1.1.21 simplify rational expression

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  • #1
karush
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simplify
$\left(\dfrac{a^2b^3-2a^{-3}b^3}{2a}\right)^2=$
OK this could get confusing quickly
but I don't think we want to square it first
sscwt.png
 
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Answers and Replies

  • #2
HOI
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Yes, it can be simplified by cancelling "a" before squaring.
(The "2" in the denominator is already simple.)
$\frac{\left(ab^3-2a^{-4}b^3\right)^2}{4}$

Of course we can also factor out that $b^3$ that is in each term. Squaring it gives $b^6$.

$\frac{b^6(a- 2a^{-4})^2}{4}$

Now it is relatively easy to square $a- 2a^{-4}$.
$(a- 2a^{-4})^2= a^2- 2(a)(2a^{-4})+ (2a^{-4})^2$
$= a^2- 4a^{-3}+ 4a^{-8}$

$\frac{b^6(a^2-4a^{-3}+4a^{-8}}{4}$.
 
  • #3
karush
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ok we still have negative exponents in the numerator??
 
  • #4
HOI
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Is that a problem? Different people have different ideas of what "simple" is. If you don't want negative exponents, remember what a negative exponent means:
$\frac{b^2- 4a^{-3}+ 4a^{-8}}{4}= \frac{b^2- 4\frac{1}{a^{3}}+ 4\frac{1}{a^{8}}}{4}$

Now, to eliminate the "fractions in fractions", multiply both numerator and denominator by $a^8$
$\frac{a^8b^3- 4a^5+ 4}{4a^8}$
 
  • #5
karush
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well the few examples I saw they brought the terms with negative exponents in the numerator to the denumerator
 
  • #6
HOI
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That is exactly what I did!
 
  • #7
karush
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That is exactly what I did!

yes but you created fractions over over a fraction $ \dfrac{a^{-3}}{4}=\dfrac{1}{4a^3}$

anyway,, just being cranky
 
  • #8
HOI
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No, [tex]a^{-3}[/tex] is not a fraction!
 
  • #9
karush
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why not the exponent is positive
 
  • #10
Prove It
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No, [tex]a^{-3}[/tex] is not a fraction!

It is when it's written as $\displaystyle \begin{align*} \frac{1}{a^3} \end{align*}$...
 
  • #11
HOI
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Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.
 
  • #12
karush
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numerals?
 
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  • #13
Prove It
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Yes, $\frac{1}{x^3}$ is a fraction. $x^{-3}$ is NOT.
It is a matter of the difference in numerals, not numbers.

Do you realize that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?
 
  • #14
karush
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they just said move the terms with negative exponents in the numerator to the denominator
the bell rang and I left
no fancy stuff
 
  • #15
HOI
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Do you realize that you are saying that two amounts that are equivalent are simultaneously a fraction and not a fraction?
Yes, I am. They are two different numerals that repreent the same number. The term "fraction" has to do with the way a number is represented, it is not a property of the number itself.
 

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