Find a degree 3 polynomial with real coefficients having zeros 2 and 2 − 2 i and a lead coefficient of 1. Write P in expanded form. Be sure to write the full equation, including P ( x ) = .

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.

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510

irrational, cannot