Factoring a difficult polynomial so that I can extend equation

AI Thread Summary
To extend the function and ensure continuity at x=2, the polynomial x^3 - 4x^2 - 11x + 30 must be factored. The correct factors are (x-5)(x-2)(x+3), but the user is struggling with the factoring process. The rational roots theorem suggests that potential factors must divide 30, leading to a guess that (x-2) is a factor. Dividing the polynomial by (x-2) yields a quadratic, which can then be factored to find the remaining factors. This approach clarifies the necessary steps to simplify the function and remove the undefined point.
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Homework Statement


Give a formula for the extended function that is continuous at the indicated point.

(x3-4x2-11x+30)/(x2-4), x=2


Homework Equations





The Attempt at a Solution


I know that I have to factor the top and bottom so that I can cancel terms that cause the function to be undefined at x=2, and I know how to do everything after that. However, I'm stuck on how to factor x3-4x2-11x+30. I know that the factors are (x-5)(x-2)(x+3), but I don't know how to go about getting there. I tried grouping, but it didn't work since there is no way that I see to factor -11x+30.
x3-4x2-11x+30

x2(x-4)-11x+30

I tried grouping other ways, but they did not work either.
 
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In general, what you want is the rational roots theorem. If (x-r) is a factor of the numerator then r must divide 30. That gives a limited number of guesses for possible factors. But in this particular case, it's a pretty good guess that (x-2) is factor. Once you know that, divide the numerator by (x-2) to get a quadratic. Then factor the quadratic to get the other two factors.
 
Oh gee, I completely forgot about that. (It's been a while since Algebra II) That makes total sense. Thank you so much for your help!
 
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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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