Question

Please help

Answer 1

The zeros of the function are 2, - 4 and (1 + 3i).

Because complex zeros occur in conjugate pairs, we know that (1 - 3i) is also a zero of the function. So, from the Linear Factorization Theorem, f(x) can be written as f(x) = a [x - 2] · [x + 4] · [x - (1 + 3i)] · [x - (1 - 3i)].

For simplicity, let a = 1 to obtain

f(x) = [x - 2] · [x + 4] · [(x - 1) - 3i)] · [(x - 1) + 3i)]

f(x) = [x² + 2x - 8] · [ (x - 1)² - 9i² ]

f(x) = [x² + 2x - 8] · [x² - 2x + 1 + 9]

f(x) = [x² + 2x - 8] · [x² - 2x + 10]

f(x) = x²(x² - 2x + 10) + 2x(x² - 2x + 10) - 8(x² - 2x + 10)

f(x) = x⁴ - 2x³ + 10x² + 2x³ - 4x² + 20x - 8x² + 16x - 80

f(x) = x⁴ - 2x² + 36x - 80.