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

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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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