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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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re: If there's only one answer...

| Comment 3 of 7 |

(In reply to If there's only one answer... by Federico Kereki)

Let a be the unknown root, then p(x)=-9x+(x-1)(x-2)(x-3)(x-a)

p(10)=9*8*7*(10-a), p(-6)=-7*-8*-9*(-6-a)=9*8*7*(6+a)

So (p(10)+p(-6))/4 = (7*8*9*(10-a+6+a)-90+54)/4=4*7*8*9-9=2007

Edited on March 16, 2007, 2:11 pm

Posted by Gamer on 2007-03-16 14:08:39

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