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We have a polynomial $p(x)=x^5 -ax^3+b$. We need to find the relationship between $a$ and $b$ such that $p(x)$ has multiple roots.

Assume that $p(x)$ has 2 roots $c$ and $d$ with multiplicities $2$ and $3$ respectively. Then 1) $3c+2d=0$ 2) $3c^2+6cd+d^2=-a$ 3) $(c^3)*(d^2)= -b$

Thus, $a=3.75c^2$ and $b=-2.25c^5$. and we can find the relationship between $a$ and $b$ then.

Is it the right approach?

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    $\begingroup$ $p$ has multiple (complex) roots iff $\gcd(p,p')\ne1$. $\endgroup$
    – lhf
    Dec 3, 2013 at 13:19
  • $\begingroup$ Note that if "$p(x)$ has 2 roots $c$ and $d$ with multiplicities $2$ and $3$ respectively," then the sum of roots is $2c+3d$, not $3c+2d$. Are you assuming only multiple roots? I read the Question as asking for at least one multiple root. $\endgroup$
    – hardmath
    Dec 3, 2013 at 14:50

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There is an algorithm which allows you to find if polynomials $f$, $g$ have a common root. You have to calculate the resultant of $f$ and $g$, $R(f,g)$. If it's $0$, then there are common roots. Now polynomial $f$ has multiple roots iff $f$ and $f'$ has common root.

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    $\begingroup$ For univariate (one-variable) polynomials, it's likely efficient just to compute the GCD of $f,g$, which is nontrivial if and only if $f,g$ have a common root. $\endgroup$
    – hardmath
    Dec 3, 2013 at 13:38
  • $\begingroup$ @hardmath Ofc you are right but i think it can lead to more complicated computations. $\endgroup$
    – user52045
    Dec 3, 2013 at 13:51
  • $\begingroup$ Can the resultant of $p,p'$ in this particular case be found more easily than $\gcd(p,p')$? The determinant of the Sylvester matrix gives the resultant, but this is a $9\times 9$ matrix determinant here. $\endgroup$
    – hardmath
    Dec 3, 2013 at 14:10
  • $\begingroup$ I dont think its worth arguing but in this case in any row there will be only 3 or 2 nonzero elements and Laplace formula gives almost trivial computation. I guess this is more matter of taste in this case tho. $\endgroup$
    – user52045
    Dec 3, 2013 at 14:17
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    $\begingroup$ Yes, I wasn't trying to be argumentative, just trying to compare. It seems to me the special structure of coefficients might help to do both approaches. $\endgroup$
    – hardmath
    Dec 3, 2013 at 14:20
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It is not so hard to show that if a polynomial $f$ has $\alpha$ as a root with multiplicity $k$ then $f'$ has $\alpha$ as a root with multiplicity $k-1$. Since $f'(x)=x^2(5x^2-3a)$ you can easily find the root with multiplicity $3$. From this you will be able to find the conditions on $a$ and $b$.

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    $\begingroup$ I doubt the “iff”: $f(x)=x^3-1$. $\endgroup$
    – Michael Hoppe
    Dec 3, 2013 at 13:43
  • $\begingroup$ The “iff“ I your unedited answer. $\endgroup$
    – Michael Hoppe
    Dec 3, 2013 at 13:47

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