Let p(x)=x^4+ax^3+bx^2+cx+d, where a, b, c, and d are constants.

If p(1)=-9, p(2)=-18, and p(3)=-27, find the value of
¼(p(10)+p(-6)).

Let q(x) = p(x) + 9x.

Then, by the problem:

q(1) = q(2) = q(3) = 0, and so:

q(x) is divisible by (x-1)(x-2)(x-3).

Since, the coefficient of x^4 is 1 while the coefficient of x^0 is d, it follows that:

q(x) = (x-1)(x-2)(x-3)(x- d/6); and so:

q(10) = 504(10-m)

q(-6) = 504(6+m)

Or, 1/4( q(10) + q(-6)) = 2016

Or, 1/4( p(10) + p(-6))

= 2016 - 9/4*(10-6)

= 2016 - 9

= 2007