Suppose that R(x) is a polynomial of degree 7 whose coefficients are real numbers. Also, suppose that R(x) has the following zeros.
-5-3i, 2i
Answer the following.
(a) Find another zero of R(x).
i
Х
(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?
Help, these questions make no sense!

R(x) is a polynomial of degree 7, so it has up to 7 distinct complex roots r₁, ..., r₇, and we can write it in terms of these roots as

R(x) = (x - r₁) (x - r₂) ... (x - r₇)

The coefficients of R(x) are all real, so the roots must all be complex numbers, and any of these roots with non-zero imaginary parts must occur along with their complex conjugates. This means if a + b i is a root, then is a - b i is also a root.

(a) We're told that -5 - 3i and 2i are roots to R(x), so we also know that -5 + 3i and -2i are roots.

There are 4 roots accounted for, leaving us with 3 unknown roots. These roots cannot all be non-real, because we can only count 2 of them as a conjugate pair. So we can have either

(b) at most 3 real roots, or

(c) at most 2 non-real roots and 1 real root.

use substitution to solve each system of equations. y = x + 5 3x + y = 25 62/87,21 y = x + 5 3x + y = 25 substitute x + 5 for y in the second equation. substitute the solution for x into either equation to find y. the solution is (5, 10)

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