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多项式a_nx~n+a_(n-1)~x~(n-1)+…a_1x+a。能被x-1整除的充要条件是a_n+a_(n-1)+…+a_1+a_0=0。根据因式定理,便可得到如下推论: “一元方程a_nx~n+a_(n-1)x~(n-1)+…+a_1x+a_0=0, x=1是它的一个根的充要条件是 a_n+a_(n-1)+…a_1+a_0=0”。在初中数学中,为了证明上述推论,可用以下方法:设x=1是方程的一个根,则得a_n+a_(n-1)+…+a_1+a_0=0,证明了条件是必要的。次设条件成立,则得a_n(x~n-1)+a_(n-1)(x~(n-1))+…+a_1(x-1)=0,可知此方程有一根是x=1,证明了条件充分。
Polynomials a_nx~n+a_(n-1)~x~(n-1)+...a_1x+a. The necessary and sufficient condition for being divisible by x-1 is a_n+a_(n-1)+...+a_1+a_0=0. According to the factorial theorem, the following inference can be obtained: "The unary equation a_nx~n+a_(n-1)x~(n-1)+...+a_1x+a_0=0, x=1 is the root of one of its charge The condition is a_n+a_(n-1)+...a_1+a_0=0”. In junior mathematics, in order to prove the above inference, the following method can be used: Let x = 1 be a root of the equation, then get a_n+a_(n-1)+...+a_1+a_0=0, which proves that the condition is necessary. When the secondary conditions are satisfied, a_n(x~n-1)+a_(n-1)(x~(n-1))+...+a_1(x-1)=0 is obtained. It can be seen that one of the equations is x. =1, it proves that the conditions are sufficient.