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    the question is, express x to the power of 2 divided by 4 - x to the power of 2 in partial fractions

    I got the answer to be

    1 divided by 2+x plus 1 divided by 2-x

    the answer has a minus one on its own in it? I dont really get how
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    (Original post by SunDun111)
    the question is, express x to the power of 2 divided by 4 - x to the power of 2 in partial fractions

    I got the answer to be

    1 divided by 2+x plus 1 divided by 2-x

    the answer has a minus one on its own in it? I dont really get how
    Remember that partial fractions can only be used when the degree of the numerator is less than the degree of the denominator. (The degree being the greatest power of x).
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    Is the question x^2/(4-x)^2 or x^2/(4-x^2)?
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    (Original post by SunDun111)
    the question is, express x to the power of 2 divided by 4 - x to the power of 2 in partial fractions

    I got the answer to be

    1 divided by 2+x plus 1 divided by 2-x

    the answer has a minus one on its own in it? I dont really get how
    (4-x)^2 is not equal to (2+x)(2-x) unless you mean 4-x^2

    If you have a squared bracket in the denominator you perform the brackets without the power in one fraction and then the entire thing as another. e.g. 2x/(x+1)^2 = A/(x+1) + B/(x+1)^2
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    (Original post by Math12345)
    Is the question x^2/(4-x)^2 or x^2/(4-x^2)?
    the first one
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    Whichever one it is:

    1. \frac{x^2}{4-x^2} = \frac{-(4-x^2)+4}{4-x^2} = -1 + \frac{4}{4-x^2}


    2. (4-x)^2=x^2-8x+16

    \frac{x^2}{(4-x)^2} = \frac{(4-x)^2+8x-16}{(4-x)^2} = 1 + \frac{8x-16}{(4-x)^2}

    (You need to use long divsion basically)

    Now use partial factions on the last fraction
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    (Original post by Vikingninja)
    (4-x)^2 is not equal to (2+x)(2-x) unless you mean 4-x^2

    If you have a squared number in the denominator you perform the brackets without the power in one fraction and then the entire thing as another.
    the power of 2, is inside, it is next to the x so its 4-x^2
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    (Original post by SunDun111)
    the power of 2, is inside, it is next to the x so its 4-x^2
    (Original post by SunDun111)
    "Is the question x^2/(4-x)^2 or x^2/(4-x^2)?"
    the first one
    You said in this quote that its (4-x)^2


    (Original post by Math12345)
    naughty boy
    Edit: One sec
    It's against the rules to give answers.
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    (Original post by Math12345)
    Whichever one it is:

    \frac{x^2}{4-x^2} = \frac{-(4-x^2)+4}{4-x^2} = -1 + \frac{4}{4-x^2}

    Edit:

    (4-x)^2=x^2-8x+16

    \frac{x^2}{(4-x)^2} = \frac{(4-x)^2+8x-16}{(4-x)^2} = 1 + \frac{8x-16}{(4-x)^2}

    (You need to use long divsion basically)

    Now use partial factions on the last fraction
    (Original post by Vikingninja)
    You said in this quote that its (4-x)^2



    It's against the rules to give answers.
    crap I didnt mean to say the power of 2 was outside the bracket, it is inside, in the answer book my answer is right but im missing a minus one? I am wondering where it comes from?
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    (Original post by SunDun111)
    crap I didnt mean to say the power of 2 was outside the bracket, it is inside, in the answer book my answer is right but im missing a minus one? I am wondering where it comes from?
    Use long division or inspection like I did. You can't use partial fractions since the degree of the numerator and denominator is the same.
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    (Original post by Math12345)
    Use long division or inspection like I did. You can't use partial fractions since the degree of the numerator and denominator is the same.
    Ok thanks
 
 
 
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