Solving Algebraic Expression with Limits

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To simplify the expression (x³ - 6x² + 12x - 8)/(x² - 4x + 4), first identify the roots of the denominator, which factors to (x - 2)². For the numerator, guessing a root, such as 2, allows for factoring out (x - 2) using polynomial long division. The numerator can be expressed as (x - 2)³ after factoring. This process helps in understanding limits as the expression can be simplified further. Polynomial long division is recommended unless the factors are easily recognizable.
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Hi guys, there's a sample problem in m textbook where they simplify an expression from x3-6x2+12x-8/x2-4x+4 to (x-2)3/(x-2)2. Can you explain how they solved this? For reference, I'm learning about limits
 
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I moved the thread to our homework section as the problem is homework-like.

I guess you mean (x3-6x2+12x-8)/(x2-4x+4). For the denominator, you should be able to find roots, once you know where the denominator gets zero you can also write it as product (here: (x-2)(x-2)). For the numerator, guess a root, then take it out as factor and compute the other factor, then do the same as for the denominator.
 
Knowledge of the fact that one root of a cubic polynomial with integer coefficients is a divisor of the constant term (here 8) will help in guessing.
 
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mfb said:
I moved the thread to our homework section as the problem is homework-like.

I guess you mean (x3-6x2+12x-8)/(x2-4x+4). For the denominator, you should be able to find roots, once you know where the denominator gets zero you can also write it as product (here: (x-2)(x-2)). For the numerator, guess a root, then take it out as factor and compute the other factor, then do the same as for the denominator.

Uh, yeah, I missed the big notice at the top... Had a couple glasses of wine :oops::H

Ok, so I guessed root 2 (lol). Would I then just divide out factor (x-2), like polynomial long division, or is there an easier way I'm overlooking?
 
Polynomial long division is the right approach - unless you directly see or guess that the numerator is (x-2)3 (possible with practice), then you can skip those steps.
 
mfb said:
Polynomial long division is the right approach - unless you directly see or guess that the numerator is (x-2)3 (possible with practice), then you can skip those steps.

Ok thank you!
 
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
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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