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How do you find a third degree polynomial given roots #-4# and #4i#?

1 Answer
Dec 27, 2017

Answer:

#x^3+4x^2+16x+64 = 0#

Explanation:

Since the question ask for a third degree polynomial, I am going to assume that you want a polynomial with real coefficients, with #-4i# as the third root, being the complex conjugate of #4i#.

Each zero #a# corresponds to a factor #(x-a)#, so we can write:

#f(x) = (x+4)(x-4i)(x+4i)#

#color(white)(f(x)) = (x+4)(x^2-(4i)^2)#

#color(white)(f(x)) = (x+4)(x^2+16)#

#color(white)(f(x)) = x^3+4x^2+16x+64#

So we can write a cubic equation:

#x^3+4x^2+16x+64 = 0# with the desired roots.