Results 1 to 7 of 7

Thread: Factoring the difference of two perfect squares

  1. #1
    Member cmf0106's Avatar
    Joined
    May 2008
    Posts
    104

    Factoring the difference of two perfect squares

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

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

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

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

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    $\displaystyle (x^{2})^{2}-(y^{3})^{2}$

    There, now it is two perfect squares.

    $\displaystyle (x^{2}-y^{3})(x^{2}+y^{3})$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member cmf0106's Avatar
    Joined
    May 2008
    Posts
    104
    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 $\displaystyle
    (x^{2})^{2}-(y^{3})^{2}
    $ are both perfect squares if you wouldst mind elaborating.
    Last edited by cmf0106; Jun 1st 2008 at 05:38 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    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 $\displaystyle
    (x^{2})^{2}-(y^{3})^{2}
    $ are both perfect squares if you wouldst mind elaborating.
    There was no error.
    $\displaystyle x^4 - y^6 = (x^2)^2 - (y^3)^2$

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

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

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member cmf0106's Avatar
    Joined
    May 2008
    Posts
    104
    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 $\displaystyle x^4 - y^6$ has nothing to do with factoring the difference of two perfect squares?

    IE
    $\displaystyle 25z^2 - 81y^2$?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by cmf0106 View Post
    So the problem at hand $\displaystyle x^4 - y^6 $ has nothing to do with factoring the difference of two perfect squares?
    It does. What I am saying is that $\displaystyle y^6 = (y^3)^2$ is a perfect square: the square of $\displaystyle y^3$.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by cmf0106 View Post
    So the problem at hand $\displaystyle x^4 - y^6 $ has nothing to do with factoring the difference of two perfect squares?

    $\displaystyle x^4 - y^6 $

    Both of these are perfect squares

    Here is a numerical expample to help clarify.

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

    So we can write both of the above as perfect squares

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

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

    I hope this helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Sep 18th 2011, 05:12 AM
  2. Perfect Squares
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Sep 6th 2011, 03:46 PM
  3. [SOLVED] Factoring Perfect Squares
    Posted in the Algebra Forum
    Replies: 8
    Last Post: Sep 2nd 2011, 03:03 AM
  4. not perfect squares
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Mar 21st 2010, 12:53 PM
  5. perfect squares
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Oct 11th 2009, 05:49 PM

Search Tags


/mathhelpforum @mathhelpforum