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

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point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x