Let $P$ be a degree $3$ polynomial with complex coefficients such that the constant term is $2010$. Then $P$ has a root Let $P$ be a degree $3$ polynomial with complex coefficients such that the constant
term is $2010$. Then $P$ has a root $a$ with $|a| > 10$.
how can I show that the above statement is true/false? Can anyone help?
 A: Consider the polynomial $2010(x+1)^3$. It has constant term $2010$, and all roots are $-1$. So the result is not correct. 
The additional condition that was presumably taken for granted but left out is that the polynomial has lead coefficient $1$ (is monic).  Then the product of the roots is $-2010$, so at least one root has norm $\ge \sqrt[3]{2010}$.   So we can do somewhat better than $10$.
A: I assume the polynomial is also a monic polynomial. Else we can scale any polynomial to have the constant term as $2010$. Let the polynomial be $p(x) = x^3 + bx^2 + cx + 2010$. If the roots of the polynomial are $\alpha, \beta$ and $\gamma$, then we have that $$\alpha \beta \gamma = 2010$$
Hence, we get that $$\vert \alpha \vert \vert \beta \vert \vert \gamma \vert = 2010$$
If we have that $\vert \alpha \vert \leq 10, \vert \beta \vert \leq 10$ and $\vert \gamma \vert \leq 10$, then we get that $$2010 = \vert \alpha \vert \vert \beta \vert \vert \gamma \vert \leq 10^3$$
Hence, a contradiction. Hence, at-least one root must have magnitude greater than $10$.
A: Well it's pretty simple. The polynomial of degree three can be written as 
$$
A(x-b_1)(x-b_2)(x-b_3)= ax^3 + \dots - ab_1 b_2 b_3.
$$
By assumption we know that $Ab_1 b_2 b_3 = 2010$. Therefore, if the statement is false, all three roots have $|b_i| \le 10$, so that $|A b_1 b_2 b_3| \le 1000 |A|$. Therefore you can prove the statement if and only if $|A| <2.01$, otherwise you can easily find counter-examples by taking an appropriate polynomial with the appropriate roots.
Hope that helps, 
