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Math Help - Factoring to reduce Questions

  1. #1
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    Factoring to reduce Questions

    In the expression
    \frac{16x^3-24x^2}{8x^4-12x^3}=\frac{8x^2(2x-3)}{4x^3(2x-3)}=\frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}

    I understand everything but the last step, how is the \frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}?

    If it were me I would think the correct thing to do in this case would be to apply the exponent property  \frac{a^m}{a^n}=a^{m-n}, thus equaling 2x^{2-3} =\frac{1}{2x}

    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by allyourbass2212 View Post
    In the expression
    \frac{16x^3-24x^2}{8x^4-12x^3}=\frac{8x^2(2x-3)}{4x^3(2x-3)}=\frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}

    I understand everything but the last step, how is the \frac{2}{1} * \frac{x^2}{x^3}= \frac{2}{x}?

    If it were me I would think the correct thing to do in this case would be to apply the exponent property  \frac{a^m}{a^n}=a^{m-n}, thus equaling 2x^{2-3} =\frac{1}{2x}

    Thanks!
    \frac 21 \cdot \frac {x^2}{x^3} = \frac 21 \cdot x^{2 - 3} = \frac 21 \cdot \frac 1x = \frac 2x

    the 2 is NOT affected by the power you have here. this is why it is important for you to get into the habit of using parentheses properly.

    if it were (2x)^{2-3} you would be correct. that is \frac 1{2x}. but in 2x^{2-3}, only the x is affected by the power. the answer is \frac 2x
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