✔ 最佳答案
f(x) = (x^2+x+1)(3x^2+2x+1) + 2x-3
令 f(x) = (x^2+1)(3x^2+ax+b) + cx+d
f( 0 ) = 1*1-3 = 1*b+d
b+d = -2 .....(1式)
f( i ) = (-1+i+1)(-3+2i+1)+2i-3 = (-1+1)(-3+ai+b)+ci+d
i ( -2+2i ) + 2i-3 = ci+d
-2i-2+2i-3 = ci+d
-5 = ci+d
故 c=0 , d = -5
代入(1式)得 b = 3
f( 1 ) = 3*6-1 = 2*(3+a+b)-5 ←因為c=0 , d = -5
17 = 2*(3+a+b)-5
22 = 2*(3+a+b)
3+a+b = 11
a = 8 - b = 8 - 3 = 5
故所求商式 = 3x^2+ax+b = 3x^2+5x+3
Ans: 3x^2+5x+3