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kaylala
 one year ago
The rational expression
a^3  a^2

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

This Question is Closed

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1What happens if you factor the numerator of this expression? Do you see any common factors after you do so?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1\[a^3a^2 = a^2(a1)\]right?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1Nothing appears to be a common factor with the denominator, does it?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1What if you now factor the denominator?

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

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1After you've factored both numerator and denominator you have the following: \[\frac{a^2(a1)}{(a1)(a+1)} = \frac{a^2\cancel{(a1)}}{(a+1)\cancel{(a1)}} = \frac{a^2}{a+1}\]

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1the original restrictions on \(a\) still apply: \(a \ne \pm 1\)

kaylala
 one year ago
Best ResponseYou've already chosen the best response.0oh ok thanks @whpalmer4 by the way, what original restrictions?

whpalmer4
 one year ago
Best 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.

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.1otherwise, it isn't equivalent — it's a function that is defined somewhere where the other one wasn't.

kaylala
 one year ago
Best ResponseYou've already chosen the best response.0ah... okay thank you so much @whpalmer4
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