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\[G \frac{ Mm }{ x^2 }-G \frac{ Lm }{(C- x)^2 }\]
factor after taking common terms out..
IDK how
G G G G G G G
haha
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Common term :d
x and m too!
m is also common
Was going to say m haha.
x? :O
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you want to factor for solving x ?
if so, u may use quadratic formula after pulling out GCF
\[G \frac{ Mm }{ x^2 }-G \frac{ Lm }{(C- x)^2 } = \dfrac{Gm}{x^2(C-x)^2}\left(M(C-x)^2 - x^2L \right)\]
oh save the quadratic formula, we use the identity : \(a^2 - b^2 = (a+b)(a-b)\)
\[G \frac{ Mm }{ x^2 }-G \frac{ Lm }{(C- x)^2 } = \dfrac{Gm}{x^2(C-x)^2}\left(M(C-x)^2 - x^2L \right) \\ = \dfrac{Gm}{x^2(C-x)^2}\left( \sqrt{M}(C-x) + x\sqrt{L}\right)\left( \sqrt{M}(C-x) - x\sqrt{L} \right)\]
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