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

C

Explanation:

Just took the test on edge

Answer 2

`f(x)=x^(3)-3x^(2) +4x-2`

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

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

Remember that

The leading coefficient is 1

so

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)]\\\\f(x)=(x-1)[x^(2) -x+xi-x-xi+2]\\\\f(x)=(x-1)[x^(2) -2x+2]\\\\f(x)=x^(3)-2x^(2) +2x-x^(2) +2x-2\\\\f(x)=x^(3)-3x^(2) +4x-2`