There is a unique polynomial p(x) of degree 4 with rational coefficients and leading coefficient 1 which has sqrt(2) + sqrt(5) as a root. What is p(1)?
+6248 \(\text{The coefficient of $x^3$ is -1 times the sum of the roots}\\ \text{The constant term is the product of all the roots}\\ \text{Thus if the coefficients are rational and $\sqrt{2}+\sqrt{5}$ is a root then so must the following be roots}\\ -\sqrt{2}+\sqrt{5},~\sqrt{2}-\sqrt{5},~-\sqrt{2}-\sqrt{5}\)
\(\text{Using those 4 roots expand out $\prod \limits_{k=1}^4 (x-r_k)$ and find the coefficient of the $x^3$ term}\\ \text{This is the negative of $p(1)$ i.e. the sum of the coefficients of the quartic}\\ \text{You should get the sum is equal to $2\sqrt{5}$}\)
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