ya want to rationalize the denomenator
if we mutliply the whole thing by (√41)/(√41), we wil still have a √ there messing things up
we need to remember the difference of 2 perfect squares
remember
(a+b)(a-b)=a²-b²
so, we can see the denomenator is 5+√41
multiply the whole fraction by [tex]\frac{5-\sqrt{41}}{5-\sqrt{41}}[/tex]
we get
[tex]\frac{3(5-\sqrt{41})}{5^2-(\sqrt{41})^2}=\frac{15-3\sqrt{41}}{25-41}=[/tex]
[tex]\frac{15-3\sqrt{41}}{-16}=\frac{-15+3\sqrt{41}}{16}[/tex]
