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

1. whpalmer4 Group Title

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

2. whpalmer4 Group Title

$a^3-a^2 = a^2(a-1)$right?

3. whpalmer4 Group Title

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

4. whpalmer4 Group Title

What if you now factor the denominator?

5. kaylala Group Title

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

6. kaylala Group Title

7. kaylala Group Title

8. whpalmer4 Group Title

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

9. whpalmer4 Group Title

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

10. kaylala Group Title

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

11. whpalmer4 Group Title

$$a\ne \pm1$$ 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.

12. whpalmer4 Group Title

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

13. kaylala Group Title

ah... okay thank you so much @whpalmer4

14. whpalmer4 Group Title

you're welcome.