Find all roots of the equation `x^4+2x^3-16x^2-22x+7=0`, if one root is `2+sqrt3`

Explanation

All coefficients are teal, irrational roots will occur in congugate pairs.
Hence another root is `2-sqrt(3)`.
`:.` Product of these roots `=(x-2-sqrt(3))(x-2+sqrt(3))`
`=(x-2)^(2)-3=x^(2)-4x+1`
On dividing `x^(4)+2x^(3)-16x^(2)-22x+7` by `x^(2)-4x+1`, then the other quadratic factor is `x^(2)+6x+y`.
Then the given equation reduce in the form
`(x^(2)-4x+1)(x^(2)+6x+7)=0`
`:.x^(2)+6x+7=0`
Then `x=(-6+-sqrt(36-28))/2=-3+-sqrt(2)`
Hence the other roots are `2-sqrt(3),-3+-sqrt(2)`