anonymous
  • anonymous
Long division for x^4 - 2x^3 + 3x^2 - 4x + 6 by x^2 + 2x -1?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[x ^{2} + 2x - 1 \div x ^{4} - 2x ^{3} + 3x ^{2} - 4x + 6\]
triciaal
  • triciaal
switch
anonymous
  • anonymous
\[x ^{4} - 2x^3 + 3x^2 - 4x + 6 \div x^2 + 2x - 1\]

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triciaal
  • triciaal
yes
calculusxy
  • calculusxy
\[\ \large \frac{ x^4 - 2x^3 + 3x^2 - 4x + 6 }{ x^2 +2x - 1 }\] when u r dividing, you are basically subtracting the exponents with the same root for example \[\large \frac{ x^6 }{ x^2 } = x^4\]
calculusxy
  • calculusxy
and if we have \[\large 2x \] then the exponent is 1 \[\large 2x^1\]
anonymous
  • anonymous
Would x^4 end up divided (subtracting, I guess) by x^2, turning into x^2?
calculusxy
  • calculusxy
yes
anonymous
  • anonymous
|dw:1441163299335:dw|
anonymous
  • anonymous
|dw:1441164156039:dw|
anonymous
  • anonymous
You also need a place holder for the x² term|dw:1441164237673:dw|
anonymous
  • anonymous
|dw:1441164097336:dw|
anonymous
  • anonymous
I have no idea where or why I put -1/4x.
anonymous
  • anonymous
|dw:1441164511713:dw| Do you follow this?|dw:1441164769424:dw|