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 f = p(10)+p(-6). The answer to the problem is then f/4.

Four equations can be formed from the information in the problem:

1 + a + b + c + d = -9

16 + 8a + 4b + 2c + d = -18

81 + 27a + 9b + 3c + d = -27

11296 + 784a + 136b + 4c + 2d = f

Solving this system results in the following:

a + d/6 = -6

b + -d = 11

c + 11d/6 = -15

0 = f - 8028

From this it is easy to see that the solution is 8028/4 = 2007. Also note that with d as a parameter, there are multiple polynomials which can generate the answer.

From the problem, three points of intersection between p(x) and q(x) = -9x are at x=1, x=2, and x=3. To find the fourth point, solve the equation p(x) - q(x) = 0:

p(x)-q(x) = 0 = (1/6)(6*x^4 - (36+d)x^3 + (66+6d)x^2 - (36+11d)x + 6d)

Factoring the equation yields:

0 = (x^3-6x^2+11x-6)*(6x-d)

Then the fourth point of intersection is x=d/6. That intersection varies with d just as the polynomial does.