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anonymous

  • 5 years ago

factor 12v^2+36vp+27p^2

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  1. angela210793
    • 5 years ago
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    3(4v^2+12vp+9p^2)

  2. anonymous
    • 5 years ago
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    can it be factored further?

  3. mathteacher1729
    • 5 years ago
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    Yes. Note that there is a perfect square hidden in \[3(4v^2+12vp+9p^2)\] \[3(4v^2+12vp+9p^2)=3(2v+3p)(2v+3p)\] So that means \[ 12v^2+36vp+27p^2 =3(4v^2+12vp+9p^2) =2(2v+3p)^2 \]

  4. anonymous
    • 5 years ago
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    3(2v+3p)^2??

  5. mathteacher1729
    • 5 years ago
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    To see if that is the answer, expand it. First compute \[(2v+3p)^2\] and then multiply that by three. :)

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