用待定系数法的题
如果f(x)=x④+2x③+ax②+bx+1是一个二次多项式的完全平方式,试用待定系数法,求a,b的值
PS:④是四次方 ③是三次方 ②是平方
人气:114 ℃ 时间:2020-05-13 14:00:36
解答
设二次多项式是:x^2+mx+n
x^4+2x^3+ax^2+bx+1=(x^2+mx+n)^2=x^4+m^2x^2+n^2+2mx^3+2nx^2+2mnx
所以:
2=2m
a=m^2+2n
b=2mn
n^2=1
即:m=1,n=1或-1
a=1+2=3或者a=1-2=-1
b=2*1*1=2或者b=2*1*[-1]=-2
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