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Math Help - Can someone explain..

  1. #1
    anx
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    Can someone explain..

    I have the answer, but don't understand how it works.

    For all, r ≠ 2, r - 5r + 6 = ?
    _______________r - 4

    r - 3
    r + 2 is the answer, but I have no idea how I am supposed to get it.

    Thanks!~
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  2. #2
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    Quote Originally Posted by anx View Post
    I have the answer, but don't understand how it works.

    For all, r ≠ 2, r - 5r + 6 = ?
    _______________r - 4

    r - 3
    r + 2 is the answer, but I have no idea how I am supposed to get it.

    Thanks!~
    We start by factoring the numerator and the denominator

    \frac{r^2-5r+6}{r^2-4}=\frac{(r-2)(r-3)}{(r-2)(r+2)}

    Now we can reduce out the common factor of (r-2) to get

    \frac{r-3}{r+2}

    Note: that even though the factor r-2 reduced out r \ne 2

    becuase we would still be dividing by zero in the original equation.

    I hope this clears it up.
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  3. #3
    anx
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    Thank you, but that is not exactly the way for me to figure it out.

    Say I don't have the answer in the first place, and simply need to answer the original question, how would i find

    r - 3
    r + 2

    I only said i had the answer in the first place because I am doing a practice test which includes the answers, which usually lets me figure out things backwards, but in this case I do not know how, and certainly will not have the answer to do so on the real test.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by anx View Post
    Thank you, but that is not exactly the way for me to figure it out.

    Say I don't have the answer in the first place, and simply need to answer the original question, how would i find

    r - 3
    r + 2

    I only said i had the answer in the first place because I am doing a practice test which includes the answers, which usually lets me figure out things backwards, but in this case I do not know how, and certainly will not have the answer to do so on the real test.
    I didn't start with the answer I started with the expression.

    \frac{r^2-5r+6}{r^2-4}

    If we factor this, as in my above post, we get

    \frac{(r-2)(r-3)}{(r-2)(r+2)}

    From here we can reduce the fraction by eliminating common factors

    \frac{(r-3)}{(r+2)}

    Do you not know how to factor the expression?

    Please be clear with your questions.
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  5. #5
    anx
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    Quote Originally Posted by TheEmptySet View Post
    Do you not know how to factor the expression?
    I suppose not. This is probably why I don't understand it. All I see is 0/0 every time.
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  6. #6
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    Quote Originally Posted by anx View Post
    I suppose not. This is probably why I don't understand it. All I see is 0/0 every time.

    I think I see your problem:

    We are not using the values r=2 or r=-2.

    In fact they are not in the domain of the function so we are not allowed to use them.

    lets look at the numerator for a second...

    r^2-5r+6

    We need to factor this expression loosely speaking un Distribute or FOIL it

    To do this we need to compare it to the general form

    ar^2+br+c

    we can identify a=1 b=-5 and c=6.

    Since a=1 when just need to find factors of c (6) that add up to b (-5)

    We start by looking at all of the factors of 6 they are

    1\cdot 6 \\\ 2\cdot 3

    We know that since c is postitive and b is negative that both of our factors need to be negative.

    (-2)(-3)=6 \\\ (-2)+(-3)=-5 So it seems that these work so we can factor r^2-5r+6=(r-3)(r-2)

    We can check that the factorization is correct by multiplying it out and seeing if it equals what we started with so lets check

    (r-3)(r-2)=r^2-2r-3r+6 when we collect the like terms we end up with r^2-5r+6 so our factorization checks.


    Note: that the denominator can be factored this way, but it is in a special form called the difference of squares that can always be factored using this formula

    a^2-b^2=(a+b)(a-b)

    Hint: for yours remember that 4=2^2

    Good luck.
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  7. #7
    anx
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    Well I understand what you did now. Hopefully I can apply it when I need to, but I don't have another sample problem like that to try it. ><
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