Question

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? f(x) = x2 – 2x + 2 f(x) = x3 – x2 + 4x – 2 f(x) = x3 – 3x2 + 4x – 2 f(x) = x2 – x + 2

Answer 1

Answer:

C. f (x) = x^3-3x^2-4x-2

Step-by-step explanation:

we know that

The conjugate root theorem states that if the complex number a + bi is a root of a polynomial P(x) in one variable with real coefficients, then the complex conjugate a - bi is also a root of that polynomial

In this problem we have that

The polynomial has roots 1 and (1+i)

so

by the conjugate root theorem

(1-i) is also a root of the polynomial

therefore

The lowest degree of the polynomial is 3

so

Remember that

The leading coefficient is 1

so

a=1