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anonymous

  • 5 years ago

find a third-degree polynomial equation with rational coefficants that has roots -4 and 2+i please explain

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  1. amistre64
    • 5 years ago
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    it wants you to find a glorified fraction: that is equal to zero when you use those values

  2. amistre64
    • 5 years ago
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    read rational in the wrong place.... its just a poly

  3. amistre64
    • 5 years ago
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    (a3/b3)x^3 + (a2/b2)x^2 + (a1/b1)x + (a0/b0)

  4. amistre64
    • 5 years ago
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    the 2 + i root means that there is a bend in the graph along the x=2 line, but that it is either above or below the graph ....

  5. amistre64
    • 5 years ago
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    f'(2) = 0 would be a part of the equation i think

  6. amistre64
    • 5 years ago
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    might be wrong tho ...

  7. amistre64
    • 5 years ago
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    (x-1)^2 + 3 is a quadratic with roots at: 1 +- i sqrt(3) ; which means that its bend, its vertex, is along the x=1 line and that it is either above or below it.... so I am assuming that is a good analogy to this

  8. amistre64
    • 5 years ago
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    the cubic formula tho is alot more complicated ..

  9. amistre64
    • 5 years ago
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    (x+4)(x+4)(x - (2+i)) or (x+4)(x - (2+i))(x - (2+i)) seem to be the possibilities; lets try to find the products of these...

  10. amistre64
    • 5 years ago
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    (x-2-i)(x-2-i) = x^2 -2x -ix -2x -ix +3 + 4i -------------------- (x^2 -4x -2ix +4i +3) (x+4) ------------------- x^3 -4x^2 -2ix^2 +4ix +3x +4x^2 -8ix -16x +16i +12 ----------------------------------- x^3 -2ix^2 -4ix -13x +16i +12 kinda gotta wonder abt my technique lol

  11. amistre64
    • 5 years ago
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    (x+4)^2 = x^2 +8x +16 (x^2 +8x +16) (x-2-i) -------------- x^3 +8x^2 +16x -2x^2 -16x -32 -ix^2 -8ix -16i -------------------------------- x^3 -6x^2 -32 -ix^2 -8ix -16i (x^3 -6x^2 -32) - i( x^2 +8x +6)

  12. amistre64
    • 5 years ago
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    (x^3 -13x +12) - 2i( x^2 +2x -8)

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