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Math Help - Stuck on quotient of two polynomials

  1. #1
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    Stuck on quotient of two polynomials

    Hello

    (x-a)(x-b) / (c-a)(c-b) + (x-a)(x-c) / (b-a)(b-c)
    = (x-a)(x-b)(b-a) - (x-a)(x-c)(c-a) / (c-a)(c-b)(b-a)

    My textbook suggests the absence of (b-c) in the denominator is accounted by the fact that b-c = -(c-b) . I understand this to imply that (c-b) and (b-c) cancel each other out. Yet the denominator is absent only of (b-c). (c-b) is still there. Would someone please help me understand what's going on here?
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  2. #2
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    Re: Stuck on quotient of two polynomials

    In other words, my text book says:

    b - c = -(c - b)

    That seems obvious enough.

    It also says:

    a / b + c / d = ad + bc / bd

    That seems clear too. It then goes onto say:

    (x-a)(x-b) / (c-a)(c-b) + (x-a)(x-c) / (b-a)(b-c)
    = (x-a)(x-b)(b-a) - (x-a)(x-c)(c-a) / (c-a)(c-b)(b-a)

    There are 2 things I don't understand about this. One is that the first expression is adding fractions and yet in the second it is subtracting them. I don't see how it turned into a subtraction.

    The second thing I don't understand is how it got the denominator (c-a)(c-b)(b-a) from (c-a)(c-b) and (b-a)(b-c).
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  3. #3
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    Re: Stuck on quotient of two polynomials

    The basic strategy is to make the two denominators the same, after which the two fractions can be combined as a single fraction.

    The first step, as you have already indicated, is to deal with the (c-b) and (b-c) terms.
    They can be made the same by writing (b-c) as -(c-b).

    Your original problem

    \frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-a)(x-c)}{(b-a)(b-c)}

    then becomes

    \frac{(x-a)(x-b)}{(c-a)(c-b)}-\frac{(x-a)(x-c)}{(b-a)(c-b)}.

    For the next bit, consider a numerical problem, say \frac{5}{7}+\frac{2}{3}.

    To make the two denominators the same, multiply top and bottom of the first fraction by 3 and top and bottom of the second fraction by 7.

    That gets you

    \frac{5\times 3}{7 \times 3}+ \frac{2 \times 7}{3 \times 7} = \frac{15}{21}+\frac{14}{21}.

    Now you can simply add the two to get \frac{(15+14)}{21} = \frac{29}{21}.

    For your algebraic problem, you do pretty much the same thing, multiply top and bottom of the first fraction by (b-a), top and bottom of the second fraction by (c-a).
    The two fractions can then be combined as a single fraction.
    Thanks from topsquark
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    Re: Stuck on quotient of two polynomials

    Thank you, BobP. My problem was not that I did not know the solution. My text book has the solution, and the main staging posts to the solution. But that I did not understand how you get between the staging posts. That is, I needed help to understand how it works.

    So I can see that (b - c) is the same as -(b - c). It is very helpful to know this is the first thing you do. It seems from your application of this ("Your original problem... then becomes...") that the minus power of the -(b - c) is multiplied through the whole right hand side of the expression, which would explain how it changes from a + to a -. Do I understand this correctly?

    But this next bit does not make sense to me. You are not multiplying the whole denominator on each side but only that part of it that is different. Why would you not multiply the whole of each denominator? I can see how it might be possible that as each denominator already has (c - b) you wouldn't need to multiply those along the bottom again. But why do you not still need to multiply it across the top of both cases?
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  5. #5
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    Re: Stuck on quotient of two polynomials

    Hi Spod

    Sorry for the delay, I've been away from home for the last week or so.

    In your second paragraph, you use the word power. There is no power involved, it's a multiplication by minus 1.
    That is, it is simply being said that (b-c) = (-1)\times(c-b), but we don't normally bother writing the 1. Multiplying either the top or the bottom of a fraction by minus 1 causes the whole fraction to change sign, which is what is happening in this example. As far as this process is concerned, the first fraction is totally separate and has no relevance, it could be anything.

    I don't understand your third paragraph. What do you mean by ' the whole denominator on each side ' ?
    Trying to guess what you mean, remember first that if several things are multiplied together, the order in which they are written is not important. So for example
    (x-a)(x-b)(x-c) is the same as (x-c)(x-b)(x-a) is the same as (x-b)(x-c)(x-a) etc. .

    Suppose now that you have two fractions, the first having denominator (x-a)(x-b)(x-c) and the second having denominator (x-b)(x-p). As far as the process of adding (or subtracting) the two fractions is concerned the numerators are not relevant, (we are assuming that each has been cancelled down to its simplest form). In order to add the two fractions, the process we are using requires us to make the two denominators identical. So ask yourself what is there in the second denominator that isn't in the first ? The answer is (x-p) so that is what we multiply top and bottom of the first fraction by, (and it doesn't matter whether you write the resulting denominator as (x-a)(x-b)(x-c)(x-p) or (x-p)(x-a)(x-b)(x-c), they're the same thing).
    Repeat the question for the second fraction and the answer is (x-a)(x-c), so that's what top and bottom of the second fraction gets multiplied by.
    The two denominators will now be the same and the two fractions can be combined as a single fraction.
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