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.

what are ur choices?

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b