Is Newton Sums Relevant?

Note that for all positive integers $n$ , we have that $n^5\equiv n\pmod{10}$ , a direct result from Fermat's Little Theorem. Since the exponents ( $1$ and $5$ ) are both odd, it follows that this holds for all integers. Therefore, $r_1+r_2+r_3+r_4+r_5$ and $r_1^5+r_2^5+r_3^5+r_4^5+r_5^5$ have the same units digit regardless of what roots they are; thus, we just need to calculate all possible $(r_1,r_2,r_3,r_4,r_5)$ such that $P(0)=2014$ .

The statement $P(0)=2014$ means that $r_1r_2r_3r_4r_5=-2014$ by Vieta's formula. We can count the number of ordered quintuplets $(r_1,r_2,r_3,r_4,r_5)$ by allotting each of $2014$ 's prime factors to a root, and then distributing positive and negative signs.

We see that $2014=2\cdot 19\cdot 53$ . Each of these prime factors can choose to be in any of the $5$ roots; thus, we have $5\times 5\times 5=125$ total ways so far.

Now, we can distribute the positive and negative signs in 3 different ways: $-----$ $---++$ $-++++$ The first case has only $\binom{5}{5}=1$ ordering, the second has $\binom{5}{3}=10$ orderings, and the third has $\binom{5}{1}=5$ orderings. Thus, there are a total of $1+10+5=16$ ways to distribute the positive and negative signs to the roots.

Thus, there is a total of $125\times 16=\boxed{2000}$ ordered pairs $(r_1,r_2,r_3,r_4,r_5)$ .