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- kaylala

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

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- kaylala

- chestercat

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- whpalmer4

What happens if you factor the numerator of this expression? Do you see any common factors after you do so?

- whpalmer4

\[a^3-a^2 = a^2(a-1)\]right?

- whpalmer4

Nothing appears to be a common factor with the denominator, does it?

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- whpalmer4

What if you now factor the denominator?

- kaylala

yes. @whpalmer4
see the image: (but there could be 2 answers; see the top and bottom image)

- kaylala

- kaylala

- whpalmer4

After you've factored both numerator and denominator you have the following:
\[\frac{a^2(a-1)}{(a-1)(a+1)} = \frac{a^2\cancel{(a-1)}}{(a+1)\cancel{(a-1)}} = \frac{a^2}{a+1}\]

- whpalmer4

the original restrictions on \(a\) still apply: \(a \ne \pm 1\)

- kaylala

oh ok thanks
@whpalmer4
by the way, what original restrictions?

- whpalmer4

\(a\ne \pm1\)
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 off-limits, just as it was in the original.

- whpalmer4

otherwise, it isn't equivalent — it's a function that is defined somewhere where the other one wasn't.

- kaylala

ah... okay thank you so much @whpalmer4

- whpalmer4

you're welcome.

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