F.4 AM

1a) (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

1b) a^3 + b^3 = (a+b)^3 - 3a^2b - 3ab^2
= p^3 - 3aq - 3bq
= p^3 - 3q(a+b)
= p^3 - 3pq
= p(p^2 - 3q)
a^5 + b^5 = (a+b)^5 - 5a^4b - 10a^3b^2 - 10a^2b^3 - 5ab^4
= p^5 - 5a^3q - 10aq^2 - 10bq^2 - 5b^3q
= p^5 - 5q(a^3 + 2aq + 2bq + b^3)
= p^5 - 5q(a^3 + b^3 +2pq)
= p^5 - 5q(p^3 - pq)
= p^5 - 5p^3q + 5pq^2

1c) (a^5+b^5) = -123 , a^5 + b^5 = p^5 - 5p^3q + 5pq^2
-123 = -233 - 5(-3)^3q + 5(-3)q^2
120 = 135 q -15q^2
0 = 15q^2 - 135q + 120
q = 8 OR 1
ab = 8 OR 1
when ab = 8 , a^3 + b^3 = p(p^2 - 3q)
=-3(9-24)
= 45
when ab = 1 , a^3 + b^3 = p(p^2 - 3q)
= -3(9 - 3)
= -18

2007-11-04 21:02:49 補充：
2) ax^2 bx c=0 delta = b^4 - 4ac BECAUSE IT HAS DOUBLE ROOT b^2 - 4ac = 0 b^2 = 4ac b^2/4c = a A B = -b/a AB = c/aBECAUSE IT HAS DOUBLE ROOTA B = 2A = -b/aAB = A^2 = c/a2A = -b/a = -b/b^2/4c = -4c / b A = -2c / b the root is -2c/b