Let p(x)=a_0+a_1 x+ldots+a_n x^n be a non-zer | vedclass

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Question

(c)

We have,

$p(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$

$p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$

$	herefore \quad \quad x=\sqrt{2}+\sqrt{3}+\sqrt{6

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(c)

We have,

$p(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$

$p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$

$	herefore \quad \quad x=\sqrt{2}+\sqrt{3}+\sqrt{6}$

$\Rightarrow \quad(x-\sqrt{6})^2=(\sqrt{2}+\sqrt{3})^2$

$\Rightarrow x^2-2 \sqrt{6} x+6=2+3+2 \sqrt{6}$

$\Rightarrow \quad\left(x^2+1ight)^2=2 \sqrt{6}(x+1)^2$

$\Rightarrow \quad x^4+2 x^2+1=24\left(x^2+2 x+1ight)$

$\Rightarrow x^4-22 x^2-48 x-23=0$

$	herefore p(x)=x^4-22 x^2-48 x-23=0$

$	ext { Minimum value of } x=4$

Let $p(x)=a_0+a_1 x+\ldots+a_n x^n$ be a non-zero polynomial with integer coefficients. If $p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$, then the smallest possible value of $n$ is

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