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As the question states, I am looking to solve the following problem:

Find a $p \neq 0$ quadratic $q(X)$ with integral coefficients such that $q(2+\sqrt{3}) = 0$

I know how to find the answer if the roots are given, but they are not here. I was also given a hint saying "try using the fact that $(x-a)(x+a)=x-a^2$". However, I'm still not sure what to do.

The second part of the question is very similar and asks for a non-zero polynomial, but hopefully with some help on this one I can do that by myself. Thanks you for any help.

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  • $\begingroup$ Are you sure you've written the problem correctly? As written, the problem tells you that one of the roots is $2+3$, which is 5; you could then pick any integer you like for the other root. $\endgroup$ – Reese Oct 12 '17 at 14:09
  • $\begingroup$ Apologies, I have edited the question to change the 3 to sqrt 3 $\endgroup$ – user416936 Oct 12 '17 at 14:11
  • $\begingroup$ What does $p$ mean in the question? $\endgroup$ – wgrenard Oct 12 '17 at 14:23
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    $\begingroup$ You should have $(x+a)(x-a)=x^2-a^2$ $\endgroup$ – Mark Bennet Oct 12 '17 at 14:28
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Hint:

You know that $q(2 + \sqrt{3}) = 0$, and that any quadratic can be factored as $$ q(x) = a (x-r) (x-s) $$ where $r$ and $s$ are roots of the quadratic. So you can plug in $r = 2 + \sqrt{3}$ and expand out that expression for $q$ to find the coefficients (in terms of the unknowns $a$ and $s$. You then need to pick $a, s$ that will make those coefficients into integers.

Note: $s$, like $r$, will probably involve a $\sqrt{3}$.

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  • $\begingroup$ Perfect, thanks very much $\endgroup$ – user416936 Oct 12 '17 at 14:28
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You are given one root so your polynomial is $(q-2-\sqrt 3)(q-r)=0$, where $r$ is the other root. If you write it in terms of $x=q-2$ you get $(x-\sqrt 3)(x-r)=0$ Now can you find $r$ so the polynomial has rational coefficients?

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  • $\begingroup$ The quadratic is not specified to be monic; also, I think tha tthe quadratic is supposed to be $q$, with the variable being $x$, but I could be misreading -- it's a little garbled, as you can see. I have no idea what the $p \ne 0$ clause at the front is all about, for instance. $\endgroup$ – John Hughes Oct 12 '17 at 14:19
  • $\begingroup$ @JohnHughes: any polynomial can be made monic by dividing by the leading coefficient.k The $p\neq 0$ is just supposed to eliminate the simple answer that the zero polynomial evaluated at $2+\sqrt 3$ is zero. I thought the shift by $2$ made it easier to see. $\endgroup$ – Ross Millikan Oct 12 '17 at 14:27
  • $\begingroup$ I agree about the shift by $2$ -- that's a nice idea. I just wasn't certain whether the polynomial was named "p" or "q". I also agree that the polynomial, whatever it may be called, can be made monic, but if it's $2x^2 + 3x + 1$, then doing so will change it from having integer coeffs to non-integer ones like $3/2$. Judging from the phrasing of the question, I suspect that the OP might well not see the relationship between the reduced polynomial having rational coeffs and the original having integer coeffs. But that's only a guess, of course. $\endgroup$ – John Hughes Oct 12 '17 at 15:52
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If you have the equation $ax^2+bx+c=0$ with integer coefficients, you can divide through by $a$ to obtain an equation $x^2+dx+e=0$ where $d$ and $e$ are rational.

Now the sum of the roots is $-d=2+\sqrt 3 +r$ which means that $r+\sqrt 3=-d-2$ must be rational i.e. $r=f-\sqrt 3$ where $f$ is rational.

Now the product of the roots is the rational number $e=(2+\sqrt 3)(f-\sqrt 3)$. The coefficient of $\sqrt 3$ in the product has to vanish (else the product is irrational), whence $f$ can be determined. And you should be able to see how this is related to the hint, which would give you a short cut to spot the solution immediately.

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