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

f(x) = x^{2} - 2x + 2, f(x) = x^{3} - x^{2} + 4x - 2, f(x) = x^{3} - 3x^{2} + 4x - 2, f(x) = x^{2} - 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)[x^{2} - (1 - i)x - (1 + i)x + (1- i)^{2}]

By further calculation,

f(x) = (x - 1)[x^{2} - x + xi - x - xi + 2]

f(x) = (x - 1)[x^{2} - 2x + 2]

So we get,

f(x) = x^{3} - 2x^{2} + 2x - x^{2} + 2x + 2

f(x) = x^{3} - 3x^{2} + 4x - 2

Therefore, the polynomial function is x^{3} - 3x^{2} + 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) = x^{3} - 3x^{2} + 4x - 2.