Divisible by?

Since $p(x)$ has no constant term, $x=0$ is another root. Also, note that all terms have $x$ to an odd power which means that the polynomial is an odd function. Therefore, we see that $r=-1,-2,-3$ are the three remaining roots.

As $p(x)$ has roots $r=-3,-2,-1,0,1,2,3$ , we can write it as $p(x)=d(x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)$ . We can determine $d$ by multiplying this exspression out and get that $d$ is the coefficient of $x^7$ which is given as $1$ .

What we got so far, is that $p(x) = (x+3)(x+2)(x+1)x(x-1)(x-2)(x-3)$ , so $p(x)$ is the product of 7 consecutive integers. This product will always contain three multiples of 2, one of them will be a multiple of 4. Also, there will always be 2 multiples of 3, 1 multiple of 5 and 1 multiple of 7 in the product.

It is therefore always divisible by $2^4 \cdot 3^2 \cdot 5^1 \cdot 7^1 = \boxed{5040}$ .

Factor out x, then you can see that the rest is a function of x^2, so for any nonzero root, the negative of that must also be a root. Therefore the roots are -3,-2,-1,0,1,2,3, and we can factor the polynomial as 7! (x+3 choose 7). For x=4, this is equal to 7! and it will always be divisible by 7! so the answer is 7! =5040

$p(x) = x(x^6 + ax^4 + bx^2 + c) = x(x - 1)(x - 2((x - 3)q(x) = (x^3 - 6x^2 + 11x - 6)q(x)$

Dividing we obtain the quotient $q(x) = x^3 + 6x^2 + (a + 25)x + 6(a + 15)$ and the remainder $25a + b + 301)x^2 + (-60a - 840)x + (36a + c + 540)$

The problem has all real roots $\implies$ the remainder must be zero $\implies 25a + b + 301 = 0, \:\ -60a - 840 = 0$ and $36a + c + 540 = 0 \implies a = -14, \:\ b = 49$ and $c = -36$

$\implies q(x) = x^3 + 6x^2 + 11x + 6 = (x + 1)(x + 2)(x + 3) \implies p(n) = (n + 3)(n + 2)(n + 1)n(n - 1)(n - 2)(n - 3)$

Using $n = 4 \implies p(n) = 7! = 2^4 * 3^2 * 5 * 7 \implies k = \boxed{5040}$

First set polynomial as

p(x)=x(x-1)(x+1)(x-2)(x+2(x-3)(x+3),

gives

p(x) = (x^7) - 14(x^5) + 49(x^3) - 36(x)

Then, p(1), p(2), p(3)=0 and p(4)=5040, p(5)=40320, etc

Further, GCD=5040 for all p(4) and beyond.

Answer=5040

For a longer solution let's first solve for a,b and c. And let's consider the 6th order polynomial by first dividing it by x

We know that x=1 is a root so the polynomial must divide by (x-1). Dividing the polynomial by (x-1) yields 1+a+b+c =0

Doing the same with (x-2) and (x-3) yield two further equations:

64+16a+4b+c=0 and 729+81a+9b+c=0.

Solving the simultaneous equations yields (a,b,c) = (-14,49,-36)

Now divide our 6th order polynomial by (x-1),(x-2) and (x-3) to leave x^3+6x^2+11x+6 = (x+1)(x^2+5x+6) = (x+1)(x+2)(x+3)

And so the original polynomial factorises as the product of 7 consecutive integers which must include factors of 16(3 even numbers including one multiple of 4),9(2 multiples of 3),5 and 7.

Since p is an odd function, the roots must be -3,-2,-1,0,1,2,3, so $p(n)= (n+3)(n+2)(n+1)(n)(n-1)(n-2)(n-3)$ For any integer n this is the product of 7 consecutive integers, so there must be at least three multiples of 2, of which at least one multiple of 4, two multiples of 3, one multiple of 5 and one multiple of 7 among its factors, and hence

$2^4×3^2×5×7=5040$ must divide it.