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How do you know which factor to use?

  1. Dec 21, 2007 #1
    How do you know which factor to use???

    I know what factoring is. Basically, you take any given number or expression, and rewrite it as a product of its factors.

    For instance in 3xy + 9y, every term is divisable by "3y", so we could rewrite it as 3y(x + 3).

    But, consider: [tex]\displaystyle{28xy^2 - 14x}[/tex]

    Every term there is divisible by "14x", so is there something at all wrong with: [tex]\displaystyle{14x(2y^2 - 1)}[/tex]???

    That's how I would factor it, but the book instead choose to use "7x" as the factor, and uses [tex]\displaystyle{7x(4y^2 - 2)}[/tex] as the answer.

    If given an exam, how do you know which factor they are going to claim is the "correct" one????

    Thanks a lot.
  2. jcsd
  3. Dec 21, 2007 #2
    Neither factorization is incorrect, but if asked to simplify an expression by factoring, the book's answer would not be acceptable.
  4. Dec 21, 2007 #3

    So basically just go with the greatest possible factor?
  5. Dec 22, 2007 #4


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    Yes in general you should pull out all the factors that you can. In that book answer there is still a factor or 2 remaining inside the brackets, so you would typically lose marks for giving that answer.

    As slider pointed out neither of the two factorizations is incorrect. Both expressions are in a factorized form and both are equal to the original expressions. There could even be instances where you may prefer the incomplete factorization. Say you had to simplify

    [tex] \frac{28 x y^2 - 14x}{4y^2 - 2} [/tex]

    In this case the incomplete factorization of the numerator might lead to a slightly quicker solution.
    Last edited: Dec 22, 2007
  6. Dec 24, 2007 #5


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    Factorizations are not unique, and what counts as "simplest" may often be a matter of taste.

    One rule of thumb, though:
    1. In general, an expression is not simplified by introducing explicit fractions, radicals etc.
    For example,
    the valid factorization [tex]28xy^{2}-14x=23\sqrt{x}y(\frac{28}{23}\sqrt{x}y-\frac{14\sqrt{x}}{23y})[/tex] is not generally regarded as a simplification.
  7. Dec 24, 2007 #6


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    one does not know which factorization to use until one knows what ring the factorization occurs in. in general one wants to factor somehting into irreducible elements and ignore units.

    in the ring Z[X,Y], the units are 1 and -1, so one would have four irreducible factors

    2(7)X(2Y^2-1), but if the ring is Q[X,Y], then all non zero costants are units, your element will only have 2 irreducible factors, and hence the 2 and the 7 do not matter. so either of your factorizations is entirely acceptable and equivalent.

    i.e. two irreducible factors whose quotient is a unit are equivalent, so 2X, 7X, X, and 14X are all equivalent, so in fact X(28Y^2-14) is just as good as the others.

    on the other hand, in the ring Q(X,Y), which is a field, there is no need to factor it at all, since the original element is a unit.

    but it seems to me the people who wrote your book are not very knowledgeable and are not following these rules for factorizations but are just following some pattern they think is "nice" and simple. so it is difficult to say what they will prefer as an answer.
    Last edited: Dec 24, 2007
  8. Dec 24, 2007 #7


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    Ideally you want to reduce it into its prime factors. Can't get anything simpler than that. So yours would be correct.
  9. Dec 24, 2007 #8


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    as i said, the meaning of the word "prime" or "irreducible", depends on the containing ring, and what are its units, since units are not prime, but define an equivalence relation on primes.

    thus you want to factor into "primes" p , but where u.p and p are considered equivalent if u is a "unit", i.e. is invertible.

    thus in Z[X,Y], 2 is a prime, but not in Q[X,Y], and in Q(X,Y) not even X is a prime.
    Last edited: Dec 24, 2007
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