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27 February, 00:20

Finding all real roots of x^4-3x^3-5x^2+13x+6=0?

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  1. 27 February, 00:26
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    Hello from MrBillDoesMath!

    Answer:

    3, - 2, 1 + / - sqrt (2)

    Discussion:

    By the "rational root theorem" the roots of the given polynomial are factors of the constant term 6. These factors are + / -1, + / -2, + / -3, or + / -6). Trial and error showed that - 2 and 3 were root so

    x^4-3x^3-5x^2+13x+6 =

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

    Next we solve for the roots of the quadratic using the quadratic equation.

    Regards,

    MrB

    P. S. I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!
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