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Polynomial Puzzle (Posted on 2007-03-16) Difficulty: 3 of 5
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
answerkevin raponi2007-03-18 12:00:52
SolutionSolution (and that fourth point)Brian Smith2007-03-17 21:52:42
SolutionSolutionK Sengupta2007-03-17 03:53:25
Some Thoughtsre(2): If there's only one answer...Federico Kereki2007-03-16 17:02:34
re: If there's only one answer...Gamer2007-03-16 14:08:39
SolutionsolutionCharlie2007-03-16 13:59:30
SolutionIf there's only one answer...Federico Kereki2007-03-16 13:10:06
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