Questions: Suppose that R(x) is a polynomial of degree 11 whose coefficients are real numbers. Also, suppose that R(x) has the following zeros. -6, 8, -2-3i, 5i Answer the following. (a) Find another zero of R(x). (b) What is the maximum number of real zeros that R(x) can have? (c) What is the maximum number of nonreal zeros that R(x) can have?

Solution Steps

Since the coefficients of the polynomial R(x) are real, any nonreal complex zeros must occur in conjugate pairs. Given that -2-3i and 5i are zeros, their conjugates, -2+3i and -5i, must also be zeros.

We are given three real zeros: -6, 8, and one other from the answer in step 1. Besides -2-3i, its conjugate, -2+3i and 5i, its conjugate -5i are also zeros. Since R(x) has degree 11, it has 11 zeros in total. We are given three real zeros. Then R(x) has at most $11-2-2 = 7$ more real zeros. Adding the given zeros -6 and 8, the maximum number of real zeros would be $3 + 7 = 9$. Since -2-3i is a zero, its conjugate -2+3i is also a zero. Since 5i is a zero, its conjugate -5i is also a zero. Thus we know 4 non-real zeros. So the maximum number of real zeros is 11-4 = 7. With -6, 8, we have three given real roots, but it says "another" zero, which implies a distinct root. Therefore the maximum number of real zeros R(x) can have is $3 + (11-4-3) = 3 + (11-7) = 3 + 4 = 7$.

Since the polynomial has degree 11, and non-real zeros come in conjugate pairs, R(x) can have at most 10 nonreal zeros. With 5i and -2-3i, this is 2 pairs of non-real zeros which is four complex zeros. Since we already have 3 real zeros, we could have at most $11 - 3 = 8$ more zeros. Nonreal zeros come in conjugate pairs, so there can be at most $8 / 2 = 4$ pairs of nonreal zeros. This adds up to $4 \cdot 2 = 8$ nonreal zeros in addition to the known 4, giving a maximum of 8 nonreal zeros. If we have 7 real zeros, the other 4 roots must be complex (non-real), i.e., -2+3i, -2-3i, 5i and -5i. So R(x) can have 4 or 6 or 8 or 10 non-real roots since the polynomial has real coefficients. Since we are given only 3 real roots, and a pair of complex roots (and thus 4 roots), at most, R(x) can have 7 real roots and 4 complex roots. So, the maximum number of non-real zeros is $2*2 = 4$.

Final Answer