# 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#