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Math Help - Factoring

  1. #1
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    Factoring

    (2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)


    If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?

    And why can't X = -3\2?
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  2. #2
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    Quote Originally Posted by Alan306090 View Post
    (2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)


    If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?

    And why can't X = -3\2?
    \frac{2x^2 - 7x - 15}{2x^3 + 13x^2 + 15x}

     = \frac{2x^2 - 10x + 3x - 15}{x(2x^2 + 13x + 15)}

     = \frac{2x(x - 5) + 3(x - 5)}{x(2x^2 + 10x + 3x + 15)}

     = \frac{(x - 5)(2x + 3)}{x[2x(x + 5) + 3(x + 5)]}

     = \frac{(x - 5)(2x + 3)}{x(x + 5)(2x + 3)}

    Can you see here that you can't let x = -\frac{3}{2}? If you did the denominator would be 0.

    Now cancelling gives

     = \frac{x - 5}{x(x + 5)}

     = \frac{x - 5}{x^2 + 5x}.
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  3. #3
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    Quote Originally Posted by Alan306090 View Post
    (2X^2 - 7X - 15)/(2X^3 + 13X^2 + 15X)


    If X is greater than 10, then why is (x-5)\(x^2 + 5x) equivalent to it?

    And why can't X = -3\2?
    \frac{2x^2-7x-15}{2x^3+13x^2+15x}=\frac{(2x+3)(x-5)}{x(2x+3)(x+5)}

    This has no value if x=0, x= -5 and 2x = -3.

    You can cancel the term (2x+3) only if (2x+3) is not zero.
    Since in that case you have 0/0.

    However, if x>0, you can cancel those terms and then we can say it's the same as

    \frac{x-5}{x(x+5)}
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