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The rational expression
a^3  a^2

a^2  1
where a ≠ ±1 can be simplified as WHAT?
TOPIC: Rational Expressions
 9 months ago
 9 months ago
The rational expression a^3  a^2  a^2  1 where a ≠ ±1 can be simplified as WHAT? TOPIC: Rational Expressions
 9 months ago
 9 months ago

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whpalmer4Best ResponseYou've already chosen the best response.1
What happens if you factor the numerator of this expression? Do you see any common factors after you do so?
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
\[a^3a^2 = a^2(a1)\]right?
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
Nothing appears to be a common factor with the denominator, does it?
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
What if you now factor the denominator?
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
yes. @whpalmer4 see the image: (but there could be 2 answers; see the top and bottom image)
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
After you've factored both numerator and denominator you have the following: \[\frac{a^2(a1)}{(a1)(a+1)} = \frac{a^2\cancel{(a1)}}{(a+1)\cancel{(a1)}} = \frac{a^2}{a+1}\]
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
the original restrictions on \(a\) still apply: \(a \ne \pm 1\)
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
oh ok thanks @whpalmer4 by the way, what original restrictions?
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
\(a\ne \pm1\) if a=1 or a = 1, the denominator of the original expression would equal 0, and the result dividing by 0 is undefined...for the new expression, in theory we could have a = 1 without any problem (the denominator would equal 2, and dividing by 2 is a commonly accepted practice :) but we are trying to make a simpler but equivalent version of the original expression, and so to make it 100% equivalent, we have to put a=1 as offlimits, just as it was in the original.
 9 months ago

whpalmer4Best ResponseYou've already chosen the best response.1
otherwise, it isn't equivalent — it's a function that is defined somewhere where the other one wasn't.
 9 months ago

kaylalaBest ResponseYou've already chosen the best response.0
ah... okay thank you so much @whpalmer4
 9 months ago
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