Find the value of $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is a root of the polynomial
$$x^3-5x^2 +ax -1$$ I got the answer and value of $a = -10 -5\sqrt{3} -1$.
Is my answer is correct or not?
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1$\begingroup$ Does it check on your calculator? If so, it is probably correct and then there is no integer $a$ that works. $\endgroup$ – Angina Seng Dec 20 '17 at 12:08
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We need $$(2+\sqrt3)^3-5(2+\sqrt3)^2+a(2+\sqrt3)-1=0,$$ which gives $$a=2-\sqrt3-(2+\sqrt3)^2+5(2+\sqrt3)=5$$
answered Dec 20 '17 at 12:07
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$\begingroup$ @ Michael Rozenberg im not getting how can directly writes which gives $$a=2-\sqrt3-(2+\sqrt3)^2+5(2+\sqrt3)=5$$...can u elaborate this more... $\endgroup$ – user476275 Dec 20 '17 at 12:16
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$\begingroup$ I just divided by $2+\sqrt3$ and $\frac{1}{2+\sqrt3}=2-\sqrt3$. Actually, I think this way is much better than to use that $a\in\mathbb Z$. $\endgroup$ – Michael Rozenberg Dec 20 '17 at 12:17
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$\begingroup$ @ Michael Rozenberg thanks a lot i got it $\endgroup$ – user476275 Dec 20 '17 at 12:19
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If $2+\sqrt3$ is a root, then so is $2-\sqrt3$
If $x$ is the third root, then by Vieta's theorem we have:
$$(2+\sqrt3)*(2-\sqrt3)*x=1$$ as $1$ is the free coefficient
Hence $x=1$
Then again, by Vieta's theorem:
$$a=(2+\sqrt3)*1+(2-\sqrt3)*1+(2+\sqrt3)*(2-\sqrt3)=5$$
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$\begingroup$ thanks asdf.... $\endgroup$ – user476275 Dec 20 '17 at 12:20
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If $a \in \mathbb Z$, then if $2+\sqrt3$ is a root, then so is $2-\sqrt3$.
Therefore, $x^3-5x^2 +ax -1$ is divisible by $x^2 - 4 x + 1$, which has $2\pm\sqrt3$ as roots.
Now, $x^3-5x^2 +ax -1 = (x^2 - 4 x + 1)(x-1)+(a-5)x$ and so $a=5$.
