Answer:
The polynomial with real coefficients having zeros 2 and 2 - 2i is
x³ - 6x² + 16x - 16 = 0
Step-by-step explanation:
Given that a polynomial has zeros at 2 and 2 - 2i, we want to write this polynomial.
We have
x - 2 = 0
x - (2 - 2i) = 0
=> x - 2 + 2i = 0
Since the polynomial has real coefficients, and 2 - 2i is a zero of the polynomial, the conjugate of 2 - 2i, which is 2 + 2i is also a polynomial.
x - (2 + 2i) = 0
=> x - 2 - 2i = 0
Now,
P(x) = (x - 2)(x - 2 + 2i)(x - 2 - 2i) = 0
= (x - 2)((x - 2)² - (2i)²) = 0
= (x - 2)(x² - 4x + 8) = 0
= x³ - 4x² + 8x - 2x² + 8x - 16 = 0
= x³ - 6x² + 16x - 16 = 0
This is the polynomial required.
