多项式x³+ax²+bx+c能被x²+3x-4整除,则
x³+ax²+bx+c = (x²+3x-4)(x-c/4)
= x³ + (3-c/4)x² + (-4-3c/4)x + c
那么
(1)
a = 3 - c/4
4a = 12 - c
4a+c = 12
(2)
b = -4 - 3c/4
4b = -16 - 3c
4b+3c+16 = 0
(4a+c) - (4b+3c+16) = 12 - 0
4a+c - 4b-3c-16 = 12
4a - 4b - 2c - 16 = 12
2a - 2b - c - 8 = 6
2a-2b-c = 14
(3)
由a,b,c 皆为整数且 c≧a>1 与 4a+c=12 ,可得：a=2 ,c=4
将a,c代入 2a-2b-c=14 得：b= -7