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

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