Degree $3$ polynomial with constant coefficient $2010$

Question

$\mathbf{Statement}$: 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.

Answer 1

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 $|\beta|\geqslant20.1>10$.

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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.