Question

Find all zeros of x^4-6x^2-7x-1=0?

Answer 1

Since x=+1 and x=-1 are both not zeroes to the equation, the roots must be irrational.
By Descartes rule of signs, there is one positive root, 3 negative roots or 1 negative root and two complex roots. It turns out that there is one each of positive and negative roots, and two complex roots.

You can find the answer by graphing (to get the approximate values of the roots, at x=-0.2, and x=3.

Use Newton's method to refine, as follows:
f(x)= x^4-6x^2-7x-1
f'(x)=4x^3-12x-7
To refine x=-0.2, take x0=-0.2, then
x1=-0.2-f(-0.2)/f'(-0.2)=-0.1651
x2=x1-f(x1)/f'(x1)=-0.166510669
x3=x2-f(x2)/f'(x2)=-0.1665129425915
x4=x3-f(x3)/f'(x3)=-0.16651294259751
which is sufficiently accurate for most purposes.

You can do the same for the other root, using x0=3 to get
x=2.918240841974453

Answer 2

that's not a choice

Answer 3

Please post all the choices so we can determine if the question itself is correct.

Answer 4

If you have all four roots in the choices, you only have to multiply together the factors to see what equations each one corresponds to.