Find a polynomial equation with real coefficients that has the given roots. 1 and -2i

Answer 1

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)(x² - 2xi + 2xi - 4i²). Note that i² = -1,

→ f(x) = (x - 1)(x² - 4(-1))

→ f(x) = (x - 1)(x² + 4). Now multiply these to together,

→ f(x) = x³ + 4x - x² - 4

→ f(x) = x³ - x² + 4x - 4

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

Michael Ehlers