Question

Suppose that R(x) is a polynomial of degree 9 whose coefficients are real numbers. Also, suppose that R(x) has the following zeros. 4+5i, -3-3i, -i 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?

Answer 1

Another zero of R(x) can be 4-5i, -3+3i, +i based on the given zeros and the Complex Conjugate Root Theorem. The maximum number of real zeros R(x) can have is 3. The maximum number of nonreal zeros it can have is 6.

Step-by-step explanation:

Finding Zeros of a Polynomial

Part (a): Another zero of R(x)

Given that R(x) is a polynomial with real coefficients and has a zero of 4+5i, by the Complex Conjugate Root Theorem, its conjugate 4-5i must also be a zero. Similarly, -3-3i implies another zero, -3+3i, and -i implies its conjugate +i as zeros of R(x).

Part (b): Maximum number of real zeros

The degree of the polynomial gives the maximum number of zeros it can have. A degree 9 polynomial can have at most 9 zeros. Since complex zeros occur in conjugate pairs, and we have identified two pairs (4+5i & 4-5i, -3-3i & -3+3i) and one pair (-i & +i), the maximum number of real zeros is 9 - (2 x 2+2 x 1) = 9 - 6 = 3.

Part (c): Maximum number of nonreal zeros

The polynomial can have at most 9 zeros, with each nonreal zero having a conjugate pair. Therefore, the maximum number of nonreal zeros is 6, as there can be three pairs of nonreal zeros (the ones already identified).

Answer 2

The polynomial R(x) has a zero at 4-5i, which is the complex conjugate of one of its given zeros, 4+5i. The maximum number of real zeros can be 6, considering the presence of the nonreal zeros. The maximum number of nonreal zeros for a 9th-degree polynomial is 8.

Step-by-step explanation:

Complex conjugate roots are a pair of zeros for any polynomial with real coefficients where one zero is complex. If a polynomial has a zero at 4+5i (a complex number), then it must also have a zero at 4-5i (the complex conjugate). Since another given zero is -3-3i, its complex conjugate and therefore another zero must be -3+3i. The last provided zero is -i, which is already its own conjugate, as i is imaginary and the real part is zero.

The maximum number of real zeros is determined by the degree of the polynomial. Since we have degree 9, at most it can have 9 real zeros. However, knowing that we already have complex zeros, the maximum number of real zeros cannot exceed 6, as there are already 3 pairs of complex zeros (4+5i and 4-5i, -3-3i and -3+3i, and -i, which counts as a pair itself).

As complex zeros come in conjugate pairs, the maximum number of nonreal zeros it can have is equal to the degree of the polynomial (in this case 9) minus the actual number of real zeros. Since the minimum number of real zeros it must have is 1 (as the number of complex zeros must be even), the maximum number of nonreal zeros is 8.