$\mathbf{Statement}$: Let $P$ be a degree $3$ polynomial with complex coefficients such that the constant term is $2010$. Then $P$ has a root $\alpha$ with $|\alpha|>10$. (TRUE OR FALSE?).

Approach: We factor the constant term $2010$. The prime factors are $2,3,5,67$ (each raised to the power $1$ in prime representation).

Let us consider the polynomial $P^*$ such that $x(x^2+b)=2010$.

We start of with (guesstimate) $x=15$. [Other numbers from $11$ to $14$ weren't chosen since the prime factors "don't match"].

Now, $x^2+b=225+b=2 \times 67=134 \implies b=-91$.

Thereby, $x(x^2-91)=2010$. From this: $(-x)^3-91(-x)+2010=0$. Taking $(-x) \mapsto x$,

$P := x^3-91x+2010$ which has a real root $\alpha=-15$, and $|\alpha|>15$.

$\mathbf{EDIT:}$ Further generalisation of the problem:

Let $P$ be an $n$-th degree polynomial with complex coefficients and with the constant term $k$, then $P$ has at least one root $\alpha$ such that $|\alpha| \geq |k|^{1/n}$

Proof: Suppose that all of the roots, say $r_1,r_2,...,r_n$ are $<|k|^{1/n}$. By Vieta's relation, $|r_1 r_2...r_n| =|k|$. But, we get $|k| <|k|$, a contradiction.

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    $\begingroup$ You have found one particular polynomial that satisfies the hypotheses (has degree $3$ and constant term $2010$) and the thesis (has a root $\alpha$ with $\lvert \alpha \rvert > 10$). But in order to prove the statement, you need to prove that for any polynomial $P$ satisfying the hypotheses, the thesis will hold. $\endgroup$ – Luca Bressan Sep 8 at 7:44
  • $\begingroup$ @LucaBressan I will try to prove it now. I actually didn't realize that I have to prove the "theorem". So I just found a "particular case". Thank you for pointing it out. $\endgroup$ – Subhasis Biswas Sep 8 at 7:46
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    $\begingroup$ Factoring $2010$ is not useful if you are not relying on the roots to be rational. $\endgroup$ – Ross Millikan Sep 8 at 7:48
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    $\begingroup$ Do you require that the polynomial be monic? Otherwise you can take a polynomial with its roots near $0$ and multiply by the factor required to make the constant term $2010$ $\endgroup$ – Ross Millikan Sep 8 at 7:52
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    $\begingroup$ Chiming in with Ross Millikan. The claim is false for example for the cubic $2010x^3+2010$. OTOH if the cubic is monic, then Vieta relations are all you need. $\endgroup$ – Jyrki Lahtonen Sep 8 at 7:54

If there are no other conditions (such as the polynomial being monic), then this is clearly false. For example, $$2010(x+1)(2x+1)(3x+1)=12060 x^3+22110 x^2+12060 x+2010$$ has roots $-1,-\frac12$ and $-\frac13$.

If, on the other hand, you require the polynomial to be monic, then it is true. Indeed, if the constant term is $2010$, then minus the product of the roots $-\alpha\beta\gamma=2010$, so that $\gamma=-\frac{2010}{\alpha\beta}$ (by Viète's formulae).

Now if $|\gamma|=|{-\frac{2010}{\alpha\beta}}|>10$ we are done, so suppose it is $\leqslant10$. It follows that $|\alpha\beta|=|\alpha||\beta|\geqslant 201$. Clearly if $|\alpha|>10$ we are done, so assume $|\alpha|\leqslant 10$. Then we get $10|\beta|\geqslant|\alpha||\beta|\geqslant201$, which implies that $|\beta|\geqslant20.1>10$.

Edit: For the general case, let $\alpha_1,\dots,\alpha_n$ be the roots. Then again by Viète's formulae, we have that $$|k|= \prod_{i=1}^n|\alpha_i|\geqslant\big(\max_{i=1,\dots,n}|\alpha_i|\big)^n,$$ which implies that $\max|\alpha_i|\geqslant |k|^{1/n}$.

After writing my edit, I read your proof by contradiction, which is also correct.

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    $\begingroup$ You could also have shorter said that by Viete $\max(|α|,|β|,|γ|)^3\ge 2010$, so that $\max(|α|,|β|,|γ|)\ge 10\sqrt[3]{2.01}>10$. $\endgroup$ – Dr. Lutz Lehmann Sep 8 at 8:13
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    $\begingroup$ Solved it. Then arrived here. When I did finally figure it out myself, I felt really dumb :( $\endgroup$ – Subhasis Biswas Sep 8 at 8:28
  • $\begingroup$ Could you please check out the edit $\endgroup$ – Subhasis Biswas Sep 8 at 8:39
  • $\begingroup$ @SubhasisBiswas Notice the shorter trick that LutzL proposed. I will edit my answer. $\endgroup$ – Luke Collins Sep 8 at 8:45
  • $\begingroup$ @LukeCollins, Noticed. That's what I have tried to do in case of arbitrary polynomials of $n$ th degree in edit. $\endgroup$ – Subhasis Biswas Sep 8 at 8:48

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