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anonymous

  • one year ago

Long division for x^4 - 2x^3 + 3x^2 - 4x + 6 by x^2 + 2x -1?

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  1. anonymous
    • one year ago
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    \[x ^{2} + 2x - 1 \div x ^{4} - 2x ^{3} + 3x ^{2} - 4x + 6\]

  2. triciaal
    • one year ago
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    switch

  3. anonymous
    • one year ago
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    \[x ^{4} - 2x^3 + 3x^2 - 4x + 6 \div x^2 + 2x - 1\]

  4. triciaal
    • one year ago
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    yes

  5. calculusxy
    • one year ago
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    \[\ \large \frac{ x^4 - 2x^3 + 3x^2 - 4x + 6 }{ x^2 +2x - 1 }\] when u r dividing, you are basically subtracting the exponents with the same root for example \[\large \frac{ x^6 }{ x^2 } = x^4\]

  6. calculusxy
    • one year ago
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    and if we have \[\large 2x \] then the exponent is 1 \[\large 2x^1\]

  7. anonymous
    • one year ago
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    Would x^4 end up divided (subtracting, I guess) by x^2, turning into x^2?

  8. calculusxy
    • one year ago
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    yes

  9. anonymous
    • one year ago
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    |dw:1441163299335:dw|

  10. anonymous
    • one year ago
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    |dw:1441164156039:dw|

  11. anonymous
    • one year ago
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    You also need a place holder for the x² term|dw:1441164237673:dw|

  12. anonymous
    • one year ago
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    |dw:1441164097336:dw|

  13. anonymous
    • one year ago
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    I have no idea where or why I put -1/4x.

  14. anonymous
    • one year ago
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    |dw:1441164511713:dw| Do you follow this?|dw:1441164769424:dw|