $f\left(x\right)$ has no multiple roots, then it has n distinct roots $\left\{a_1,a_2,........a_n\right\}$.

If z is a non-real complex root, then by theorem 4.29, $\overline{z}$is also a root. Thus, the non-real complex roots always come in pairs $\sout{which}\underline{that}\underline{are}\sout{is}$even.

Since, $f\left(x\right)$is a polynomial of odd degree n$\underline{,}$then there must be an odd number of real roots.

Therefore, a polynomial of odd degree in $\square \left[x\right]$with no multiple roots has an odd number of real roots.