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Cutiepo0 Group Title

How to factor (x^3 + 1)

  • one year ago
  • one year ago

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  1. Brittni0605 Group Title
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    \[(x^{3}+1)\]

    • one year ago
  2. Cutiepo0 Group Title
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    mmhm!

    • one year ago
  3. bandicoot12 Group Title
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    no, (x+1)(x^2-x+1)

    • one year ago
  4. Hero Group Title
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    \[(a + b)^3 = (a + b)(a^2 - ab + b^2)\]

    • one year ago
  5. Brittni0605 Group Title
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    \[(x+1)(x^{2}-x+1)\]

    • one year ago
  6. Hero Group Title
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    That's the sum of cubes formula I posted above.

    • one year ago
  7. bandicoot12 Group Title
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    ya basically what they said

    • one year ago
  8. Cutiepo0 Group Title
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    ok but I just established that (x^3+1) and (x+1)^3 weren't the same thing :S

    • one year ago
  9. Hero Group Title
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    No, they are not. One is simply x cubed plus one. The other is \((x + 1) \times (x + 1) \times (x + 1)\)

    • one year ago
  10. Cutiepo0 Group Title
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    right, so then how can I use the sum of cubes formula? b/c isn't that for (a+b)3

    • one year ago
  11. Hero Group Title
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    Because \((x^3 + 1) = (x^3 + 1^3)\)

    • one year ago
  12. Hero Group Title
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    Remember, \(1 = 1 \times 1 \times 1 = 1^3 \)

    • one year ago
  13. Cutiepo0 Group Title
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    = (x + 1) ^3 ?? but wouldnt that make it the same??

    • one year ago
  14. Hero Group Title
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    No, I just explained to you earlier that \((x^3 + 1) \ne (x + 1)^3\)

    • one year ago
  15. Hero Group Title
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    \((x^3 + 1) = (x^3 + 1^3)\) \((x + 1)^3 = (x + 1)(x + 1)(x + 1)\)

    • one year ago
  16. Cutiepo0 Group Title
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    ok, so (a^3 + b^3) doesn't equal (a+b)^3, ever right? Sorry, just trying to wrap my head around it

    • one year ago
  17. Hero Group Title
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    The only way is if x = 0 or x = 1

    • one year ago
  18. Hero Group Title
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    But in general, they will not be equal

    • one year ago
  19. Cutiepo0 Group Title
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    ok, so now though with the sum of cubes formula, the whole point was to find the zeroes of the equation f(x)= (x^3 + 1)(x-17)

    • one year ago
  20. Cutiepo0 Group Title
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    and the answer is x=17 and x = -1

    • one year ago
  21. Cutiepo0 Group Title
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    so how do I get from the sum of cubes to -1

    • one year ago
  22. Hero Group Title
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    Basically, the question is asking: What x values would make \((x^3 + 1)(x - 17) = 0\) You know that anything times zero equals zero. So when you expand \((x^3 + 1)\) you get \((x + 1)(x^2 - x + 1)\) Thus \((x^3 + 1)(x - 17) = (x + 1)(x^2 - x + 1)(x - 17) = 0\)

    • one year ago
  23. Hero Group Title
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    If you look closely you'll realize that if (x + 1) = 0 or if (x - 17) = 0, then the product of the whole thing will be zero. There's only one way to make (x + 1) = 0 There's only one way to make (x - 17) = 0.

    • one year ago
  24. Hero Group Title
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    What I say doesn't matter if you don't understand anything.

    • one year ago
  25. Hero Group Title
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    The silence is deafening

    • one year ago
  26. Cutiepo0 Group Title
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    Oh, I got it, thanks a lot for your help with this :) b/c (x2−x+1) doesn't have any roots, so it's just the other two roots

    • one year ago
  27. Hero Group Title
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    Oh goodie :D

    • one year ago
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