Question

How do you find all the zeros of `f(x)=x^4-x^3-3x^2-x+1`?

Answer 1

`f(x)` has zeros:

`x_(1,2)=(1+sqrt(21)+-sqrt(6+2sqrt(21)))/4`

`x_(3,4)=(1-sqrt(21))/4+-sqrt(2sqrt(21)-6)/4i`

Explanation:

`f(x) = x^4-x^3-3x^2-x+1`

Notice the symmetry of the coefficients: `1, -1, -3, -1, 1`

So:

`f(x)/x^2 = x^2-x-3-1/x-1/x^2=(x+1/x)^2-(x+1/x)-5`

Let `t = x+1/x`

Then:

`0 = t^2-t-5`

`= (t-1/2)^2-1/4-5`

`= (t-1/2)^2-(sqrt(21)/2)^2`

`= (t-1/2-sqrt(21)/2)(t-1/2+sqrt(21)/2)`

So:

`x+1/x = t = 1/2+-sqrt(21)/2`

Multiply both ends by `x` and rearrange slightly to get:

`x^2-(1/2+-sqrt(21)/2)x+1 = 0`

Writing the solutions of these two possibilities separately, we have solutions given by the quadratic formula:

`x_(1,2) = (1/2+sqrt(21)/2+-sqrt((1/2+sqrt(21)/2)^2-4))/2`

`=(1+sqrt(21)+-sqrt((1+sqrt(21))^2-16))/4`

`=(1+sqrt(21)+-sqrt(6+2sqrt(21)))/4`

`x_(3,4) = (1/2-sqrt(21)/2+-sqrt((1/2-sqrt(21)/2)^2-4))/2`

`=(1-sqrt(21)+-sqrt((1-sqrt(21))^2-16))/4`

`=(1-sqrt(21)+-sqrt(6-2sqrt(21)))/4`

`=(1-sqrt(21))/4+-sqrt(2sqrt(21)-6)/4i`