Polynomial and Rational Functions Linear factors theorem and conjugate zeros theorem Suppose that $R(x)$ is a polynomial of degree 8 whose coefficients are real numbers. Also, suppose that $R(x)$ has the following zeros. \[ 2,-4,5,-5+3 i \] 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?

Step 1 ：We are given that the polynomial $R(x)$ has degree 8 and has the following zeros: 2, -4, 5, and -5+3i.

Step 2 ：From the conjugate zeros theorem, we know that if a polynomial has real coefficients and a complex number a + bi is a zero, then its conjugate a - bi is also a zero. Therefore, the conjugate of -5+3i, which is -5-3i, is also a zero of $R(x)$.

Step 3 ：Thus, another zero of $R(x)$ is \(-5-3i\).

Step 4 ：A polynomial of degree n can have at most n zeros. Since the degree of $R(x)$ is 8, it can have at most 8 zeros.

Step 5 ：Since we already know 5 zeros (2, -4, 5, -5+3i, -5-3i), the maximum number of real zeros that $R(x)$ can have is 5.

Step 6 ：Complex zeros always come in conjugate pairs, so the maximum number of nonreal zeros that $R(x)$ can have is the degree of the polynomial divided by 2, which is 4. However, since we already have a pair of nonreal zeros (-5+3i, -5-3i), the maximum number of nonreal zeros that $R(x)$ can have is 3.

Step 7 ：Final Answer: (a) Another zero of $R(x)$ is \(\boxed{-5-3i}\). (b) The maximum number of real zeros that $R(x)$ can have is \(\boxed{5}\). (c) The maximum number of nonreal zeros that $R(x)$ can have is \(\boxed{3}\).