x4+2x3+3x2+2x+1 (02年河南赛题)

设x4+2x3+3x2+2x+1≡(x2+ax+b)(x2+cx+d)

≡x4+(a+c)x3+(d+ac+b)x2+(ad+bc)x+bd，比较两边同次幂的系数，得

∴ 原式=(x2+x+1)2

待定系数法太麻烦
=x4+2x2+1+2x3+x2+2x
=(x2+1)2+2x(x2+1)+x2
=(x2+1+x)2

写清楚点啊