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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

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  1. whpalmer4
    • 2 years ago
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    What happens if you factor the numerator of this expression? Do you see any common factors after you do so?

  2. whpalmer4
    • 2 years ago
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    \[a^3-a^2 = a^2(a-1)\]right?

  3. whpalmer4
    • 2 years ago
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    Nothing appears to be a common factor with the denominator, does it?

  4. whpalmer4
    • 2 years ago
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    What if you now factor the denominator?

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

  6. kaylala
    • 2 years ago
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  7. kaylala
    • 2 years ago
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  8. whpalmer4
    • 2 years ago
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    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
    • 2 years ago
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    the original restrictions on \(a\) still apply: \(a \ne \pm 1\)

  10. kaylala
    • 2 years ago
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    oh ok thanks @whpalmer4 by the way, what original restrictions?

  11. whpalmer4
    • 2 years ago
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    \(a\ne \pm1\) 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
    • 2 years ago
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    otherwise, it isn't equivalent — it's a function that is defined somewhere where the other one wasn't.

  13. kaylala
    • 2 years ago
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    ah... okay thank you so much @whpalmer4

  14. whpalmer4
    • 2 years ago
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    you're welcome.

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