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.
Solution (and that fourth point)
