Question

Can someone please give me an example of a couple of polynomials and predict the number of complex roots for each.
Also can you tell me how you found the number of complex roots for each?

Answer 1

Basically, the number of roots that a polynomial function has depends on the degree of the polynomial, that is, a linear polynomial has 1 root, a quadratic has 2 roots and so on.

Answer 2

So how does it make it complex?

Answer 3

You cannot really tell the number of complex roots a polynomial function has. But what I can say about it is that it must always be an even number. I think you should first know the number of real roots a function has before you can actually know the number of complex roots

Answer 4

Okay but my teacher says she wants examples of polynomial functions and their complex roots, which is why I'm pretty stuck

Answer 5

x^2= -16 would be an example since its roots are 4i and -4i

Answer 6

You can come up with a polynomial with complex roots in this fashion:

Say you want a polynomial with complex roots at 1 + 2i and 3 +4i, then you can write:

(x - (1 + 2i) ) (x - (3+4i))=0, which means that at x = 1 + 2i or x = 3 +4i, this polynomial will be zero, that is, these are the roots of this polynomial. It is of polynomial of degree 2 because when you multiply it out, it will have the highest degree of 2:

x^2-(4+6 i) x-(5-10 i) = 0

If the coefficients would have been real, rather than complex, then the roots would be complex conjugates of each other, and you would have an even number of complex roots. For example, at 1+2i, 1-2i, 3+4i, 3-4i. Then the polynomial would be:

(x- (1+2i))(x- (1-2i))(x- (3+4i))x- (3-4i))=0

Which expands to, x^4-8 x^3+42 x^2-80 x+125 = 0 and all coefficients are real.