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anonymous

  • one year ago

Divide Polynomials? (27x^4-18x^3+9x^2)/3x

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  1. 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

  2. anonymous
    • one year ago
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    can you show me how?

  3. Nnesha
    • one year ago
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    |dw:1441648360041:dw|

  4. Nnesha
    • one year ago
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    so divide first terms \[\huge\rm \frac{ 27x^4 }{ 3x } =?\]

  5. anonymous
    • one year ago
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    9x but what is the exponent rule?

  6. anonymous
    • one year ago
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    ok page works now. 9x3?

  7. anonymous
    • one year ago
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    ?

  8. Nnesha
    • one year ago
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    when we divide same bases we should subtract their exponents or x^4 is same as x times x times x times \[\huge\rm \frac{ 27 ⋅x ⋅x⋅x⋅x }{3⋅x}\]

  9. Nnesha
    • one year ago
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    yes that's right now multiply the divisor by 9x^3

  10. Nnesha
    • one year ago
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    |dw:1441649011288:dw|

  11. anonymous
    • one year ago
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    so the final answer is 9x^3+6x^2+3x

  12. Nnesha
    • one year ago
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    9x^3 times 3x = 27x^4 and then change the sign first term would cancel out the repeat those steps |dw:1441649027500:dw|

  13. Nnesha
    • one year ago
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    show ur work plz :=0

  14. Nnesha
    • one year ago
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    :=)*

  15. anonymous
    • one year ago
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    Ok I see! Thanks!!! I assume multiplying is adding exponents so just the division in reverse?

  16. Nnesha
    • one year ago
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    yes right!

  17. anonymous
    • one year ago
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    thank you!

  18. Nnesha
    • one year ago
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    multiply same bases `ADD` exponents \[\rm x^m \times x^n =x^{m+n}\] divide same bases `subtract` exponents \[\rm \frac{ x^m }{x^n }=x^{m-n}\]

  19. Nnesha
    • one year ago
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    yw!

  20. anonymous
    • one year ago
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    btw, what if there's more after the variable in the divisor. say 3x-2?

  21. Nnesha
    • one year ago
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    then you have to multiply by both terms like 9x^3(3x-2) 9x^3 times 3x - 9x^3 times -2

  22. anonymous
    • one year ago
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    what would 9x^3 times -2 look like

  23. Nnesha
    • one year ago
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    lets say |dw:1441649570148:dw|there is 3x-2

  24. Nnesha
    • one year ago
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    just multiply the coefficients\[\huge\rm -9x^3 \times -2=(-9 \times -2)x^3\]

  25. anonymous
    • one year ago
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    oh so just 18x3

  26. Nnesha
    • one year ago
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    yes right!

  27. anonymous
    • one year ago
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    but where do you go from there? how does it get to the next step?

  28. Nnesha
    • one year ago
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    just repeat the steps divide first terms multiply the result by the divisor change signs

  29. anonymous
    • one year ago
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    but what's the point of the -2 then? idk im confused. i guess where do you put the 18x3?

  30. Nnesha
    • one year ago
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    below the 2nd term

  31. Nnesha
    • one year ago
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    coefficients should be the same so 9x^3 times -2 = -18x^3 <--3rd degree so put under the x^3 term

  32. anonymous
    • one year ago
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    oh i see, it cancels out? so it's 9x^2 / 3x-2 now?

  33. anonymous
    • one year ago
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    ok i think i got it now.

  34. Nnesha
    • one year ago
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    no just divide first terms !

  35. Nnesha
    • one year ago
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    first term of polynomial by first term of divisor so it would be \[\huge\rm \frac{ 9x^2 }{ 3x }\]

  36. anonymous
    • one year ago
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    ok give me a second

  37. Nnesha
    • one year ago
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    ok!

  38. anonymous
    • one year ago
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    i have to go be back in a while

  39. Nnesha
    • one year ago
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    alright same here cya

  40. anonymous
    • one year ago
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    @nnesha are you still here? ok i was using 9x^3 as an example. the real problem is 6x^3+11x^2-4x-4 / 3x-2

  41. Nnesha
    • one year ago
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    same steps ;=)

  42. Nnesha
    • one year ago
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    so first terms are 6x^3 and first term of divisor is 3x divide 6x^3/3x

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