How to read this problem from Dummit-Foote's "Abstract Algebra"? Problem 14.6.1 on page 617 says

Show that a cubic with a multiple root has a linear factor. Is the same true for quartics?

Let $f \in F[x]$ be a cubic. If $f$ has a root in $F$, let alone a multiple root in $F$, then it has a linear factor in $F[x]$. So, I suppose we should really be assuming that $f$ has a multiple root in an extension field $E \supseteq F$. Obviously $f$ has a linear factor in $E[x]$, so I suppose we should be arguing that $f$ has a linear factor in $F[x]$.  But, I don't think that is even true. Consider $f(x) = x^3 - t$ in $F_3(t)[x]$. This has a root of multiplicity $3$ in the splitting field, but no root in $F_3(t)$, right?
This problem is discussed here, but I wondered if anyone had a suggestion for how to interpret this problem, so that the question becomes sensible.
 A: If the cubic $f$ has a multiple root in some extension field  $E$, and $F$ does not have characteristic $3$, then the cubic and its derivative have a common root in the extension field, so they are not relatively prime in that field. 
It follows $f$ and $f'$ are not relatively prime in $F$. If $f'$ divides $f$, then $f$ has a linear factor. If $f'$ does not divide $f$, then they have a linear gcd, and again $f$ has a root in $F$.
A: The usual meaning of “having a multiple root” includes “in some extension field”. I can only interpret “has a linear factor” assuming “in $F[x]$” or it would make no sense.
I don't have Dummit-Foote available, so I guess that the exercise has some context, because the statement can be false in case the field has characteristic $3$.
A polynomial $f(x)\in F[x]$ has a multiple root (in some extension field) if and only if $f$ and $f'$ (the formal derivative) have a common factor. This can be checked by just working over $F$, because the greatest common divisor can be computed with the Euclidean algorithm.
The statement is true provided $F$ doesn't have characteristic $3$. Indeed, your example is good: the polynomial $x^3-t$ with coefficients in the field of rational functions $\mathbb{F}_3(t)$ has a multiple root in $\mathbb{F}_3(\sqrt[3]{t})$ because it factors as $(x-\sqrt[3]{t})^3$, but has no linear factor in $\mathbb{F}_3(t)$.
If the characteristic is not $3$, the derivative of a cubic polynomial is not $0$; so if $g=\gcd(f,f')\ne 1$, it has degree $1$ or $2$. In the case $\deg g=1$, we found a linear factor of $f$. In the case $\deg g=2$, $f/g$ is a linear factor.
For the quartic case, consider $(x^2+1)^2$, with $F$ the real field.
