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Math Help - Factorising a complicated expression

  1. #1
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    Thumbs up Factorising a complicated expression

    thanks. how about this?
    factor completely
    y^2(x^4-z^4) - x^2(y^4-z^4) +z^2(y^4-x^4)
    i really have a hard time in factoring long expressions. is there any technique? thanks
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    i really have a hard time in factoring long expressions. is there any technique?
    Yes ...PRACTISE (I hope the spelling is right)
    Last edited by mr fantastic; January 13th 2009 at 11:25 PM.
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  3. #3
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    i didn't get the answer..

    my answer is (x-y)(x+y)(x-z)(x+z)(y-z)(y+z)
    is this correct??
    Last edited by mr fantastic; January 14th 2009 at 04:45 AM. Reason: Merge
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  4. #4
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    Quote Originally Posted by princess_21 View Post
    my answer is (x-y)(x+y)(x-z)(x+z)(y-z)(y+z)
    is this correct??
    Please reply showing the steps you used to arrive at the above, so that we can try to help you find the error(s).

    Thank you!
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  5. #5
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    Question

    Quote Originally Posted by princess_21 View Post
    i didn't get the answer..

    my answer is (x-y)(x+y)(x-z)(x+z)(y-z)(y+z)
    is this correct??

    x^4 y^2 - y^2 z^4 - x^2 y^4 + x^2 z^4 + y^4 z^2 - x^4 z^2

    x^2y^2 (x^2-y^2) + z^4 (x^2-y^2) - z^2 (x^4-y^4)
    (x^2 -y^2) (x^2y^2 -z^4 - x^2z^2 - y^2z^2)
    (x^2 -y^2) x^2(y^2-z^2) - z^2 (y^2-z^2)
    (x^2 -y^2) (x^2-z^2) (y^2-z^2)

    (x-y)(x+y)(x-z)(x+z)(y-z)(y+z)
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  6. #6
    Like a stone-audioslave ADARSH's Avatar
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    Its correct beyond any doubt and I must say You did well
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  7. #7
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    You started with y^2(x^4\, -\, z^4)\, -\, x^2(y^4\, -\, z^4)\, +\, z^2(y^4\, -\, x^4)

    Quote Originally Posted by princess_21 View Post
    x^4 y^2 - y^2 z^4 - x^2 y^4 + x^2 z^4 + y^4 z^2 - x^4 z^2
    In the above, you multiplied everything out to get:

    . . . . . x^4 y^2\, -\, y^2 z^4\, -\, x^2 y^4\, +\, x^2 z^4\, +\, y^4 z^2\, -\, x^4 z^2

    There appear to be some steps missing between the above and your next step:

    Quote Originally Posted by princess_21 View Post
    x^2y^2 (x^2-y^2) + z^4 (x^2-y^2) - z^2 (x^4-y^4)
    I think you regrouped and then factored out of pairs of terms, as:

    . . . . . x^4 y^2\, -\, x^2 y^4\, +\, x^2 z^4\, -\, y^2 z^4\, -\, x^4 z^2\, +\, y^4 z^2

    . . . . . x^2 y^2 (x^2\, -\, y^2)\, +\, z^4 (x^2\, -\, y^2)\, -\, z^2 (x^4\, -\, y^4)

    Then you factored the difference of squares in the last set of parentheses above, and factored " x^2\, -\, y^2" out front:

    . . . . . x^2 y^2 (x^2\, -\, y^2)\, +\, z^4 (x^2\, -\, y^2)\, -\, z^2 \left[(x^2\, -\, y^2)(x^2\, +\, y^2)\right]

    . . . . . (x^2\, -\, y^2)\left[x^2 y^2\, +\, z^4\, -\, z^2(x^2\, +\, y^2)\right]

    . . . . . (x^2\, -\, y^2)(x^2 y^2\, +\, z^4\, -\, x^2 z^2\, -\, y^2 z^2)

    You then paired and factored inside the second parentheses:

    . . . . . (x^2\, -\, y^2)(x^2 y^2\, -\, x^2 z^2\, -\, y^2 z^2\, +\, z^4)

    . . . . . (x^2\, -\, y^2)\left[x^2(y^2\, -\, z^2)\, -\, z^2(y^2\, -\, z^2)\right]

    . . . . . (x^2\, -\, y^2)\left[(y^2\, -\, z^2)(x^2\, -\, z^2)\right]

    Then you factored the differences of squares:

    . . . . . (x\, -\, y)(x\, +\, y)\left[(y\, -\, z)(y\, +\, z)(x\, -\, z)(x\, +\, z)\right]

    This wasn't the factored form I'd expected, but it looks to be correct. Good job!
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  8. #8
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    thanks, then what did you expect?
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  9. #9
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    Quote Originally Posted by princess_21 View Post
    what did you expect?
    I'd expected a sum of squares, at some point. But the "plus" and "minus" signs worked out such that this was avoided.
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