A Biquadratic Polynomial Problem (Posted on 2006-11-20)

	Submitted by K Sengupta
Rating: 3.0000 (2 votes)
Solution:	(Hide)
Let x_1, x_2, x_3 and x_4 be the roots of the polynomial. Writing Vieta’s relations in terms of s = x_1+x_2; t = x_3+x_4; p=x_1*x_2 and q = x_3*x_4, we obtain: s+t = 0; p+q+st = 0; pt+qs = -12; pq =5 Substituting t=-s into the second and the third equalities, we obtain: p+q = s^2; -q+p = -12/s The above simultaneous equations yield: p = (s^2 + 12/s)/2; q = (s^2 - 12/s)/2 From pq =-5, it follows that: (s^2 + 12/s)(s^2 - 12/s)/4 = -5 Or, s^6 +20*s^2 – 144 =0 Hence, s = x_1+x_2 is a root of the polynomial: L(x) = x^6 + 20*s^2 -144 Using similar arguments, we deduce that its other five roots are x_1+x_3, x_1+x_4, x_2+x_3, x_2+x_4 and x_3+x_4. Now, P(2) = 64 + 20*4 -144 = 0 Consequently, it follows that two of four roots of the polynomial: x4+12x -5 indeed add up to 2.

	Subject	Author	Date
	Analytical Factoring	Gamer	2006-11-21 00:20:32
	Keeping it real	Bernie Hunt	2006-11-20 17:51:56
	Solution	Bractals	2006-11-20 15:40:33