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sam (.sam.):
You can factor the x from the denominator
sam (.sam.):
Which x will be cancelled with numerator and denominator
OpenStudy (anonymous):
so it would be something without any xes?
sam (.sam.):
There will be an x left over, look at this
\[\frac{x}{4x+x^2} \\ \\ =\frac{x}{x(4+x)} \\ \\ =\frac{1}{4+x}\]
OpenStudy (anonymous):
Okay, but could it be 1/x+4 as well?
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sam (.sam.):
yes
OpenStudy (anonymous):
And that is the answer? Ia m so bad at this. I am doing review for my test and just trying to get some help.
sam (.sam.):
Yea you need to practice more of these especially factoring, you have to make use of cancelling terms in numerators and denominators
OpenStudy (anonymous):
Alright, thanks so much! Do you think you can help me with one more?
OpenStudy (anonymous):
@.Sam.
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sam (.sam.):
ok
OpenStudy (anonymous):
Simplify. -14x^3/x^3-5x^4
sam (.sam.):
Again, factor x^3 from denominator, then cancel x^3 for top and bottom
OpenStudy (anonymous):
I get -14/-5x^4?
sam (.sam.):
When you factor you gotta remember that both of the terms in the denominator must be factored out
\[\frac{-14x^3}{x^3-5x^4} \\ \\ =\frac{-14x^3}{x^3(1-5x)} \\ \\ =\frac{-14}{1-5x}\]
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