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Math Help - Factoring the difference of two perfect squares

  1. #1
    Member cmf0106's Avatar
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    Factoring the difference of two perfect squares

    The author starts by giving the example 25z^2 - 81y^2, which equals (5z + 9y) (5z - 9y).

    Both 81, and 25 are perfect squares so this makes sense.

    \sqrt{25} = 5*5, \sqrt{9} = 3*3

    Then the author uses this example factoring x^4 - y^6
    Which equals (x^2 + y^3)(x^2 - y^3).

    But this does not make sense to me
    \sqrt{4} = 2 * 2, \sqrt{6} = 2.45...

    Can some one please help clarify what I am doing wrong and how the author is pulling y^3 out of y^6 in the second example?

    Many thanks
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  2. #2
    Eater of Worlds
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    (x^{2})^{2}-(y^{3})^{2}

    There, now it is two perfect squares.

    (x^{2}-y^{3})(x^{2}+y^{3})
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  3. #3
    Member cmf0106's Avatar
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    Galactus forgive me for being math dense, but are you rearranging the problem because the original problem had an error?

    Also I do not see how <br />
(x^{2})^{2}-(y^{3})^{2}<br />
are both perfect squares if you wouldst mind elaborating.
    Last edited by cmf0106; June 1st 2008 at 05:38 PM.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cmf0106 View Post
    Galactus forgive me for being math dense, but are you rearranging the problem because the original problem had an error?

    Also I do not see how <br />
(x^{2})^{2}-(y^{3})^{2}<br />
are both perfect squares if you wouldst mind elaborating.
    There was no error.
    x^4 - y^6 = (x^2)^2 - (y^3)^2

    This follows from the general rule
    (z^p)^q = z^{pq}

    Perhaps one more clarification: We are not taking the square root of the exponent, we are dividing it by 2.

    -Dan
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  5. #5
    Member cmf0106's Avatar
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    Quote Originally Posted by topsquark View Post
    Perhaps one more clarification: We are not taking the square root of the exponent, we are dividing it by 2.

    -Dan
    So the problem at hand x^4 - y^6 has nothing to do with factoring the difference of two perfect squares?

    IE
    25z^2 - 81y^2?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cmf0106 View Post
    So the problem at hand x^4 - y^6 has nothing to do with factoring the difference of two perfect squares?
    It does. What I am saying is that y^6 = (y^3)^2 is a perfect square: the square of y^3.

    -Dan
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  7. #7
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    Quote Originally Posted by cmf0106 View Post
    So the problem at hand x^4 - y^6 has nothing to do with factoring the difference of two perfect squares?

    x^4 - y^6

    Both of these are perfect squares

    Here is a numerical expample to help clarify.

    lets say that x=2 then x^4=16
    but x^2=4 and (x^2)^2=x^4=16=4^2

    So we can write both of the above as perfect squares

    (x^2)^2 \\\ (y^3)^2

    Remember that if you have (x^a)^b=x^{a \cdot b}

    I hope this helps.
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