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已知多项式2X^3-X²+M有一个因式X-1,求常数M值.
人气:363 ℃ 时间:2020-04-06 13:08:53
解答
2x3-x2+ x3 = x3- x2+ x3+M
= x2(x-1)+ x3+(M1/3)3
= x2(x-1) +(x+M1/3)(x2-x•M1/3+ M2/3)
要从中提出因子(x-1),则(x+M1/3)=(x-1)
则M1/3=-1 则M=-1
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