Write a polynomial f(x) that satisfies the given conditions. Degree 3 polynomial with integer coefficients with zeros -2i and 8/5

Answer 1

First, find the factors. For one complex zero, you also have to have the complex conjugate as another zero. So 2i would be the 3rd zero.

(x+2i)(x-2i)(5x-8) Now multiply. Will FOIL the first 2 terms.

(x²-2i+2i-4i²)(5x-8) Simplify

(x²+4)(5x-8) Now FOIL again

f(x) = 5x³-8x²+20x-32

Answer 2

If x = -2i is a zero, then x = 2i is also a zero. Thus, our three zeros are:

x = - 2i , x = 2i and x = 8/5

Now, if we get them into factor form, we have:

x + 2i = 0, x - 2i = 0 and x - 8/5 = 0

(We multiply both sides of the last factor by 5 to get rid of the fraction)

x + 2i = 0, x - 2i = 0 and 5x - 8 = 0

Thus, we can start creating our polynomial:

f(x) = a(x + 2i)(x - 2i)(5x - 8)

We can let a = 1 for simplicity and start foiling the factors:

f(x) = a(x + 2i)(x - 2i)(5x - 8)

f(x) = (x + 2i)(x - 2i)(5x - 8)

f(x) = (x² + 4)(5x - 8)

f(x) = 5x³ - 8x² + 20x - 32