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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 | Comment 5 of 7 |

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

 Posted by K Sengupta on 2007-03-17 03:53:25
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