I have this problem:
$$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}$$
I'm able to get to this part by myself:
$$\frac{15\sqrt{2}+2}{2\sqrt{2}}$$
But that's when I get stuck. The book says that the next step is:
$$\frac{15\sqrt{2}}{2\sqrt{2}}+ \frac{2}{2\sqrt{2}}$$
But I don't understand why you can take the 2 out of the original fraction, make it the numerator of its own fraction and having root of 2 as the denominator of said fraction.
On
To me the next obvious step is to remove the radical in the denominator so
$$\frac{15\sqrt{2}+2}{2\sqrt{2}} = \frac{15+\sqrt{2}}{2}.$$
Personally I would stop there, but you could separate this into $\dfrac{15}{2}+\dfrac{\sqrt{2}}{2}$ or into $\dfrac{15}{2}+\dfrac{1}{\sqrt{2}}$ if you really wanted to.
On
$$\frac{\sqrt{18}+\sqrt{98}+\sqrt{50}+4}{2\sqrt{2}}= \frac{3\sqrt{2}+7\sqrt{2}+5\sqrt{2}+4}{2\sqrt{2}}= \frac{15\sqrt{2}+4}{2\sqrt{2}} $$ Now the standard procedure is to remove the radical in the denominator: $$ \frac{15\sqrt{2}+4}{2\sqrt{2}}= \frac{15\sqrt{2}+4}{2\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}= \frac{15\sqrt{2}\cdot\sqrt{2}+4\sqrt{2}}{2\sqrt{2}\cdot\sqrt{2}}= \frac{30+4\sqrt{2}}{4}=\frac{15+2\sqrt{2}}{2} $$ One can do it differently: set $a=\sqrt{2}$, so you can write
$$ \frac{15\sqrt{2}+4}{2\sqrt{2}}= \frac{15a+a^4}{a^3}= \frac{15+a^3}{a^2}= \frac{15+2\sqrt{2}}{2} $$
The final result can also be written by using $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$ so $$ \frac{15+2\sqrt{2}}{2}=\frac{15}{2}+\sqrt{2} $$ Whether you want to do this last transformation depends on what you have to do with this number.
Why shouldn't it be $+4$ at the end of the second expression? That should carry down to the third. From the second to the third you are just splitting one fraction into two.