what is wrong with $p(x)=x^n+a_{n-1}x^{n−1} +...+a_0$ ?

$p(x)=x^n+a_{n-1}x^{n-1}+...+a_0$

$p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0$ Your minus in the exponant ... (And, to be perfect, you may use \ldots )

How about $p(x) = x^{2} - 2014x$? Then $q(x)$ just has at most $2$ zeros. Something's wrong. rightways wrote: Let $n > 1$ and $p(x)=x^n+a_{n-1}x^{n-1} +...+a_0$ be a polynomial with $n$ real roots (counted with multiplicity). Let the polynomial $q$ be defined by $$q(x) = \prod_{j=1}^{2015} p(x + j)$$. We know that $p(2015) = 2015$. Prove that $q$ has at least $1970$ different roots $r_1, ..., r_{1970}$ such that $|r_j| < 2015$ for all $ j = 1, ..., 1970$.

Let $x_1,x_2 , \dots , x_n$ be the roots of $p(x)$ and W.L.O.G let $x_1 \leq x_2 \leq \dots \leq x_n$ $\Longleftrightarrow p(x)=(x-x_1)(x-x_2) \dots (x-x_n)$ On the other hand $$q(x) = \prod_{j=1}^{2015} p(x + j)=\prod_{j=1}^{2015}(x-x_1+j)(x-x_2+j) \dots (x-x_n+j) $$Wich means all the roots of $q(x)$ are $r=x_i-j$ for $i \in \{1,2,\dots ,n\} $ and $j \in \{1,2, \dots ,2015\}$ (counted with multiplicity) Claim : If $x_i \in (0,2015]$ then there exist $2015$ roots of $q(x)$ such that $|r-2015| \leq 2015$ Proof: Trivial , just see that $|x_i-j|\leq 2015$ The condition $p(2015)=2015$ is equivalent to $ \prod_{i=1}^{n} (2015-x_i)=2015 $ Now we split in two cases First case : $n$ is odd From the bounding above we have that $$(2015-x_n)^n \leq 2015 \Longleftrightarrow x_n\geq 2015-\sqrt[n]{2015}$$Assume that $x_n >2015$ because otherwise we would be done by the first claim This means that $2015-x_n<0$ , and assume that the number of $x_i $'s such that $x_i>2015 $ is odd , but that would mean $ \prod_{i=1}^{n} (2015-x_i) <0$ , a clear contradiction. So there is an even number of $x_i$'s such that $x_i>2015$ and an odd such that $x_i<2015$ For all $x_i$'s such that $x_i<2015$ assume that $x_i\leq 2015-\sqrt[n]{2015}$ because again by the first claim we would be done . Let $k$(odd) be the number of $x_i$'s such that $x_i <2015$ , then let $A$ the product of $2015-x_i$ 's of such $x_i$'s ,and by the above inequality $A>(\sqrt[n]{2015})^k$ Claim : If $x_i\in [2015,2015+\sqrt[n]{2015}]$ then there exists more than $1970$ roots such that $|r|<2015$ Proof : this is trivial too , see that $x_i< 2059$ and the number of $|x_i-j|<2015$ is $>2015-(2059-2015)>1970$ . So , by the above claim if the $x_i$'s $>2015$ are also less than $2015+\sqrt[n]{2015}$ , we are done . So assume that these $x_i$'s are greater than $2015+\sqrt[n]{2015}$, that means $x_i-2015>\sqrt[n]{2015} $, that would mean $B$ the product of such $2015-x_i$ is greater than$( \sqrt[n]{2015} )^{n-k}$ This means $\prod_{i=1}^{n} (2015-x_1)=A \cdot B >2015$ , contradiction . This proves for $n$ odd. Second case : If $n$ is even. Like above we have that $(2015-x_n)^n \leq 2015$ and since $n$ even we have that $$2015+\sqrt[n]{2015} \geq x_n \geq 2015-\sqrt[n]{2015}$$wich is true by the two claims. $\blacksquare$ A polynomial with exactly $1970$ roots such that conditions are fullfilled is $p(x)=(x-2015-\sqrt[2]{2015})^2$