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

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

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