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anonymous

  • 5 years ago

how do you solve these problems: 2n(squared) - 11n + 15 and -3n(squared) - 75

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  1. anonymous
    • 5 years ago
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    are you dealing with imaginary numbers?

  2. anonymous
    • 5 years ago
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    no i am dealing with factoring polynomials

  3. anonymous
    • 5 years ago
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    so 2n^2-11n+15 = 2n^2 -6n-5n+15

  4. anonymous
    • 5 years ago
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    thank you so much!!

  5. anonymous
    • 5 years ago
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    -3n^2-75=0--------> n^2=-25

  6. anonymous
    • 5 years ago
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    thank you soo much!!

  7. anonymous
    • 5 years ago
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    you are welcome. note that for the second equation, you will have imaginary roots.

  8. radar
    • 5 years ago
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    Since we are tasked to factor the polynomials, take a close look at the second equation.\[-3n ^{2}-75\]

  9. radar
    • 5 years ago
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    It is a factor problem, extract the -3 which is present in both terms.\[-3(n ^{2}+25)\]

  10. radar
    • 5 years ago
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    Thats is about as far as you can go with this.

  11. anonymous
    • 5 years ago
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    That's true, unless the right hand side is specified in the equation.

  12. radar
    • 5 years ago
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    I didn't notice any but if it is zero, we would have the situation that you mentioned and probably doehrman hasn't got into complex number yet.

  13. anonymous
    • 5 years ago
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    yes, you are right

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