Question 395529
We could simply multiply the whole thing out, rewriting as {{{((x-2)-i)((x-2)+i)}}} as the other tutor did (so that the difference of two squares can be carried out). This method uses a set of formulas called Vieta's formula:


The polynomial is in the form {{{ax^2 + bx + c}}}. Right off the back we deduce {{{a = 1}}} since we can tell that the x^2 coefficient is 1 without even multiplying. Vieta's formulas say that:


The sum of the roots of the polynomial is -b/a = -b. Since the roots are 2+i, 2-i, their sum is 4, so b = -4.


The product of the roots is c/a = c. Since {{{(2+i)(2-i) = 4 + 1 = 5}}} (applying difference of two squares) we get c = 5.


Therefore the polynomial is {{{x^2 - 4x + 5}}}.


Here are a couple articles on Vieta's formulas, just in case (I rarely see these formulas taught in school, but they're pretty useful to know):
http://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te
http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_Formulas