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

Hello from MrBillDoesMath!

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!

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743