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

Accepted Solution

Answer:[tex]f(x)=x^{3}-3x^{2} +4x-2[/tex]Step-by-step explanation:we know thatThe 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 polynomialIn this problem we have thatThe polynomial has roots 1 and (1+i)soby the conjugate root theorem(1-i) is also a root of the polynomialthereforeThe lowest degree of the polynomial is 3so[tex]f(x)=a(x-1)(x-(1+i))(x-(1-i))[/tex]Remember thatThe leading coefficient is 1soa=1[tex]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[/tex]