Michael E. answered • 11/05/18

College and High School Math for Classes and Test Prep

Hello Sarah,

Since we are looking for 'real coefficients', complex roots will have to come in pairs.

So with the root of -2i given, we want its conjugate root of 2i.

So the roots are

x = 1

→ x - 1 = 0,

x = - 2i

→ x + 2i = 0, and

x = 2i

→ x - 2i = 0

→ f(x) = (x - 1)(x + 2i)(x - 2i),

which I will expand. Multiply the quantities with the complex roots together first, as terms will cancel, and make the final multiplication easier,

→ f(x) = (x - 1)(x^{2} - 2xi + 2xi - 4i^{2}). Note that i^{2} = -1,

→ f(x) = (x - 1)(x^{2} - 4(-1))

→ f(x) = (x - 1)(x^{2} + 4). Now multiply these to together,

→ f(x) = x^{3} + 4x - x^{2} - 4

→ f(x) = x^{3} - x^{2} + 4x - 4

Thank you for posting the question with Wyzant, and have a great night!

Michael Ehlers