Forcing a Polynomial Factor (Posted on 2013-08-03)

	Submitted by Brian Smith
Rating: 4.0000 (1 votes)
Solution:	(Hide)
x^10/(x^2+2x+3) has a remainder of 22x-219, then x^10-22x+219 has x^2+2x+3 as a factor. x^5/(x^2+2x+3) has a remainder of -11x-12, then x^5+11x+12 has x^2+2x+3 as a factor. Now to find a and b: the expression x^10-22x+219 + a(x^5+11x+12) must have the linear term drop out. a=2 does that. Then x^10-22x+219 + a(x^5+11x+12) = x^10 + 2x^5 + 243. The answer is a=2, b=243. To check (x^10+2x^5+243)/(x^2+2x+3) = x^8-2x^7+x^6+4x^5-11x^4+12x^3+9x^2-54x+81.

	Subject	Author	Date
	re(2): Another approach	Steve Herman	2013-08-04 23:21:39
	re: Another approach	Larry	2013-08-04 20:05:43
	Another approach	Steve Herman	2013-08-04 15:34:09
	solution	Daniel	2013-08-03 22:45:59