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Kim
Mathematics
27 February, 00:20
Finding all real roots of x^4-3x^3-5x^2+13x+6=0?
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Myla Mcdowell
27 February, 00:26
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
P. S. I'll be on vacation from Friday, Dec 22 to Jan 2, 2019. Have a Great New Year!
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