Suppose that P(x) is a polynomial with rational coefficients that has a root at x = 8 + sqrt(3). Find another root of P(x).
+505 Somebody correct me if I am wrong, but another root would be 8 - sqrt(3). My reason is when using the quadratic formula, you get the root and its conjugate.
+118608 I think the main point here is that the polynomial has rational coefficients.
is [x-8 - sqrt3 ] is one factor then [x-8 + sqrt3] would have to be another.
So when they are multiplied the factor will be [(x-8)^2 - 3] or [x^2-16x+61] only rational coefficients are left.
