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Math Help - Horrible Factorization

  1. #1
    Junior Member BobBali's Avatar
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    Horrible Factorization

    Again, to factorize; not solve..

     <br />
x^(3a+2) + 3x^(2a+2) - 5x^2<br />
     <br />
-5x^2 + x^(3a+2) + x^(2a+2)<br />

     <br />
x^-0.5( -5x^(2.5) + 3^(2a+2.5) + 1^(3a+2.5))<br />
*Not too sure about this step, went off-track here, i think..
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  2. #2
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    Quote Originally Posted by BobBali View Post
    Again, to factorize; not solve..
     x^{(3a+2)} + 3x^{(2a+2)} - 5x^2
     -5x^2 + x^(3a+2) + x^(2a+2)<br />

     x^{-0.5}( -5x^{(2.5)} + 3^{(2a+2.5)} + 1^{(3a+2.5)})<br />
*Not too sure about this step, went off-track here, i think.
    If you have more than one character in an exponent, set off the whole exponent in braces.
    [tex] x^{(3a+2)} + 3x^{(2a+2)} - 5x^2[/tex] gives  <br />
x^{(3a+2)} + 3x^{(2a+2)} - 5x^2<br />
    Please edit your post to correct exponents.
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  3. #3
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    Hello, BobBali!

    Didn't you PREVIEW your post?
    You didn't see that it didn't come out clearly?

    And how did you manage to factor out x^{-\frac{1}{2}} ? . . . And why?

    Like Plato, I'll guess at what you meant.


    \text{Factor: }\;x^{3a+2} + 3x^{2a+2} - 5x^2

    Take another look at what we have: . x^{3a}\!\cdot\!x^2 + 3\!\cdot\!x^{2a}\!\cdot\!x^2 - 5\!\cdot\!x^2


    Factor out x^2\!:\;\;x^2\left(x^{3a} + 3x^{2a} - 5)

    . . and that's the best we can do.

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