Polynomial (degree 3) with only 1 root

Discussion in 'Algebra' started by TheOAP, May 18, 2022.

  1. TheOAP

    TheOAP

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    Assume a degree 3 polynomial. Assume the graph (and my program) indicate there is only 1 Root.
    According to Dr Richard Feynman there must be 2 solution. So the other 2 must be a pair of complex "roots" ?
    Assume equation : x3 +bx2+ c7x + 12
    and assume the "root" found was x = 3. To make the ("root" times constant) equal to 12 then the Complex values (surely!) must be (x + 2i)(x - 2i)
    I hope this makes sense as I don't see otherwise how the Complex values can be calculated.
    Pete
    Regards
     
    TheOAP, May 18, 2022
    #1
  2. TheOAP

    MathLover1

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    x^3 +bx^2+ cx + 12=0

    assume the "root" found was x = 3

    as a product of factors, polynomial is

    f(x)=(x-x1)(x-x2)(x-x3)

    so, one factor is (x-3)
    assuming, as you state, other two factors are (x + 2i) and (x - 2i)

    f(x)=(x-3)(x + 2i)(x - 2i)
    f(x)=x^3 - 3x^2 + 4x - 12

    =>b=3 and c=4

    check the graph:
    [​IMG] as you can see, there is only one real root at x=3
     
    MathLover1, May 18, 2022
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    nycmathguy likes this.
  3. TheOAP

    nycmathguy

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    Don't forget to say thank you to MathLover1 for answering YOUR thread.
     
    nycmathguy, May 19, 2022
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  4. TheOAP

    nycmathguy

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    I love it. Thank you. This is precalculus, right?
     
    nycmathguy, May 19, 2022
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