Find a monic quartic polynomial f(x) with rational coefficients whose roots include x = 1 - sqrt(2) and x = 2 + sqrt(7).

Guest Apr 17, 2021

#1**0 **

Since we wish to get rid of the irrational numbers in the polynomial, we set the other 2 roots to be 1+sqrt(2) and 2-sqrt(7), since they are the conjugates of those roots, and when an irrational number multiplies the conjugate of that number, it becomes rational. So the monic polynomial is:

\((x-(1-\sqrt{2}))(x-(1+\sqrt{2}))(x-(2+\sqrt{7}))(x-(2-\sqrt{7}))=\\ (x^2-x(1+\sqrt{2})-x(1-\sqrt{2})+(1-\sqrt{2})(1+\sqrt{2}))(x^2-x(2+\sqrt{7})-x(2-\sqrt{7})+(2+\sqrt{7})(2-\sqrt{7}))=\\ (x^2-2x-1)(x^2-2x-3)=\\ \boxed{x^4-6x^3+4x^2+10x+3}\)

textot Apr 17, 2021