## anonymous one year ago What is the polynomial function of lowest degree with lead coefficient 1 and roots 1 and 1 + i? Also whoever answers this question and more I will give that person a medal. I NEED HELP FAST!!!!!!!!!!!!! a) f(x) = x2 – 2x + 2 b)f(x) = x3 – x2 + 4x – 2 c)f(x) = x3 – 3x2 + 4x – 2 d)f(x) = x2 – x + 2

• This Question is Open
1. anonymous

if $$1+i$$ is a root and the polynomial has strictly real coefficients, then $$1-i$$ must be a root as well. so the minimal polynomial with roots $$1+i,1-i,1$$ must be cubic of the form $$a(x-1)(x-(1+i))(x-(1-i))=a(x^3+\dots)=ax^3+\dots$$ for some real constant $$a$$ corresponding to the coefficient of the leading term in the expansion.

2. anonymous

so it follows $$a=1$$ and we have \begin{align*}(x-1)(x-(1+i))(x-(1-i))&=(x-1)(x^2-(1+i+1-i)x+(1+i)(1-i))\\&=(x-1)(x^2-2x+2)\\&=x^3-2x^2+2x-x^2+2x-2\\&=x^3-3x^3+4x-2\end{align*}