Again a polynomial

This problem stands on an observation.

We first notice that when $P(x)$ is divided by $(x-a)$ , it leaves a remainder of $\large{(2a^{2}+1)}$

Therefore, we can re-write the polynomial as

$\large{P(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)+2x^{2}+1}$

The reason for doing the above step:

whenever a polynomial completely divides another polynomial of higher degree, it implies that the higher degree polynomial has the other polynomial as a factor. or more importantly, both polynomials have common roots.

so, when a polynomial leaves a remainder when divided by another polynomial, it implies that when the roots of the polynomial of lower degree are plugged into the other polynomial, the polynomial will take the value of the remainder . in this case, the roots of the divisor are $1,2,3,4,5,6$ respectively, and the remainder is $2a^{2}+1$ . thus, when we plug $a$ into $P(x)$ we should get the remainder. Hence the result

Doing the above step, we can easily calculate $P(0)$ by plugging $0$ in it:

$P(0)=(0-1)(0-2)(0-3)(0-4)(0-5)(0-6)+2(0)^{2}+1=6!+1=\boxed{721}$