Evaluate the number of possible combinations of roots for a given polynomial equation. For example for quadratic, you can have:

  • 2 imaginary roots (conjugate)
  • two real distinct roots
  • or 1 repeated root.
    In this case, we have only 3 possible combinations For cubic you can have:
  • 3 distinct real roots
  • 3 repeated real roots
  • 2 repeated and 1 distinct real roots
  • 2 distinct imaginary (conjugate) and 1 real
  • 1 repeated imaginary (conjugate) and 1 real and in this case we have 5 possible combinations:

can you find a formula that gives the total number combinations, for example, 3 for quadratics and 5 for cubic equations, for any given polynomial of degree n?

  • $\begingroup$ Note that if you have a quadratic polynomial in one variable with real coefficients, you are right about the number of possibilities, but if you allow complex coefficients as well as complex roots, you can have one real and one complex root. So I think you need to be just a little more precise about the question. [You could be referring also to integer or rational coefficients] $\endgroup$ – Mark Bennet Oct 20 '19 at 11:26

I guess that here you are considering polynomial with real coefficients. In that case, for $n=3$, we can't have 1 repeated imaginary (conjugate) and a real root and the number of possibilities is only $4$.

In general when the degree is $n$, we can have $k=n-2j$ real roots with $j=0,\dots,\lfloor n/2\rfloor$ with non-negative multiplicities $m_1,m_2,\dots,m_k$ such that $$m_1+2m_2+3m_3+\dots+km_k=k.$$ The number of non-negative integer solutions of this equation is $p(k)$ the number of partitions of $k$. Hence the total number of cases is given by the formula $$\sum_{j=0}^{\lfloor n/2\rfloor}p(j)\cdot p(n-2j)$$ which gives the sequence A002513: $1, 3, 4, 9, 12, 23, 31, 54, 73, 118,\dots$.

P.S. Reading the comments at the OEIS's link, it turns out that this question appeared as Problem 2055 in the American Mathematical Monthly.


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