Question

Suppose that R(x) is a polynomial of degree 13 whose coefficients are real numbers. also, suppose that R(x) has the following zeros. answer the following.edit: if possible please double check the answers just to be safe.

Answer 1

(a) Complex zeros of a polynomial come in pairs.

If a + bi is a zero of a polynomial then its conjugate a - bi is also a zero of the polynomial.

The given complex zeros of R(x) are 1 + 3i and -2i.

1 - 3i is the conjugate of 1 + 3i.

Hence, another zero of R(x) is 1 - 3i

b)

Since the polynomial R(x) is of order 13 then R(x) must have 13 zeros.

The given complex zeros of R(x) are 1 + 3i and -2i. We also know that the conjugates of 1 + 3i and -2i are zeros of R(x). Hence, R(x) has at least 4 complex roots

Hence, the maximum number of real zeros of R(x) is (13 -4).

The maximum number of real zeros of R(x) is 9

c) Let the maximum number of nonreal zeros (complex roots) be n

Complex roots come in pairs. Therefore, n must be even.

Hence, n ≤ 13 - 1 = 12

n ≤ 10

We have been given a real zero of R(x), 3 ( With the multiplicity of 4).

12 - 4 = 8

Therefore,

n ≤ 8.

Hence the maximum number of nonreal zeros of R(x) is 8