Question

Using real and imaginary as types of roots, list all possible combinations of root type for a fourth-degree polynomial equation.

Answer 1

I just want to check:
all reals
2 reals, 2 imaginary
all imaginary
Is that correct?

Answer 2

Yeah

Answer 3

since complex roots always occur as conjugates, there can be either 0,2 or 4 complex roots, the remaining roots are purely real

Answer 4

yup, now repeat the whole process for 5th degree polynomial equations and odd-degree polynomial equations

Answer 5

Note that what you wrote:
all reals
2 reals, 2 imaginary
all imaginary

isn't exactly right.

It's:

all real
2 real, 2 complex
4 complex

...because there can be roots of the form a + bi where a is not zero.

Answer 6

but isn't a+bi an imaginary number?

Answer 7

complex numbers are reals + imaginary together.
using complex numbers we can solve any polynomial equations so all the solutions to polynomial equations are complex numbers

Answer 8

If we proceed in the same manner for 5 roots, we have 0 complex, 5 real
2 complex, 3real
4 complex and one real root

Answer 9

so how will you generalize for all odd-degree polynomial equations?

Answer 10

For a (2n+1) degree equation we have n+1 combinations which is equal to the number of conjugate pairs of complex roots. As before the remaining roots can be filled by reals. Also this does NOT account for repeated real roots.

Answer 11

Aaaaand, for the record, this is all assuming that the coefficients of the polynominal are real.

If any of the coefficients are complex, all bets are off.

Answer 12

Yeah, thought that was obvious.