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Precalculus

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Hello,

Thanks for all of the help guys. I really appreciate it.

I was woudnering if you could show me how to do this step by step and the thought process that goes through your head while you do this. This way I will be able to try to solve some of these problems on my own becasue I really have no idea how to do them.

Perform the indicated operation and simplify the ersult. Leave your anser in factored form.

1/x - 2/(x^2 + x) + 3/(x^3 - x^2)

THANK YOU!!!

  • Precalculus -

    remember to add/subtract fractions we need a common denominator.

    let's write our expression in factored form
    1/x - 2/(x^2 + x) + 3/(x^3 - x^2)
    = 1/x - 2/[x(x+1)] + 3/[x^2(x-1)]

    so the common denominator is x^2(x+1)(x-1)

    first term: x(x+1)(x-1)/[x^2(x+1)(x-1)]
    notice if we cancel we get 1/x

    second term: -2x(x-1)/[x^2(x+1)(x-1)]
    again, see what you get when you cancel

    third term: 3(x+1)/[x^2(x+1)(x-1)]

    so we get
    [x(x+1)(x-1) - 2x(x-1) + 3(x+1)]/[x^2(x+1)(x-1)]

    I will let you finish it,

    (I got (x^3 - 2x^2 + 4x + 3)/[x^2(x^2 - 1)]

  • Precalculus -

    I bet you have a sign wrong but will do as shown.
    1/x
    -2/[x(x+1)]
    +3/[x^2(x-1)]
    we need a common denominator
    need LCD = x^2 (x-1)(x+1) = x^2(x^2-1)
    so
    1[(x)(x+1)(x-1)]/LCD
    -2 [ (x)(x-1) ] /LCD
    +3 [ (x+1 ] / LCD

    then
    [1 (x(x^2-1) -2 (x^2-x) +3 (x+1) ]/LCD
    [ x^3-x -2x^2 +2x +3x+3 ]/LCD
    (x^3 -2x^2 +4x +3)/[x^2(x^2-1)]

    I could not simplify, suspct mistake but that is the idea.

  • Precalculus -

    Oh, I see Reiny got the same answer. If you typed it right the answer is right.

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