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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)).

 Submitted by Dennis Rating: 4.0000 (5 votes) Solution: (Hide) Let q(x)=p(x)+9x. So q(1)=q(2)=q(3)=0. Since q(x) is a fourth degree polynomial, q(x)=(x-1)(x-2)(x-3)(x-r) --> p(x)=(x-1)(x-2)(x-3)(x-r)-9x --> p(10)+p(-6)=9*8*7*(10-r)-90+(-7)(-8)(-9)(-6-r)+54 --> p(10)+p(-6)=10*9*8*7+6*7*8*9-36=8028 --> .25(p(10)+p(-6))=2007.

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 Subject Author Date answer kevin raponi 2007-03-18 12:00:52 Solution (and that fourth point) Brian Smith 2007-03-17 21:52:42 Solution K Sengupta 2007-03-17 03:53:25 re(2): If there's only one answer... Federico Kereki 2007-03-16 17:02:34 re: If there's only one answer... Gamer 2007-03-16 14:08:39 solution Charlie 2007-03-16 13:59:30 If there's only one answer... Federico Kereki 2007-03-16 13:10:06

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