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在中学数学课本里有一条定理叫余数定理“多项式f(x)除以x-b所得的余数等于f(b)”,证明时引用了下列恒等式 f(x)=(x-b)·Q(x)+R (1)当x=b时,f(b)=R。现在我们把关系式(1)引伸一下,设 f(x)=B(x)·Q(x)+R(x) (2)是一个关于x的恒等式,如果当x=b时,我们有B(b)=0,则得f(b)=R(b)。由于我们所讨论的是一元多项式,当B(x)的次数低于f(x)的次数时,R(x)的次数将低于B(x)的次数而更低于f(x)的次数,因此,求函数f(x)当x=b的值时,可以不直接代入f(x)计算而可以代入较为简单的式子R(x)里去计算,这样就方便得多了。例1. 已知 x=1/(3~(1/2)+2~(1/2)),求 f(x)=x~5+x~4-10x~3-10x~2+2x+1的值。解:x=1/(3~(1/2)+2~(1/2))=(3~(1/2)-2~(1/2)。
In the middle school mathematics textbook, there is a theorem called the remainder theorem “The remainder of the polynomial f(x) divided by xb is equal to f(b)”, and the following quotient is quoted when f(x)=(xb)Q(x)+ R (1) When x=b, f(b)=R. Now that we extend the relation (1), let f(x) = B(x) · Q(x) + R(x) (2) be an identity about x. If x = b, we have If B(b) = 0, then f(b) = R(b). Since we are talking about a univariate polynomial, when the number of B(x) is lower than the number of f(x), the number of R(x) will be lower than the number of B(x) but lower than that of f(x). The number of times, therefore, to find the function f (x) when the value of x = b, can not directly into the f (x) calculation can be substituted into the simpler formula R (x) to calculate, so much more convenient. Example 1. Given x=1/(3~(1/2)+2~(1/2)), find f(x)=x~5+x~4-10x~3-10x~2+2x The value of +1. Solution: x=1/(3~(1/2)+2~(1/2))=(3~(1/2)-2~(1/2)).