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zzr0ck3r

  • 2 years ago

how do I get from here 2x^3-3x+1 to (x-1)(2x^2+2x-1)

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  1. Mimi_x3
    • 2 years ago
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    maybe long division

  2. Mimi_x3
    • 2 years ago
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    long division works!

  3. Mimi_x3
    • 2 years ago
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    |dw:1347405240429:dw| hopefully you can understand that..

  4. zzr0ck3r
    • 2 years ago
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    note at all, I want get from the first one to the second one, say I was given to2x^3-3x+1, how do I get to (x-1)(2x^2+2x-1). I dont see how what you did applies

  5. Mimi_x3
    • 2 years ago
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    wait a mistake

  6. asnaseer
    • 2 years ago
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    there is an alternative way:\[\begin{align} 2x^3-3x+1&=x(2x^2-3)+1\\ &=x(2x^2-3)-(2x^2-3)+(2x^2-3)+1\\ &=(x-1)(2x^2-3)+2x^2-3+1\\ &=(x-1)(2x^2-3)+2x^2-2\\ &=(x-1)(2x^2-3)+2(x^2-1)\\ &=(x-1)(2x^2-3)+2(x-1)(x+1)\\ &=(x-1)(2x^2-3+2(x+1))\\ &=(x-1)(2x^2-3+2x+2)\\ &=(x-1)(2x^2+2x-1) \end{align}\]

  7. zzr0ck3r
    • 2 years ago
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    this is all to see where 2x^3-3x+1>0

  8. zzr0ck3r
    • 2 years ago
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    great ty

  9. asnaseer
    • 2 years ago
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    yw :)

  10. zzr0ck3r
    • 2 years ago
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    ty also @Mimi_x3

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