perplexus.info :: Just Math : Polynomial Puzzle

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.

Comments: ( You must be logged in to post comments.)

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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