Suppose that \( R(x) \) is a polynomial of degree 8 whose coefficients are real numbers. Also, suppose that \( R(x) \) has the following zeros. Answer the following. (a) Find another zero of \( R(x) \). \( \square \) (b) What is the maximum number of real zeros that \( R(x) \) can have? \( \square \) (c) What is the maximum number of nonreal zeros that \( R(x) \) can have? (

If \( R(x) \) has a real zero at \( r \), then it must also have a zero at \( \overline{r} \) due to the polynomial having real coefficients. This means that if you have a zero that is complex, its conjugate must also be a zero of the polynomial. Thus, if complex zeros are present, they come in pairs. Given that \( R(x) \) is a degree 8 polynomial, it can have a maximum of 8 real zeros, all of which can be distinct or some repeated. If all the zeros are real, then that's the maximum number of real zeros. If the polynomial has nonreal zeros, these will correspond to pairs; hence, for instance, if there are 2 nonreal zeros (1 pair), this leaves 6 possible places for real zeros. In total, the maximum number of nonreal zeros that a polynomial can have is 8, but they must be an even number since they come in pairs.