Degree $3$ polynomial with constant coefficient $2010$

$$\mathbf{Statement}$$: 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.

• 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. – Luca Bressan Sep 8 at 7:44
• @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. – Subhasis Biswas Sep 8 at 7:46
• Factoring $2010$ is not useful if you are not relying on the roots to be rational. – Ross Millikan Sep 8 at 7:48
• 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 $2010$ – Ross Millikan Sep 8 at 7:52
• 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. – 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 $$\max|\alpha_i|\geqslant |k|^{1/n}$$.

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