Question

用带余除法计算 （2x的2次方＋5x－1）／（x＋2）

Answer 1

(2x^2+5x-1)/(x+2)

=[(2x^2+8x+8)-3x-9]/(x+2)

=[2(x^2+4x+4)-3x-9]/(x+2)

=[2(x+2)^2-(3x+9)]/(x+2)

=2(x+2)^2/(x+2)-(3x+9)/(x+2)

=2(x+2)-(3x+9)/(x+2)