# Find all $a \in \Bbb {C}$ such that $F$ has at least one multiple root.

Let $F=X^{18}-8X^9+4A$.

Find all $A \in \Bbb {C}$ such that $F$ has at least one multiple root. For each $A$ found determine how many different roots $F$ has and their multiplicity.

My attempt:

$F$ has a multiple root then $\deg( \gcd(F,F')) \geq1$ so I proceed to find that gcd, after all the calculations I arrived at:

$$(F:F')=X^8 \forall A\neq4$$ and

$$(F:F')= X^9-4, A=4$$

So that means that when $A\neq4, F$ has $11$ different roots and one with multiplicity $8$ and all the others with multiplicity $1$ (and something analogous for the case $A=4$) ?

Is this correct?

E: I screwed up when typing the problem, it was supposed to say $4A$, not $A$.

• Hint: Let $y=x^9$, then consider $y^2-8y+A=0$. For multiple roots $64-4A=0$ – Anurag A Jun 16 '15 at 23:39

A multiple root is a root also of the derivative. Since $$F'(X)=18X^{17}-72X^8=18X^8(X^9-4)$$ the roots of the derivative are $0$ and the ninth roots of $4$.
Now $0$ is a root of $F$ if and only if $A=0$. If $b^9=4$, then $$b^{18}-8b^9+4A=16-32+4A$$ so the condition is $A=4$.
The roots of $F$ when $A=0$ are easy to find they are $0$ (with multiplicity $9$) and the roots of $X^9-8$, which are simple. For $A=4$, $$F(X)=X^{18}-8X^9+16=(X^9-4)^2$$ and every root is double.
No, it is not. Note that $$F(X) = (X^9)^2 - 8 X^9 + A$$ and define the polynomial $G(X) = X^2 - 8X + A$. Then observe that the roots of $F$ are precisely the roots of $X^9 - \beta$ where $\beta$ varies through the roots of $G$.
Now, $X^9 - \beta$ has multiple roots if and only if $\beta = 0$, which happens if and only if $A = 0$. Otherwise $F$ has multiple roots if and only if $G$ has a double root, which happens if and only if $A = 16 = (-4)^2$. In every other case $F$ has exactly one root.