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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 $A$.

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

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  • $\begingroup$ I'm sorry I had copied the problem wrongly. I editted accordingly. $\endgroup$ – YoTengoUnLCD Jun 17 '15 at 0:08
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    $\begingroup$ @YoTengoUnLCD Fixed $\endgroup$ – egreg Jun 17 '15 at 6:26
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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.

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