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Math Help - powers problems

  1. #1
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    powers problems

    Hi all, working on this problem and have hit a roadblock, I think there is a rule I can use to make it simpler and I am sure I am just not seeing it, any help?

    \frac {q^2}{56}(56-15q+q^2)-[\frac{q}{56}(56-15q+q^2)]^2

    can I pull the preceding \frac{q^2}{56} and put that together with the second terms \frac{q}{56} and then put that in front of the whole or is that not allowed? Thanks for the help!
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  2. #2
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    Hello, Raezputin!

    I hope I read the problem correctly . . .


    Smplify: . \frac{q^2}{56}(56-15q+q^2) - \bigg[\frac{q}{56}(56-15q+q^2)\bigg]^2

    We have: . \frac{q^2}{56}\left(56 - 15q + q^2\right) - \frac{q^2}{56^2}\left(56 - 15q + q^2\right)^2


    We can factor out: . \frac{q^2}{56^2}(56-15q + q^2)

    . . \frac{q^2}{56^2}(56-15q+q^2)\cdot \bigg[56 - (56 - 15q + q^2)\bigg] \;\;=\;\;\frac{q^2}{56^2}(56 -15q-q^2)\cdot\left(15q - q^2\right)

    . . = \;\;\frac{q^2}{56^2}(56-15q+q^2)\cdot q(15-q) \;\;=\;\;\frac{q^3}{56^2}(56-15q+q^2)(15-q)

    . . = \;\;\frac{q^3}{3136}(7-q)(8-q)(15-q)

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