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

Answer 1

Complex roots comes in "conjugate" pairs

(a+bi) and (a-bi)

It is because when you multiply (x - (a+bi) )*(x - (a-bi) ) the imaginary "i" will disappear and you will be left just with real numbers.

If R is the root of the polynomial - it means the (x-R) is a factor.

So our roots are 2, 2i, -2i and the factorized form of the polynomial is

(x - 2)(x - 2i)(x + 2i)

Opening parenthesis and replacing i² with (-1):

(x-2)(x²-2ix+2ix-4i²) = (x-2)(x²+4) = x³-2x²+4x-8

Notice how the terms -2ix and +2ix cancel out and i² turns out to be (-1)