I'm thinking that p(x) must have zeros near -1 and +1, in fact (maybe?) symmetrically such that the sum of those 2 roots is 0.  It is possible that the function is symmetrical such that p(-x) = -p(x), in which case the integral from -10 to 10 would be zero.

p(x) = (ax^2+bx+c)((x-1)^3 - 1)

p(x) = (dx^2+ex+f)((x+1)^3 + 1)

(ax^2+bx+c)(x^3-3x^2+3x-2)

(dx^2+ex+f)(x^3+3x^2+3x+2)

ax^5 + (-3a+b)x^4  + (3a-3b+c)x^3  

     + (-2a+3b-3c)x^2  + (-2b+3c)x - 2c

dx^5 + (3d+e)x^4  + (3d+3e+f)x^3  

     + (2d+3e+3f)x^2  + (2e+3f)x + 2f 

Clearly d = a and f = -c

Setting the other coefficients of x terms equal:

-3a+b = 3a+e

3a-3b+c = 3a+3e-c

-2a+3b-3c = 2a+3e-3c

-2b+3c = 2e-3c

move all terms to the left

-6a + 1b + 0c - 1e = 0

 0a - 3b + 2c - 3e = 0

-4a + 3b + 0c - 3e = 0

 0a - 2b + 6c - 2e = 0

Take these two

[-6a + 1b + 0c - 1e = 0 ] * -3 then add

-4a + 3b + 0c - 3e = 0

14a = 0 ....   a=0

Take these two

[-6a + 1b + 0c - 1e = 0 ] * -3 then add

0a - 3b + 2c - 3e = 0

12a + 2c = 0 .... c=0

Using Excel matrix solver, I get an inverse matrix:

-3/14 0 1/14 0

-1/7 -3/14 3/14 1/14

0 -1/7 0 3/14

1/7 -3/14 -3/14 1/14

Multiplying this by a 1x4 array of zeros shows a solution of 

a=b=c=d=e=f = 0.

So either I made a mistake, or not only does the integral equate to 0 but p(x)=0.