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

1 Answer
Sep 11, 2016

Answer:

#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#