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

1 Answer
Sep 11, 2016

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