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    https://youtu.be/pOH13db9qi0?t=5m41s

    I know this is pretty basic stuff but could someone explain why on the numerator he gets  4(x-2)(2x+5) and not  4(x-2)^2(2x+5) and so on?
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    (Original post by AxSirlotl)
    https://youtu.be/pOH13db9qi0?t=5m41s

    I know this is pretty basic stuff but could someone explain why on the numerator he gets  4(x-2)(2x+5) and not  4(x-2)^2(2x+5) and so on?
    Because theres already an (x-2) in the denominator, so you have to multiply top and bottom by a factor (same as multiplying by 1) that makes the common denominator essentially.
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    The first term is:

     \dfrac{4}{x-2}.

    To place over a common denominator:

     (x-2)^{2} (2x+5),

    we need:

     \dfrac{4}{(x-2)}\, \ast \, \dfrac{(x-2)(2x+5)}{(x-2)(2x+5)}.

    Does this help? :-)
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    (Original post by NotNotBatman)
    Because theres already an (x-2) in the denominator, so you have to multiply top and bottom by a factor (same as multiplying by 1) that makes the common denominator essentially.
    (Original post by simon0)
    The first term is:

     \dfrac{4}{x-2}.

    To place over a common denominator:

     (x-2)^{2} (2x+5),

    we need:

     \dfrac{4}{(x-2)}\, \ast \, \dfrac{(x-2)(2x+5)}{(x-2)(2x+5)}.

    Does this help? :-)
    Ok thanks, I think I understand now
 
 
 
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