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 

Solution:

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 question the polynomial has roots 1 and (1 + i)

Using conjugate root theorem, (1 - i) is a root of the polynomial

3 is the lowest degree of the polynomial

f(x) = a(x - 1)(x - (1 + i))(x - (1 - i))

1 is the leading coefficient

a = 1

f(x) = (x - 1)(x - (1 + i))(x - (1 - i))

f(x) = (x - 1)[x2 - (1 - i)x - (1 + i)x + (1- i)2]

By further calculation,

f(x) = (x - 1)[x2 - x + xi - x - xi + 2]

f(x) = (x - 1)[x2 - 2x + 2]

So we get,

f(x) = x3 - 2x2 + 2x - x2 + 2x + 2

f(x) = x3 - 3x2 + 4x - 2

Therefore, the polynomial function is x3 - 3x2 + 4x - 2.

What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i?

Summary:

The polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i is f(x) = x3 - 3x2 + 4x - 2.

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