Simplifying x / [ (x^-3) - (2x^2) ] into Positive Indices

Z

Zed55

New member
Joined
Apr 15, 2016
Messages
2
Hey guys,

Been reading the forums for a little while but this is my first post. Hope I do it right!

Basically I have been doing some simplifying and expressing in terms of positive indices but one particular question I am having trouble expressing as positive indices. Hopefully someone can help!

Here it basically is:

x / [ (x^-3) - (2x^2) ]

Basically my first step has been to take the x on top out to simplify, leaving me with:

1 / [ (x^-4) - (2x) ]

But obviously this isn't positive. How do I get first part of denominator positive while keeping the second half positive? Thanks in advance.
 
Zed55 said:
Hey guys,

Been reading the forums for a little while but this is my first post. Hope I do it right!

Basically I have been doing some simplifying and expressing in terms of positive indices but one particular question I am having trouble expressing as positive indices. Hopefully someone can help!

Here it basically is:

x / [ (x^-3) - (2x^2) ]

Basically my first step has been to take the x on top out to simplify, leaving me with:

1 / [ (x^-4) - (2x) ]

But obviously this isn't positive. How do I get first part of denominator positive while keeping the second half positive? Thanks in advance.
Click to expand...

What do you get - if you multiply the numerator and the denominator by x3.
 
Zed55 said:
I have been doing some simplifying and expressing in terms of positive indices....

x / [ (x^-3) - (2x^2) ]
Click to expand...
You have:

. . . . .\(\displaystyle \dfrac{x}{x^{-3}\, -\, 2x^2}\)

A good first step might be to convert the one negative exponent ("index") to positive:

. . . . .\(\displaystyle \dfrac{x}{\dfrac{1}{x^3}\, -\, 2x^2}\)

Then work with this as a "complex" fraction (here) or just work with the fractions:

. . . . .\(\displaystyle \dfrac{x}{\dfrac{1}{x^3}\, -\, \dfrac{2x^5}{x^3}}\)

. . . . .\(\displaystyle \dfrac{x}{\left(\dfrac{1\, -\, 2x^5}{x^3}\right)}\)

. . . . .\(\displaystyle \dfrac{\left(\dfrac{x}{1}\right)}{\left(\dfrac{1\, -\, 2x^5}{x^3}\right)}\)

Then flip and multiply, etc, etc. ;)
 
stapel said:
You have:

. . . . .\(\displaystyle \dfrac{x}{x^{-3}\, -\, 2x^2}\)

A good first step might be to convert the one negative exponent ("index") to positive:

. . . . .\(\displaystyle \dfrac{x}{\dfrac{1}{x^3}\, -\, 2x^2}\)

Then work with this as a "complex" fraction (here) or just work with the fractions:

. . . . .\(\displaystyle \dfrac{x}{\dfrac{1}{x^3}\, -\, \dfrac{2x^5}{x^3}}\)
Click to expand...
Denis said:
...or, at this point, multiply numerator and denominator by x^3:

x^4 / (1 - 2x^5)
Click to expand...
Subhotosh Khan said:
Or at the beginning!!!!
Click to expand...
Doing this "at this point" or "at the beginning" is, of course, the "'complex' fraction method" mentioned in the prior reply, the method for which an informational link was provided. The quoted portion of my reply was showing the other method. ;)
 
Zed55 said:
Here it basically is:

x / [ (x^-3) - (2x^2) ]
Click to expand...
Why not just: \(\displaystyle \left( {\dfrac{x}{{{x^{ - 3}} - 2{x^2}}}} \right)\left( {\dfrac{{{x^3}}}{{{x^3}}}} \right) = \dfrac{{{x^4}}}{{1 - 2{x^5}}}~?\)
 
Well that's so simple looking at it from both ways explained above. I should of really got that myself but thanks heaps guys!
 
You must log in or register to reply here.
Top