Solving Algebraic Expression with Limits

  • Thread starter Thread starter krbs
  • Start date Start date
  • Tags Tags
    Algebra
AI Thread Summary
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.
krbs
Messages
51
Reaction score
3
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
 
Last edited:
Physics news on Phys.org
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.
 
  • Like
Likes krbs
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 picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top