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f(x)=x^5-4x^3-x^2+4?

the real factors are (x+2), (x-1), and (x-2)

how do I find the imaginary factors using long polynomial division

3 Answers

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  • 2 months ago

    (x+2)(x-1)(x-2) = x^3 - x^2 - 4x + 4

    From the first and last term the quadratic yielding the complex solutions is x^2 + ax + 1.

    (x^3 - x^2 - 4x + 4)(x^2 + ax + 1) = x^5 + (a-1)x^4 + .... + 4. a - 1 = 0,  so the quadratic is x^2 + x + 1, which factors to (x + ½ + ½√3 i)(x + ½ - ½√3 i).

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  • ted s
    Lv 7
    2 months ago

    either multiply those factors together , then divide f by that result...OR divide f by ' x + 2 ' , that result by ' x - 1 ' , that result by x - 2...to get a quadratic function....use quadratic equation to find the last 2 zeros...( x - r_1) (x - r_2)

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  • Jim
    Lv 7
    2 months ago

    Since you have a quadratic left, I would simply use Quadratic formula or Completing the squares.

     (x^2 + x + 1) = 0

    Subtract 1 from both sides: x^2 + x = -1

    Add 1/4 to both sides: x^2 + x + 1/4 = -3/4

    Write the left hand side as a square: (x + 1/2)^2 = -3/4

    Take the square root of both sides: x + 1/2 = (i sqrt(3))/2 or x + 1/2 = -(i sqrt(3))/2

    Subtract 1/2 from both sides: x = (i sqrt(3))/2 - 1/2 or x + 1/2 = -(i sqrt(3))/2

    Subtract 1/2 from both sides:

    x = (i sqrt(3))/2 - 1/2 or x = -(i sqrt(3))/2 - 1/2

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