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?
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.