Cubic with repeated roots has a linear factor If $f$ is a cubic polynomial with a repeated root over a field then $f$ has a linear factor.
I think that is true in perfect fields but I don't know how to prove it.
 A: In a perfect field, an irreducible polynomial has distinct roots. Since you know the polynomial has a repeated root, it is not irreducible. Since the polynomial is of degree 3, there are not many ways for it to factor, and any reduction requires a linear factor.
A: If we don't assume perfectness, we find the following:
Assume $f(X)=X^3+aX^2+bX+c\in F[X]$ and $f$ has a repeated root $\alpha\in E\supseteq F$. Then $\alpha$ is a common root of $f(X)$ and $f'(X)=3X^2+2aX+b$, hence root of of $3f-Xf'=aX^2+2bX+3c$ and of 
$$ 3(2f-Xf')-af' = (6b-2a^2)X+(9c-ab).$$ 
If $6b\ne 2a^2$, we conclude that $\alpha=\frac{ab-9c}{6b-2a^2}\in F$ and $(X-\alpha)^2\mid f$.
Assume therefore that $6b=2a^2$. 


*

*In characteristic $3$ this means $a=0$, so that already $3f-Xf'=aX^2+2bX+3c=2bX$. We conclude that $\alpha=0\in F$ or $b=0$. In the latter case $f(X)=X^3+c=(X-\alpha)^3$ is the factorization in $E[X]$, but we do not necessarily have a linear factor in $F[X]$. In fact, that $c$ is a third power for all $c\in F$ is equivalent to $F$ being perfect.

*In characteristic $\ne 3$ we may asume wlog. that $a=0$ (substitute $X\leftarrow X-\frac a3$). So we already find that $\alpha$ is a root of $2bX+3c$. if $2b\ne 0$, we find $\alpha=-\frac{3c}{2b}\in F$ and are done like above. If $2b=0$ then also $3c=0$, i.e. $c=0$ and $f$ has $X$ as linear factor.

