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Math Help - Factor an expression containing rational exponents

  1. #1
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    Factor an expression containing rational exponents

    Factor the expression. Express your answer so that only positive exponents occur.

    <br />
4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)} \qquad x>= -\frac{1}{2}<br />

    The answer:
    <br />
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(38x+13)<br />

    so far I've done:
    <br />
(24x+12)^{(1/2)}(5x+1)^{(11/7)}+(18x+9)^{(3/2)}(5x+1)^{(4/7)}<br />

    <br />
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(4x+4)(5x+1)^{(7/7)}(3x+3)<br />

    Is that right so far and if so, what do I do next?

    Thanks.
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  2. #2
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    Quote Originally Posted by desiderius1 View Post
    Factor the expression. Express your answer so that only positive exponents occur.

    <br />
4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)} \qquad x>= -\frac{1}{2}<br />

    The answer:
    <br />
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(38x+13)<br />

    so far I've done:
    <br />
\textcolor{red}{(24x+12)^{(1/2)}}(5x+1)^{(11/7)}+\textcolor{red}{(18x+9)^{(3/2)}}(5x+1)^{(4/7)}<br />

    these two factors are incorrect ... you cannot distribute the constants into factors w/ exponents not equal to 1.


    4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)}<br />

    common factors are (6x+3)^{1/2}(5x+1)^{4/7} ...

    (6x+3)^{1/2}(5x+1)^{4/7}\left[4(5x+1) + 3(6x+3)\right]

    (6x+3)^{1/2}(5x+1)^{4/7}\left[(20x+4) + (18x+9)\right]

    (6x+3)^{1/2}(5x+1)^{4/7}\left[38x + 13\right]<br />

    ...
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  3. #3
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    Another factor question:

    <br />
8x^{(-1/2)}+\frac{4}{7}x^{(1/2)}<br />

    Answer:
    <br />
\frac{4(14+x)}{7x^{(1/2)}}<br />

    Wouldn't 8x^{(-1/2)} = \frac{1}{\sqrt{8x}}?

    I don't see how they got the 7x^{(1/2)} on the bottom.. when the exponent isn't negative.
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  4. #4
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    Quote Originally Posted by desiderius1 View Post
    Another factor question:

    <br />
8x^{(-1/2)}+\frac{4}{7}x^{(1/2)}<br />

    Answer:
    <br />
\frac{4(14+x)}{7x^{(1/2)}}<br />

    Wouldn't 8x^{(-1/2)} = \frac{1}{\sqrt{8x}}?

    I don't see how they got the 7x^{(1/2)} on the bottom.. when the exponent isn't negative.
    <br />
8x^{-1/2}+\frac{4}{7}x^{1/2}<br />

    \frac{8}{x^{1/2}} + \frac{4x^{1/2}}{7}

    common denominator is 7x^{1/2} ...
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  5. #5
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    <br />
8 + 4x^{1/2}<br />

    How do you get 4(14 + x)?
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  6. #6
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    oh I see, you multiply 8 and 7 to get 56.
    56/14 = 4.
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