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x^2-9/x^2+2x * x+2/x-3 @kc_kennylau
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Alot tougher huh :)
\[\Large\frac{x^2-9}{x^2+2x}\cdot\frac{x+2}{x-3}\] Why can't you just put parentheses lol
ok factorize \(x^2-9\) first by using \(a^2-b^2=(a-b)(a+b)\) :)
Then factorize \(x^2+2x\) by taking out the common factor :)
Oh the perfect square formula
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oh you call it the perfect square formula? interesting name :p
(x+3)(x-3)
right
then factorize \(x^2+2x\) by taking out the common factor :)
x(x+2)
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ahhhh
now they cancel!
yep :D So now we got \(\dfrac{(x+3)(x-3)}{x(x+2)}\cdot\dfrac{x+2}{x-3}\)
yep they cancel :D
AWESOME!
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i love math!
good :D keep up this passion :D
alirghty! college algebra final here i come!
Ill tag you if i need any more help.
ok :D
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