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Math Help - Factorisation

  1. #1
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    Factorisation

    Can someone tell me if i have factorized correctly:

    Equation is this: x^4-6x^2+1
    First i completed the square:
    x^4-6x^2+1
    x^4-6x^2+1-9x^2
    (x^2-3)^2-(3x)^2
    (x^2-3-3x)(x^2-3+3x)
    Now i use the difference of two squares where a^2-b^2=(a-b)(a+b)
    x^2-3x-3
    (x-3)^2-(\sqrt3)^2
    (x-3-\sqrt3)(x-3+\sqrt3)
    x^2+3x-3
    (x+3)^2-(\sqrt3)^2
    (x+3-\sqrt3)(x+3+\sqrt3)
    Therefore answer is:
    (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)


    P.S
    Last edited by Paymemoney; December 4th 2009 at 12:11 AM.
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  2. #2
    Super Member Bacterius's Avatar
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    Use the \sqrt{number} command to make a square root
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  3. #3
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    Quote Originally Posted by Bacterius View Post
    Use the \sqrt{number} command to make a square root
    oh ok
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  4. #4
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    i got this

    x^4-6x^2+1=(x^2-3)^2-{\sqrt{8}}^2 = (x^2-3-\sqrt{8})(x^2-3+\sqrt{8})
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  5. #5
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    Quote Originally Posted by Paymemoney View Post
    Can someone tell me if i have factorized correctly:

    Equation is this: x^4-6x^2+1
    First i completed the square:
    x^4-6x^2+1
    x^4-6x^2+1-9x^2
    (x^2-3)^2-(3x)^2
    (x^2-3-3x)(x^2-3+3x)
    Now i use the difference of two squares where a^2-b^2=(a-b)(a+b)
    x^2-3x-3
    (x-3)^2-(\sqrt3)^2
    (x-3-\sqrt3)(x-3+\sqrt3)
    x^2+3x-3
    (x+3)^2-(\sqrt3)^2
    (x+3-\sqrt3)(x+3+\sqrt3)
    Therefore answer is:
    (x-3-\sqrt3)(x-3+\sqrt3)(x+3-\sqrt3)(x+3+\sqrt3)


    P.S
    I'd use a dummy variable X = x^2 to turn it into a quadratic.

    So x^4 - 6x^2 + 1 = X^2 - 6X + 1

     = X^2 - 6X + (-3)^2 - (-3)^2 + 1

     = (X - 3)^2 - 8

     = (X - 3)^2 - (\sqrt{8})^2

     = (X - 3)^2 - (2\sqrt{2})^2

     = (X - 3 + 2\sqrt{2})(X - 3 - 2\sqrt{2})


    Now remembering that X = x^2

     = (x^2 - 3 + 2\sqrt{2})(x^2 - 3 - 2\sqrt{2}).


    You could now use the difference of two squares again.


     = [x^2 - (3 - 2\sqrt{2})][x^2 - (3 + 2\sqrt{2})]

     = \left[x^2 - \left(\sqrt{3 - 2\sqrt{2}}\right)^2\right]\left[x^2 - \left(\sqrt{3 + 2\sqrt{2}}\right)^2\right]

     = \left(x + \sqrt{3 - 2\sqrt{2}}\right)\left(x - \sqrt{3 - 2\sqrt{2}}\right)\left(x + \sqrt{3 + 2\sqrt{2}}\right)\left(x - \sqrt{3 + 2\sqrt{2}}\right).


    Ugly, yes, but fully factorised, yes.
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  6. #6
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    ^
    i understand the first part, however i don't understand why 3-2\sqrt2 is square rooted?
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  7. #7
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    Quote Originally Posted by Paymemoney View Post
    ^
    i understand the first part, however i don't understand why is = \left[x^2 - \left(\sqrt{3 - 2\sqrt{2}}\right)^2\right]\left[x^2 - \left(\sqrt{3 + 2\sqrt{2}}\right)^2\right]<br />
is square rooted?
    Because 3 - 2\sqrt{2} = \left(\sqrt{3 - 2\sqrt{2}}\right)^2 and 3 + 2\sqrt{2} = \left(\sqrt{3 + 2\sqrt{2}}\right)^2.

    We had to do this to create another difference of two squares.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Because 3 - 2\sqrt{2} = \left(\sqrt{3 - 2\sqrt{2}}\right)^2 and 3 + 2\sqrt{2} = \left(\sqrt{3 + 2\sqrt{2}}\right)^2.

    We had to do this to create another difference of two squares.
    ok thanks
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