If P(x)=0 is a polynomial equation of least d | vedclass

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(C) Let $\alpha = \sqrt{2}+\sqrt{3} i$. Since the coefficients are integers,the conjugate $\bar{\alpha} = \sqrt{2}-\sqrt{3} i$ must also be a root. Fu

Vedclass · Accepted Answer

$x^4+2 x^2+25=0$

Vedclass · Answer

(C) Let $\alpha = \sqrt{2}+\sqrt{3} i$. Since the coefficients are integers,the conjugate $\bar{\alpha} = \sqrt{2}-\sqrt{3} i$ must also be a root. Furthermore,since $\sqrt{2}$ is irrational,$-\sqrt{2}+\sqrt{3} i$ and $-\sqrt{2}-\sqrt{3} i$ must also be roots to ensure integer coefficients. The polynomial is given by: $(x-(\sqrt{2}+\sqrt{3} i))(x-(\sqrt{2}-\sqrt{3} i))(x-(-\sqrt{2}+\sqrt{3} i))(x-(-\sqrt{2}-\sqrt{3} i)) = 0$ Grouping the terms: $((x-\sqrt{2})-\sqrt{3} i)((x-\sqrt{2})+\sqrt{3} i) 	imes ((x+\sqrt{2})-\sqrt{3} i)((x+\sqrt{2})+\sqrt{3} i) = 0$ $((x-\sqrt{2})^2 + 3)((x+\sqrt{2})^2 + 3) = 0$ $(x^2 - 2\sqrt{2}x + 2 + 3)(x^2 + 2\sqrt{2}x + 2 + 3) = 0$ $(x^2 + 5 - 2\sqrt{2}x)(x^2 + 5 + 2\sqrt{2}x) = 0$ $(x^2+5)^2 - (2\sqrt{2}x)^2 = 0$ $x^4 + 10x^2 + 25 - 8x^2 = 0$ $x^4 + 2x^2 + 25 = 0$

If $P(x)=0$ is a polynomial equation of least degree with integer coefficients and $\sqrt{2}+\sqrt{3} i$ is one of its roots,then that equation is

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