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Thread: Reducing to lowest terms

  1. #1
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    Reducing to lowest terms

    Could someone help explain these problems? They've both stumped me.

    (z^2-11z+24)/(5z^2-25z-120)

    (48-3z^2)/(z^2-7z+12)


    Not sure how to use the <math> coding...

    Thanks so much
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  2. #2
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    Quote Originally Posted by tony_rose88 View Post
    Could someone help explain these problems? They've both stumped me.

    (z^2-11z+24)/(5z^2-25z-120)

    (48-3z^2)/(z^2-7z+12)


    Not sure how to use the <math> coding...

    Thanks so much
    Hints:
    1)$\displaystyle z^2-11z+24 = (z-8)(z-3)$
    2)$\displaystyle 5z^2-25z-120 = 5(z^2 - 5z - 24) = 5(z-8)(z+3)$
    3)$\displaystyle z^2-7z+12 = (z-3)(z-4)$
    4)$\displaystyle 48-3z^2 = 3(16 - z^2) = 3(4-z)(4+z)$

    EDIT:Thank you Danshader. Edited my mistake
    Last edited by Isomorphism; Apr 28th 2008 at 07:36 AM. Reason: sloppy computation
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  3. #3
    Member Danshader's Avatar
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    there is a small mistake there iso o.o" in 3)
    $\displaystyle
    z^2-7z+12 = (z-5)(z-2)
    $

    shouldn't it be:
    $\displaystyle
    z^2-7z+12 = (z-4)(z-3)
    $
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  4. #4
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    Quote Originally Posted by Isomorphism View Post
    Hints:
    1)$\displaystyle z^2-11z+24 = (z-8)(z-3)$
    2)$\displaystyle 5z^2-25z-120 = 5(z^2 - 5z - 24) = 5(z-8)(z+3)$
    3)$\displaystyle z^2-7z+12 = (z-3)(z-4)$
    4)$\displaystyle 48-3z^2 = 3(16 - z^2) = 3(4-z)(4+z)$

    EDIT:Thank you Danshader. Edited my mistake

    So I've got $\displaystyle (z-8)(z-3)/5(z-8)(z+3)$ on the first one. Cancel out those z-8's cancel out so it leaves me with $\displaystyle (z-3)/(5z+3)$?

    And on the second: $\displaystyle 3(z-4)(z+4))/(z-3)(z-4)$. Cancel out the z-4's, I get $\displaystyle (3z+4)/(z-3)$

    Are those correct?
    Last edited by tony_rose88; Apr 28th 2008 at 08:06 AM.
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