+118658 Thanks Guest.
how to fully factor x^2-6x+7
This does not have rational roots so I will use the quadratic theory.
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {6 \pm \sqrt{6^2-4*1*7} \over 2*1}\\ x = {6 \pm \sqrt{36-28} \over 2}\\ x = {6 \pm \sqrt{8} \over 2}\\ x = {6 \pm 2\sqrt{2} \over 2}\\ x = {3 \pm \sqrt{2}}\\ \mbox{So the factors are}\\ x-(3 + \sqrt{2})\qquad and \qquad x-(3- \sqrt{2})\\ (x-3- \sqrt{2})\qquad and \qquad (x-3+ \sqrt{2})\\~\\ x^2-6x+7=(x-3- \sqrt{2})(x-3+ \sqrt{2})\\~\\ \mbox{This is the same as guests answer} \)
