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jaersyn

  • 2 years ago

find all solutions of the polynomial equation: x^4 - x^3 - 6x^2 + 4x + 8 = 0

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  1. jenn0814
    • 2 years ago
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    x4−x3−6x2+4x+80 is your answer

  2. cammyabbo
    • 2 years ago
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    you want to use grouping to factor this polynomial

  3. cammyabbo
    • 2 years ago
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    for the first 3 terms, you can factor out an X^2

  4. jaersyn
    • 2 years ago
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    how should i group them?

  5. jaersyn
    • 2 years ago
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    i got x^2(x^2-x-6)+4(x+2)

  6. jaersyn
    • 2 years ago
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    x^2(x-3)(x+2)+4(x+2) (x+2)(x^2(x-3)+4) :/

  7. cammyabbo
    • 2 years ago
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    lets start over

  8. cammyabbo
    • 2 years ago
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    x^4 - x^3 - 6x^2 + 4x + 8 = 0

  9. matricked
    • 2 years ago
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    use factor and remainder theorem to solve it

  10. cammyabbo
    • 2 years ago
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    x^2(x^2 - x - 6) + 4(x + 2) = 0

  11. jaersyn
    • 2 years ago
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    do you have to use special factors?

  12. matricked
    • 2 years ago
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    see by plugging x=-2 the expression x^4 - x^3 - 6x^2 + 4x + 8 becomes 0 hence x+2 is one of the factor for getting other factors divide x^4 - x^3 - 6x^2 + 4x + 8 by x+2

  13. cammyabbo
    • 2 years ago
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    no... (x^2+4)(x-3)(x+2)^2=0 now solve each equal to zero

  14. cammyabbo
    • 2 years ago
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    no to special factors, matricked is right also

  15. jaersyn
    • 2 years ago
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    how'd you get there..

  16. cammyabbo
    • 2 years ago
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    x=-2 and x=2 x=3

  17. jaersyn
    • 2 years ago
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    you skipped a couple steps from x^2(x^2 - x - 6) + 4(x + 2) = 0

  18. cammyabbo
    • 2 years ago
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    i further factored (x^2-x-6) and grouped (x^2+4)

  19. jaersyn
    • 2 years ago
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    can you show me how, from there, you got (x^2+4)(x-3)(x+2)^2=0?

  20. cammyabbo
    • 2 years ago
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    now set each equation equal to 0

  21. cammyabbo
    • 2 years ago
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    the x^2 and +4 you pulled out go in their own equation

  22. jaersyn
    • 2 years ago
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    wait so from x^2(x-3)(x+2) + 4(x+2) = (x^2+4)(x+2)^2 (x-3)?

  23. cammyabbo
    • 2 years ago
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    yes

  24. cammyabbo
    • 2 years ago
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    now set each equal to zero by themselves and solve for x

  25. jaersyn
    • 2 years ago
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    hold on

  26. cammyabbo
    • 2 years ago
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    x^2+4=0

  27. jaersyn
    • 2 years ago
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    the rest i know how, but h\where's the logic in that move, where you can take x^2+4

  28. cammyabbo
    • 2 years ago
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    you factored it out by grouping

  29. jaersyn
    • 2 years ago
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    it is my understanding that in order to factor by grouping both pieces need to have common factors like, "4(x+2)" doesn't have a (x-3)

  30. cammyabbo
    • 2 years ago
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    are you learning imaginary numbers?

  31. cammyabbo
    • 2 years ago
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    right, but x^2-x-6 did

  32. cammyabbo
    • 2 years ago
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    you further factored

  33. matricked
    • 2 years ago
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    x^4 - x^3 - 6x^2 + 4x + 8=0 x^4 +2x^3 -3x^3 - 6x^2 + 4x + 8=0 X^3(x+2) -3x^2(x+2) +4(x+2)=0 (x+2)(x^3-3x^2+4)=0 (x+2)(x^3-2x^2-x^2+4)=0 (x+2)(x^2(x-2) - (x+2)(x-2))=0 (x+2)(x-2)(x^2-x-2)=0 (x+2)(x-2)(x^2-2x+x-2)=0 (x+2)(x-2)(x(x-2)+1(x-2))=0 or(x+2)(x-2)(x-2)(x+1)=0 now set each equal to 0 thus the roots are -1,-2,2,2

  34. cammyabbo
    • 2 years ago
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    wow, nice job

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