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anonymous

  • one year ago

What is the quotient: (3x^3 + 14x^2 + 13x - 6) / (3x - 1) ?

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  1. AbdullahM
    • one year ago
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    Do you know group factoring?

  2. anonymous
    • one year ago
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    No

  3. AbdullahM
    • one year ago
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    Group factoring is like: \[x^3 + x^2 + x + 1 = x^2(x+1) + 1(x+1) = (x+1)(x^2+1)\]

  4. anonymous
    • one year ago
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    I don't really remember learning about that, but I think I understand a little bit

  5. AbdullahM
    • one year ago
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    so you need to group factor the numerator

  6. AbdullahM
    • one year ago
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    oh, wait. It can't be group factored >.<

  7. AbdullahM
    • one year ago
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    The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction p/q, where p is a factor of the trailing constant and q is a factor of the leading coefficient. and so using that, these could be possible roots \[\pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 1 }{ 3 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 2 }{ 3 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 3 }{ 3 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 6 }{ 3 } ~\]

  8. AbdullahM
    • one year ago
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    solve 3x-1 = 0 x = 1/3 and since that is one of the possible roots, we can use that since it might be divisible by it.

  9. AbdullahM
    • one year ago
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    can someone explain? ;-; this is hard to show online compared to in-person with a piece of paper ;-;

  10. AbdullahM
    • one year ago
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    someone else*

  11. Nnesha
    • one year ago
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    long division

  12. AbdullahM
    • one year ago
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    ik, but latex isn't that easy to use to show that ;-; and the draw tool isn't that easy either ;-;

  13. Nnesha
    • one year ago
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    for long division 5 steps) \(\huge\color{green}{{1:}}\) Divide the first terms \(\huge\color{green}{{2:}}\) multiply(distribute) \(\huge\color{green}{{3:}}\) subtract all terms \(\huge\color{green}{{4:}}\) carry down \(\huge\color{green}{{5:}}\) repeat

  14. Nnesha
    • one year ago
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    haha true that !

  15. Nnesha
    • one year ago
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    |dw:1442023891316:dw| divide first terms \[\huge\rm \frac{ 3x^3 }{ 3x } = ???\] then multiply the divisor with that

  16. Nnesha
    • one year ago
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    ^ divide first term of polynomial by first term of the divisor

  17. Nnesha
    • one year ago
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    wait for `ookay` :=)

  18. AbdullahM
    • one year ago
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    thanks :)

  19. Nnesha
    • one year ago
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    yw :=)

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