Question

Find a degree 3 polynomial whose coefficient of x^3 equal to 1. The zeros of this polynomial are -5, -4i, and 4i. Simplify your answer so that it has only real numbers as coefficients.
I got an answer by solving (x^3+5x^2-16x-80) but it says thats incorrect.​

Answer 1

Answer:

x^3+5x^2+16x+80

Step-by-step explanation:

You know that you can write your polynom this way : (x-r1)(x-r2)(x-r3) with r1,r2 and r3 the roots so you get :

(x+5)(x+4i)(x-4i)

Simplify (x+4i)(x-4i) with the formula (a-b)(a+b)=a²-b²

so you have (x^2+16)=(x+4i)(x-4i)

Your polynom looks like this :

(x^2+16)(x+5) just expand it

and you get x^3+5x^2+16x+80