Question

How do you find a fourth degree polynomial given roots `sqrt2` and `2-sqrt3`?

Answer 1

`x^4-4x^3-x^2+8x-2`

Explanation:

If a polynomial with all rational coefficients, then if it has one irrational root, its conjugate will also be a root.

So here, since we have a root of `sqrt2`, we also know that `-sqrt2` is a root. This goes the same for `2-sqrt3`. We know that `2+sqrt3` also must be a root. (Note these these four roots will give us a fourth degree polynomial.)

Recall that if `x=a` is a root of a polynomial, one of the polynomial's factors will be `(x-a)`. Taking our four roots in mind, we see that the polynomial function is:

`p(x)=(x-sqrt2)(x-(-sqrt2))(x-(2-sqrt3))(x-(2+sqrt3))`

`p(x)=(x-sqrt2)(x+sqrt2)(x-2+sqrt3)(x-2-sqrt3)`

Note that `(x-sqrt2)(x+sqrt2)=x^2-(sqrt2)^2=x^2-2`, as a difference of squares:

`p(x)=(x^2-2)((x-2)+sqrt3)((x-2)-sqrt3)`

Note that `((x-2)+sqrt3)((x-2)-sqrt3)` is also a difference of squares, with the terms being `(x-2)` and `sqrt3`, that is:

`((x-2)+sqrt3)((x-2)-sqrt3)=(x-2)^2-(sqrt3)^2`

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

So:

`p(x)=(x^2-2)(x^2-4x+1)`

Expanding:

`p(x)=x^2(x^2-4x+1)-2(x^2-4x+1)`

`p(x)=x^4-4x^3+x^2-2x^2+8x-2`

`p(x)=x^4-4x^3-x^2+8x-2`