Find an nth degree polynomial function with real coefficients satisfying the given conditions. n = 3; 2 and -2 + 3i are zeros; leading coefficient is 1. Find the expanded and simplified polynomial

Solution:

Given n = 3 and 2, -2 + 3i are zeros

Since -2 + 3i is an imaginary number then -2 - 3i must also be one of the zeros.

So, f(x) = A(x - 2)(x - (-2 + 3i))(x - (-2 - 3i))

= A(x - 2)(x + 2 - 3i)(x + 2 + 3i)

= A(x - 2)(x² + 4x + 13)

f(x) = A(x³ + 4x² + 13x - 2x² - 8x - 26)

= A(x³ + 2x² + 5x - 26)

After expansion, the leading coefficient is A, which is 1.

Therefore, the 3rd degree polynomial is x³ + 2x² + 5x - 26.

Find an nth degree polynomial function with real coefficients satisfying the given conditions. n = 3; 2 and -2 + 3i are zeros; leading coefficient is 1. Find the expanded and simplified polynomial

Summary:

The nth degree polynomial function with real coefficients satisfying the given conditions. n = 3; 2 and -2 + 3i are zeros; leading coefficient is 1 is x³ + 2x² + 5x - 26.