0
$\begingroup$

I want to find out the 11 possible combinations of real (distinct and equal) and complex roots for a quintic function.

For example a quintic function with...

2 real distinct and 3 real equal or 1 real distinct and 4 complex

I know there are 11 combinations but I'm missing a few

$\endgroup$
0
$\begingroup$

The number of partitions of $5$ is $7$:

  • into $5:~1,1,1,1,1$

  • into $4:~1,1,1,2$

  • into $3:~1,2,2~$ and $~1,1,3$

  • into $2:~1,4~$ and $2,3$

  • into $1:~5$.

~ In the first case, all five roots are distinct. Since complex roots can only come in pairs, we have three such possibilities, corresponding to the number of $0,~1$, and respectively $2$ complex conjugate pairs.

~ In the second case, we have one double root, and the other three are distinct. Two possibilities. Obviously, the double root cannot be complex, since that would automatically imply the existence of another complex conjugate double root.

~ In the third case, we have two distinct double roots, and a fifth root, also distinct. Two possible situations: either all five roots are real, or a double root is complex, implying that the other double root is its complex conjugate.

~ In the fourth case, we have one triple root, and the other two are distinct. Two possible options, depending on whether the latter two are real, or complex conjugates.

~ In the fifth case, we have a quadruple root, and a distinct root, all real, for obvious reasons.

~ In the sixth case, we have a double root, and a triple root, both real, for obvious reasons.

~ In the seventh and last case, we have a quintuple real root $($again, for obvious reasons$)$.

So there are twelve possible cases, in total.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.