How does $G\frac{\frac{1}{2}m_{1}\cdot m_{2}}{\frac{1}{4}r^2}$ become $8\left(G\frac{m_{1}\cdot m_{2}}{r^2}\right)$? I'm busy doing a highschool physics exercise which involves some algebra, and I'm having trouble seeing how they got from one step to the next. Here's what it looks like:
$$
...\\
F_{g}=G\frac{\frac{1}{2}m_{1}\cdot m_{2}}{\frac{1}{4}r^2}\\
F_{g}=8\left(G\frac{m_{1}\cdot m_{2}}{r^2}\right)\\
...
$$
I just can't figure out how they got from the first line to the next? Like, I can factor out the $\frac{1}{2}$ from the numerator, ending up with $F_{g}=\frac{1}{2}(G\frac{m_{1}\cdot m_{2}}{\frac{1}{4}r^2})$, but I'm stuck on how to get the $\frac{1}{4}$ out of the denominator.
Here's a screenshot of it from my book. I noticed that the numerator doesn't actually have the second mass, like how I copied it over to my question, but I'm handling that as a typo lol.

P.S. I'm not sure how to tag this question. Feel free to retag it, if there's better ones.
 A: \begin{eqnarray}
F &=& G \frac{\frac{1}{2}M}{\frac{1}{4}r^2} \nonumber \\
&=& 1 \times G\frac{\frac{1}{2}M}{\frac{1}{4}r^2} \nonumber \\
&=& \frac{16}{16} \times G\frac{\frac{1}{2}M}{\frac{1}{4}r^2} \nonumber \\
&=& ...
\end{eqnarray}
and you won't get $8(G\frac{M}{r^2})$. Then try with $F = G \frac{\frac{1}{2}M}{(\frac{1}{4}r)^2}$.
Personally  I don't know where the error occurs, depends on the problem.
A: You should multiply top and bottom by $4$ to get $F_g = 2G\frac {m_1m_2}{r^2}$ so that answer seems in error.
However if the equation where $F_g =G\frac {\frac 12m_1m_2}{(\frac 14r)^2}$ you multiply top and bottom by $4^2$ to get $F_g=8\frac {m_1m_2}{r^2}$.
But the text you quote seems to imply that Beeble is a planet with have the mass of Earth.  This would imply by $r^3\sim mass$ (mass increases cubically as radius increase) so if $M_B= \frac 12 M_E$ then $r_B$ ought to be equal to $\sqrt[3]{\frac 13}r_E$ and we'd get $F_g =G\frac {M_B}{r_B^2} = G\frac {\frac 12 M_E}{(\sqrt[3]{\frac 12}r_E)^2}=G\frac{\frac 12M_E}{(\frac 12)^{\frac 23}r_E^2}=(\frac 12)^{\frac 13}\frac {M_e}{r_E^2}$.
But maybe Beeble does not have the same density as earth.  If Beeble has half the mass of Earth and half the radius as Earth (then Beeble is $8$ times as dense as Earth) then wed have $F_g =G\frac {\frac 12m_E}{(\frac 12r)^2}= 2G\frac {m_E}{r_E^2}$.
But if Beeble has half mass of Earth and $\frac 14$ the radius of Earth (then Beeble is $32$ times as dense as Earth) then we would have $F_q = G\frac {\frac 12 m_E}{(\frac 14r)^2} = 8G\frac {m_E}{r_E}$.
I imagine this last bit is what the book had in mind.  (Although I've never heard of Beeble so I don't know.)
Anyway the book is in error, but probably because it meant $(\frac 14r)^2$ instead of $\frac 14r^2$.  Parenthesis matter.... a lot.
