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.

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18