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 Polynomial Puzzle (Posted on 2007-03-16)
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)).

 See The Solution Submitted by Dennis Rating: 4.0000 (5 votes)

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 Solution (and that fourth point) | Comment 6 of 7 |

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.

 Posted by Brian Smith on 2007-03-17 21:52:42
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