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Math Help - Simplifying expressions

  1. #1
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    Question Simplifying expressions

    I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions.

    1. x^3-8
    x-2

    2. 1 - 1
    x + h x

    3. 2
    x^2
    10
    x^5

    4. 2x - 1 - 8
    x^2-6x+9 x+1 x^2-2x-3
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  2. #2
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    Quote Originally Posted by journeygirl280 View Post
    1. x^3-8
    x-2
    Hint: Difference two two cubes in numerator.

    2. 1 - 1
    x + h x
    With common denominator,
    \frac{x}{x(x+h)} - \frac{x+h}{x(x+h)}.
    Now continue.


    The last two do not make any sense.
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  3. #3
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    Hello, journeygirl280!

    I think I deciphered the last two problems . . .


    3)\;\;\frac{\frac{2}{x^2}}{\frac{10}{x^5}}
    We have: . \frac{2}{x^2} \div \frac{10}{x^5} \;= \;\frac{2}{x^2}\cdot\frac{x^5}{10} \;=\;\frac{x^3}{5}



    4)\;\;\frac{2x}{x^2-6x+9} - \frac{1}{x+1} - \frac{8}{x^2-2x-3}
    We have: . \frac{2x}{(x-3)^2} - \frac{1}{x+1} - \frac{8}{(x+1)(x-3)}


    Get a common denominator:

    . . \frac{2x}{(x-3)^2}\cdot{\color{blue}\frac{x+1}{x+1}} \,-\,\frac{1}{x+1}\cdot{\color{blue}\frac{(x-3)^2}{(x-3)^2}} \,-\,\frac{8}{(x+1)(x-3)}\cdot{\color{blue}\frac{x-3}{x-3}}

    . . = \;\frac{2x(x+1)-(x-3)^2 - 8(x-3)}{(x-3)^2(x+1)} \;= \;\frac{2x^2 + 2x - x^2 + 6x - 9 - 8x + 24}{(x-3)^2(x+1)}<br />

    . . = \;\frac{x^2+15}{(x-3)^2(x+1)}

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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by journeygirl280 View Post
    I can't remember how to do this. Can someone please help me refresh my memory? I have to simplify these expressions.

    1. x^3-8
    x-2
    Different hint: x^3-8=0 when x=2 so x-2 is a factor of x^3-8.

    (well really the same hint in disguise)

    RonL
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