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Please help me with this problem.

Prove that if p(x) is polynomial with integer coefficients and p(√2)=0 then p( -√2 )=0.

Note by David Mcmillan
2 years, 1 month ago

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david mcmillan see this

Same thing Sandeep Bhardwaj said Megh Choksi · 2 years, 1 month ago

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@Megh Choksi But \(p(2) = 0\) doesn't imply that \(p(-2) = 0\). How else should I be reading "\(p(2)\) then \(( -2 ) = 0\)" ? Brian Charlesworth · 2 years, 1 month ago

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@Brian Charlesworth Sir he meant - \(p(\sqrt{2})\) Megh Choksi · 2 years, 1 month ago

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@Brian Charlesworth Sorry for the typo. I meant \( p(-\sqrt{2})\) David Mcmillan · 2 years, 1 month ago

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@David Mcmillan O.k., thanks for the update. However, the fact that \(p(2) = 0\) won't imply that \(p(-\sqrt{2}) = 0\) either. On the other hand, if \(p(x)\) has integer coefficients then \(p(\sqrt{2}) = 0 \Leftrightarrow p(-\sqrt{2}) = 0\). This comes as a result of the irrational conjugate roots theorem, which states that for any polynomial \(p(x)\) with rational coefficients, if \(a + b\sqrt{c}\) is a root of \(p(x)\), where \(a,b\) are rational and \(\sqrt{c}\) is irrational, then \(a - b\sqrt{c}\) must also be a root of \(p(x)\).

To save some time, I'm just going to give you a link to a proof of the general theorem, (which was surprisingly hard to find). Scroll down to theorem 16 on page 14 for the desired proof. You can adapt this proof for the special case \(a = 0, b = 1\) to simplify it a bit. Good question; I don't think I've ever come across a proof of this theorem before. :) Brian Charlesworth · 2 years, 1 month ago

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Do you mean that if \(p(2) = 0\) then \(p(-2) = 0\)? If so, then this is not the case. If \(p(x) = x^{2} - x - 2\) then \(p(2) = 0\) but \(p(-2) = 4\). Perhaps you meant something else with the terminology "( -2 ) = 0"?

What can be said is that if a complex number \(z\) is a root of a polynomial then so is its conjugate \(\bar {z}\), i.e., \(p(z) = 0 \Leftrightarrow p(\bar {z}) = 0\). Brian Charlesworth · 2 years, 1 month ago

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@Calvin Lin @brian charlesworth @Sreejato Bhattacharya Please help. David Mcmillan · 2 years, 1 month ago

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@David Mcmillan @Sudeep Salgia @Sandeep Bhardwaj Also try and help David Mcmillan · 2 years, 1 month ago

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@David Mcmillan This is the basic property of a polynomial equation. In fact, this is true for all rational coefficients. You try to prove it by taking a quadratic equation and then try to generalize that. Sandeep Bhardwaj · 2 years, 1 month ago

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