Hi alwaysalillost! Hopefully I'll find you in this explanation.
1. \(\displaystyle \L \;3\sqrt{2}\,+\,\sqrt{50}\)
\(\displaystyle 3\sqrt{2}\) is already simplified. But \(\displaystyle \sqrt{50}\) can be written as \(\displaystyle \sqrt{5^2\,\cdot\,2\)
Pull out the pair of fivess in the square root: \(\displaystyle \sqrt{50}\,\to\,5\sqrt{2}\)
So since we have the same number in the square root for both terms, we add the number in front of the same square roots and keep the same square root:
\(\displaystyle \L \;3\sqrt{2}\,+\,5\sqrt{2}\,=\,(5\,+\,3)\sqrt{2}\,=\,8\sqrt{2}\)
2. \(\displaystyle \L \;(\sqrt{2}\,-\,3)(\sqrt{2}\,+\,3)\)
This is a difference of squares. That means: \(\displaystyle \;(a\,-\,b)(a\,+\,b)\,=\,a^2 - b^2\)
3. \(\displaystyle \L \;(\sqrt{3}\,-\,sqrt{5})(\sqrt{3}\,-\,sqrt{5})\)
You have the same binomial. So: \(\displaystyle (a\,+\,b)(a\,+\,b)\,=\,a^2\,+\,2ab\,+\,b^2\)
