Solving Algebraic Expression with Limits

  • Thread starter krbs
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In summary, the problem involves simplifying the expression (x3-6x2+12x-8)/(x2-4x+4) to (x-2)3/(x-2)2. This can be done by finding the roots of the denominator and using polynomial long division to divide out the common factor (x-2). Alternatively, with practice, one can directly see or guess that the numerator is (x-2)3 and skip the steps of finding the roots and using polynomial long division.
  • #1
krbs
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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.
 
  • #3
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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  • #4
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?
 
  • #5
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.
 
  • #6
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!
 

Related to Solving Algebraic Expression with Limits

1. What is simple algebra?

Simple algebra is a branch of mathematics that deals with the manipulation and solving of equations using letters and symbols to represent unknown quantities.

2. What are the basic operations in algebra?

The basic operations in algebra are addition, subtraction, multiplication, and division. These operations are used to manipulate equations and solve for unknown variables.

3. How do I solve a simple algebraic equation?

To solve a simple algebraic equation, you must isolate the variable on one side of the equation by using inverse operations. This means performing the opposite operation to both sides of the equation until the variable is alone on one side.

4. What is the order of operations in algebra?

The order of operations in algebra is the same as in regular arithmetic: parentheses, exponents, multiplication and division (from left to right), and addition and subtraction (from left to right).

5. How is algebra used in real life?

Algebra is used in many real-life situations, such as calculating discounts, determining the dosage of medicine, and managing finances. It is also used in many fields of science and engineering to solve complex problems and make predictions.

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