Answer:
\((3-\sqrt{6}) and (2+\sqrt{5})\)
Step-by-step explanation:
If a polynomial equation with rational coefficients is \(ax^2+bx+c=0\), then by quadratic formula
\(x=\dfrac{b\pm\sqrt{b^2-4ac}}{2a}\)
where, a,b and c are rational number.
If \(\sqrt{b^2-4ac}\) is an irrational number, then we have ± sign before the irrational number.
It is given that \((3+\sqrt{6}) and (2-\sqrt{5})\) are two roots of a polynomial equation with rational coefficients. Here, \(\sqrt{6} and \sqrt{5}\) are irrational numbers.
So, the root of the polynomial equation are
\((3\pm \sqrt{6}) and (2\pm \sqrt{5})\)
Therefore, the remaining roots of the polynomial equation are \((3-\sqrt{6}) and (2+\sqrt{5})\).
