Question

How do you find all the zeros of `f(x)=x^4-2x-1`?

Answer 1

Reduce to solving two quadratics...

Explanation:

`f(x) = x^4-2x-1`

Since `f(x)` has no term in `x^3` it will factor as the product of a pair of quadratics with opposite middle terms:

`x^4-2x-1`

`= (x^2-ax+b)(x^2+ax+c)`

`= x^4+(b+c-a^2)x^2+(b-c)ax+bc`

Equating coefficients and rearranging a little we get:

`{ (b+c = a^2), (b-c = -2/a), (bc=-1) :}`

Then:

`(a^2)^2 = (b+c)^2 = (b-c)^2+4bc = (-2/a)^2-4 = 4/((a^2))-4`

Multiply both ends by `(a^2)` and rearrange slightly to get:

`(a^2)^3 + 4(a^2) - 4 = 0`

Use Cardano's method to solve this cubic in `(a^2)`...

Let `a^2 = u+v`

`u^3+v^2+(3uv+4)(u+v)-4 = 0`

Add the constraint `v = -4/(3u)` to eliminate the `(u+v)` term:

`u^3-64/(27u^3)-4 = 0`

Multiply through by `27u^3` and rearrange slightly to get:

`27(u^3)^2-108(u^3)-64 = 0`

Use the quadratic formula to find:

`u^3 = (108+-sqrt((-108)^2-4(27)(-64)))/(2*27)`

`=(108+-sqrt(11664+6912))/54`

`=(108+-sqrt(18576))/54`

`=(108+-12sqrt(129))/54`

`=(54+-6sqrt(129))/27`

Hence Real root:

`a^2 = 1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))`

We can use the positive root as the value of `a`:

`a = sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))`

Then:

`b = 1/2(a^2-2/a)`

`= 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))-1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))`

`c = 1/2(a^2+2/a)`

`= 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))+1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))`

This leaves us with two quadratics to solve:

`x^2-(sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))))x+( 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))-1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))) = 0`

`x^2+(sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))))x+( 1/6(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129)))+1/sqrt(1/3(root(3)(54+6sqrt(129))+root(3)(54-6sqrt(129))))) = 0`

We can find the roots of these quadratics using the quadratic formula, which are the zeros of the original quartic `f(x)`.