利用待定系数法求常数p、q,使得x^4+px²+q能被x²+2x+5整除
题目
利用待定系数法求常数p、q,使得x^4+px²+q能被x²+2x+5整除
答案
设(x^2+2x+5)(x^2+ax+b)=x^4+px^2+q
x^4+px+q=x^4+(a+2)x^3+(2a+b+5)x^2+(5a+2b)x+5b
所以a+2=0,5a+2b=0
得到a=-2,b=5
所以p=2a+b+5=6,q=5b=25
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