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Math Help - Reordering terms affect the factorization of a square trinomial?

  1. #1
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    Smile Reordering terms affect the factorization of a square trinomial?

    I have the square trinomial
    25 - 10x + x^2

    If I factor it I get:
    (5 - x)^2

    but If I rearrange the addends in an order in which literal addend are to precede numerial addends,
    x^2 -10x + 25,

    and factor it I get:
    (x-5)^2


    aren't 25 - 10x + x^2 and x^2 -10x + 25 equal in value?

    Why do I get different square binomials?
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  2. #2
    Senior Member Peritus's Avatar
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    <br />
\left( {x - 5} \right)^2  = \left[ {\left( { - 1} \right)\left( {5 - x} \right)} \right]^2  = \left( { - 1} \right)^2 \left( {5 - x} \right)^2  = \left( {5 - x} \right)^2
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    They are not so different...

    with some factoring we can show they are the same!

    (5-x)^2=[(-1)(-1)5+(-1)x]^2=[(-1)((-1)5+x)]^2=(-1)^2[(-1)5+x]^2=

    (1)[x-5]=(x-5)^2
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  4. #4
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    thanks!
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  5. #5
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    Quote Originally Posted by Peritus View Post
    <br />
\left( {x - 5} \right)^2  = \left[ {\left( { - 1} \right)\left( {5 - x} \right)} \right]^2  = \left( { - 1} \right)^2 \left( {5 - x} \right)^2  = \left( {5 - x} \right)^2
    And in general, (a - b)^2 = (b - a)^2.

    Even more generally:

    (a - b)^n = (b - a)^n if n is even

    (a - b)^n = -(b - a)^n if n is odd.
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    And in general, (a - b)^2 = (b - a)^2.

    Even more generally:

    (a - b)^n = (b - a)^n if n is even

    (a - b)^n = -(b - a)^n if n is odd.
    Thanks.
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