Question

How do you find all the zeros of `f(x)=x^4-8x^3+17x^2-8x+16`?

Answer 1

Zeros at `x in {4,i,-i}`

Explanation:

Given
`color(white)("XXX")color(red)(1)x^4-8x^3+17x^2-8x+color(blue)(16)`

By the rational root theorem, we know that any rational roots must be factors of `color(blue)(16)/color(red)(1) =16`
which implies they must be in the set `{+-1,+-2,+-4,+-16}`

Testing each of these possibilities give a zero at (an only at) `x=4`

If we divide `x^4-8x^3+17x^2-8x+15` by `(x-4)` (using synthetic or long polynomial division)
we get factors `(x-4)(x^3-4x^2+x-4)`

Again, applying the factor theorem of `x^3-4x^2+x-4)`
we find the only zero at `x=4` (again).

Dividing `x^3-4x^2+x-4` by `(x-4)`
we can re-factor as `(x-4)(x-4)(x^2+1)`

`(x^2+1)` has no Real zeros but can be factored as `(x+i)(x-i)` to get complex zeros at `+-i`