How do you find a fourth degree polynomial given roots sqrt2 and 2-sqrt3?
1 Answer
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
Recall that if
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
p(x)=(x^2-2)((x-2)+sqrt3)((x-2)-sqrt3)
Note that
((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