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

1. whpalmer4

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

2. whpalmer4

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

3. whpalmer4

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

4. whpalmer4

What if you now factor the denominator?

5. kaylala

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

6. kaylala

7. kaylala

8. whpalmer4

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

9. whpalmer4

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

10. kaylala

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

11. 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.

12. whpalmer4

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

13. kaylala

ah... okay thank you so much @whpalmer4

14. whpalmer4

you're welcome.