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

Solution:

Given zeros -2, 2 + 3i

N = 3 (degree = 3)

Leading coefficient is 1

Since, one of the zeros is a conjugate pair, the other one would be 2 - 3i

Let us write it in factored form

= (x - (-2)){x - (2 - 3i)}{x - (2 + 3i)}

= (x + 2){x² - (2 - 3i)x - (2 + 3i)x +(2 + 3i)(2 - 3i)}

= (x + 2){x² - 2x + 3ix - 2x - 3ix (4 - (3i)²}

= (x + 2){x² - 4x + (4 + 9)}

= (x + 2){x² - 4x + 13}

= x³ - 4x² + 13x + 2x² - 8x + 26

= x³ - 2x² + 5x + 26

f(x) = x³ - 2x² + 5x + 26

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

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
f(x) = x³ - 2x² + 5x + 26.