求多项式f(x)=x‘5-4x’3+2x‘2+3x-2的有理根
答案:1 悬赏:0 手机版
解决时间 2021-03-26 21:56
- 提问者网友：遮云壑
- 2021-03-25 23:19
求多项式f(x)=x‘5-4x’3+2x‘2+3x-2的有理根
最佳答案
- 五星知识达人网友：西风乍起
- 2021-03-26 00:11
f(x)=x^5-x³-3x³+3x²-x²+x+2x-2
=x³(x²-1)-3x²(x-1)-x(x-1)+2(x-1)
=(x-1)[x³(x+1)-3x²-x+2)
=(x-1)[x³(x+1)- (3x-2)(x+1)]
=(x-1)(x+1)(x³-3x+2)
=(x-1)(x+1)(x³-x-2x+2)
=(x-1)(x+1)[x(x²-1)-2(x-1)]
=(x-1)(x+1)(x-1)(x²+x-2)
=(x-1)(x+1)(x-1)(x+2)(x-1)
=(x-1)³(x+1)(x+2)
有理根为x=1, -1, -2
=x³(x²-1)-3x²(x-1)-x(x-1)+2(x-1)
=(x-1)[x³(x+1)-3x²-x+2)
=(x-1)[x³(x+1)- (3x-2)(x+1)]
=(x-1)(x+1)(x³-3x+2)
=(x-1)(x+1)(x³-x-2x+2)
=(x-1)(x+1)[x(x²-1)-2(x-1)]
=(x-1)(x+1)(x-1)(x²+x-2)
=(x-1)(x+1)(x-1)(x+2)(x-1)
=(x-1)³(x+1)(x+2)
有理根为x=1, -1, -2
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
大家都在看
推荐资讯