# Has the polynomial distinct roots? How can I prove it?

I want to prove that the polynomial

$$f_p(x) = x^{2p+2} - cx^{2p} - dx^p - 1$$

,where $c>0$ and $d>0$ are real numbers, has distinct roots. Also $p>0$ is an even integer. How can I prove that the polynomial $f_p(x)$ has distinct roots for any $c$,$d$ and $p$.

PS: There is a similar topic that How to prove that my polynomial has distinct roots?

• Hint: A polynomial has distinct roots if and only if it (the polynomial) and its first derivative do not have any roots in common. See if you can use it here. – mathguy Apr 16 '16 at 15:53
• Thank you for response. Of course I know this. But for this polynomial I didn't achieve. I would be grateful if you could give me the solution. – drxy Apr 16 '16 at 16:02
• It seems that for $p=2$, $c=1$, $d\approx 2.6107186132760393498186490083840586275$ a root of $36d^4 + 125d^3 - 133d^2 - 801d - 899$, the polynomial does have multiple roots at $\pm 0.81073923436477156272173034309719150582 i$ (i.e., at the roots of $(6d + 2)x^2 + (d + 9)$, which is the $\gcd(f,f')$ for such $p,c,d$ – Hagen von Eitzen Apr 16 '16 at 16:24
• @HagenvonEitzen how did you find the $d$ value ? Have you any program or any other tool for calculation this? Further, how can I prove it theoretically? – drxy Apr 20 '16 at 13:24
• @HagenvonEitzen Is there any multiple roots for higher $p$ values? For example $p=4,6,8, \ldots$. – drxy Apr 25 '16 at 7:33

Here is a partial solution that shows there can be no multiple REAL roots. The proof doesn't work for complex numbers though (and I am not sure the result is even true in complex numbers).

$f' = (2p+2) x^{2p+1} - 2pcx^{2p-1} - pdx^{p-1}$. If $x$ is a multiple root of $f$, then both $f$ and $f'$ vanish at $x$. But $(2p+2) f - xf' = -2cx^{2p}-(p+2)dx^p-(2p+2) < 0$ for all $x \in \mathbb R$ because $p$ is even and $c$, $d$ and $p$ are all positive.

• Thank you to answer. But I must prove it mathematical way. how can I prove it for complex numbers? – drxy Apr 20 '16 at 13:25
• I don't know what you mean by "mathematical way" - what I gave you is a complete, "mathematical way" proof that the polynomial does not have multiple roots that are real numbers. For complex numbers, I said I don't even believe the result is true (so you can't prove it, if it's not true!) and in a comment to your post, Hagen von Eltzen demonstrated that my suspicion was correct: the result is FALSE in complex numbers. Please see Hagen's comment to your question. – mathguy Apr 20 '16 at 13:30
• Mathematical way means that "I must prove it completely" both real and complex numbers as in this topic math.stackexchange.com/questions/1740673/…. Your solution includes real numbers. Besides, it's so important for me that the proof must include the complex numbers. Thank you so much. – drxy Apr 20 '16 at 13:34
• Can you read what I said? In complex numbers the result is FALSE, how do you want to "prove" something that is not true? Hagen gave you a counter-example! – mathguy Apr 20 '16 at 13:50
• Sorry my bad English. Thank you for all answers. Have you got any idea for $p=4,6,8$ or higher values. How can I be sure that the polynomial has multiple roots for any $p$ value. – drxy Apr 20 '16 at 13:55

‘All’ you have to do is computing the g.c.d. of $f_p(x)$ and $f'_p(x)$ via the Euclidean algorithm. A polynomial has only simple roots (in $\mathbf C$) if and only if this polynomial and its derivative are coprime.

• This should be the best and most familiar way to go. Moreover, you should expect the two polynomials to be relatively prime. I.e you should expect the gcd to be a constant. – user244643 Apr 17 '16 at 1:28
• @Bernard Since I couldn't achieve it, I'm writing here. Thank you very much for valuable answer. – drxy Apr 20 '16 at 13:22
• Not very helpful though, eh? HOW do you compute the gcd? Note that the problem is subtle; as shown by Hagen, the result is false in general, but as I have shown, it is true about real roots; so the proof definitely is not based on gcd alone. – mathguy Apr 20 '16 at 14:00