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The polynomial $P(x)$ has integer coefficients

$p+\sqrt{q}$ is a root of the polynomial, where $p, q$ are rational

Prove that $p-\sqrt{q}$ must also be a root of the polynomial.

(Assume that q is not the square of a rational number).

please delete me
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    Are you familiar with abstract algebra? If so, apply the nontrivial field automorphism $\pm\sqrt{q}\mapsto\mp\sqrt{q}$ to the factorization $P(x)=\big(x-(p+\sqrt{q})\big)Q(x)$ over ${\bf Q}(\sqrt{q})$. – anon May 04 '13 at 08:22
  • I'm a high school student and I don't think my teacher would want me to use that in a homework question. He said this is a challenge question :( – please delete me May 04 '13 at 08:29
  • What class is it for? This would be still just as easy if you had access to polynomial gcds and the properties of minimal polynomials, but I doubt these are accessible. Thus, it's not clear to me what sort of mathematical "technology" is desired here. I suppose it's still possible for a student to crank out the nuts and bolts of the machinery I mention in my first comment, given that students would understand the analogous argument with complex conjugation, but this may be a stretch. – anon May 04 '13 at 08:33
  • Grade 12 polynomials/complex numbers. I will try to learn abstract algebra from scratch and do the question, thanks. – please delete me May 04 '13 at 08:35
  • oh, don't approve my edit, I made a wrong edit with the thought I'm asking my question. – HyperGroups May 28 '13 at 08:55

1 Answers1

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if $ap+b\sqrt{q}=0 \to ap=0 , b\sqrt{q}=0 \to -b\sqrt{q}=0 \to ap-b\sqrt{q}=0$,

for example: $a_3x^3+a_2x^2+a_1x+a_0=0$ have the root $p+\sqrt{q}$,we must have :

$a_3(p+\sqrt{q})^3+a_2(p+\sqrt{q})^2+a_1(p+\sqrt{q})+a_0=0 \to $

$a_3(p^3+3pq)+a_2(p^2+q)++a_1p+a_0=0$ ......<1> and

$a_3(3p^2\sqrt{q}+q\sqrt{q})+a_2(2p\sqrt{q})+a_1\sqrt{q}=0$......<2>

<2> $\to a_3(3p^2(-\sqrt{q})+q(-\sqrt{q})+a_2(-2p\sqrt{q})+a_1(-\sqrt{q})=0$ and

$q=(-\sqrt{q})^2 \to a_3(p-\sqrt{q})^3+a_2(p-\sqrt{q})^2+a_1(p-\sqrt{q})+a_0=0$

now you can do $a_{n}x^n+a_{n-1}x^{n-1}+...a_0=0$ by your self.

chenbai
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  • How "he can do now" the general case? Induction? It doesn't look simple at all, to say the best. any other method? – DonAntonio May 04 '13 at 09:18
  • it is a hint, for general case, you only need to say $P(x)=0 $ will be $Q_1(a_{i},p,q)+Q_2(a_{i},p,(\sqrt{q})^{2k+1})=0$, then rest is similar. of course you need to explain $x^{n}$ will only be $ Q(p,\sqrt{q})$ which I think it is the key that teacher want him to do. – chenbai May 04 '13 at 12:32