Question

A leading coefficient of I, and zeros 3i and 2-i

Answer 1

Answer:

f(x) = $x^{4}$ - 4x³ + 14x² - 36x + 45

Step-by-step explanation:

Given zeros x = a, x = b, then the factors are

(x - a) and (x - b)

and the polynomial is the product of the factors

Note that complex zeros occur in conjugate pairs

3i is a zero then - 3i is a zero

2 - i is a zero then 2 + i is a zero

The corresponding factors are

(x - 3i), (x - (- 3i), (x - (2 - i)) and (x - (2 + i)), that is

(x - 3i), (x + 3i), ((x - 2) + i), ((x - 2) - i)

Thus the polynomial is

f(x) = (x - 3i)(x + 3i)((x - 2) + i )((x - 2) - i) ← expand factors in pairs

     = (x² - 9i² )((x - 2)² - i² ) → i² = - 1

    = (x² + 9)(x² - 4x + 4 + 1)

   = (x² + 9)(x² - 4x + 5) ← distribute

   =  $x^{4}$ - 4x³ + 5x² + 9x² - 36x + 45 ← collect like terms

   = $x^{4}$ - 4x³ + 14x² - 36x + 45