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# help

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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 $$P(1)$$?

Apr 16, 2019

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There is a unique polynomial  P(x) of degree 4 with rational coefficients and leading coefficient 1

which has  as a root  $$\sqrt{2}+\sqrt{5}$$

What is P(1) ?

$$\text{Let the polynomial be }\\ P(x)=x^4+bx^3+cx^2+dx+e\\ P(\sqrt2+\sqrt5)=0\\ (\sqrt2+\sqrt5)^4+b(\sqrt2+\sqrt5)^3+c(\sqrt2+\sqrt5)^2+(\sqrt2+\sqrt5)=0\\ 89+28\sqrt{10}+b(2\sqrt{10}+17\sqrt2+7\sqrt5)+c(7+2\sqrt{10})+d(\sqrt2+\sqrt5)+e=0\\ (89+7c+e)+\sqrt{10}(28+2b+2c)+\sqrt2(17b+d)+\sqrt5(7b+d)=0\\ \text{This means that}\\ 7b+d=0\qquad and \qquad17b+d=0\\ 7b=17b\\ b=0\\ d=0\\ also\\ 28+2b+2c=0\\ 28+2c=0\\ c=-14\\ and\\ 89+7c+e=0\\ 89-98=-e\\ e=9\\ \text{So the polynomial is} \qquad P(x)=x^4-14x^2+9\\ p(1)=1-14+9=-4$$

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Apr 18, 2019