Rational Expressions: Solving Homework with Difficulty

AI Thread Summary
The discussion focuses on solving the rational expression x² + x - 30 over x² - 3x + 2 and identifying when it is undefined. The key point is that a rational expression is not defined when the denominator equals zero. Participants emphasize the importance of factoring the denominator, x² - 3x + 2, to find the values of x that make it zero. The solution involves determining the roots of the trinomial, which can be found through factoring. Understanding these concepts is crucial for solving similar problems effectively.
Annie_D
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Homework Statement


x (squared) + x - 30
_____________________
x (sqaured) -3x + 2

Find all numbers for which the rational expression is not defined.

i know this is really easy stuff, but i have a hard time staying focused in math, and i have to always teach myself with the book, but sometimes the book isn't very informative.

I have the answer, but don't know how to go about doing these problems. I know you like people to attempt them, but i don't even know where to begin after factoring the 30 .
 
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When is a rational expression (a fraction) not defined? For what value is division impossible?
 
0, i got in now though. thanks anyway.
 
So for what x will x2 - 3x + 2 = 0? You need to factor this trinomial to answer that question.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
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