Write the dividend with 0s for the missing terms:
#color(white)( (2x+4)/color(black)(2x+4))color(white)((x^2+0x+0))/(")" color(white)(x)x^2+0x + 0)#
Please observe that #x^2/(2x) = 1/2x#, therefore, we put #1/2x# in the quotient:
#color(white)( (2x+4)/color(black)(2x+4))(1/2xcolor(white)(0x+0))/(")" color(white)(x)x^2+0x + 0)#
We multiply #1/2x(2x+4) = x^2+2x# and then we subtract this underneath:
#color(white)( (2x+4)/color(black)(2x+4))(1/2xcolor(white)(0x+0))/(")" color(white)(x)x^2+0x + 0)#
#color(white)(".............")ul(-x^2-2x)#
#color(white)(".....................")-2x#
Please observe that #(-2x)/(2x) = -1#, therefore, we add #-1# in the quotient:
#color(white)( (2x+4)/color(black)(2x+4))(1/2x-1color(white)(+0))/(")" color(white)(x)x^2+0x + 0)#
#color(white)(".............")ul(-x^2-2x)#
#color(white)(".....................")-2x#
We multiply #-1(2x+4) = -2x-4# and then we subtract this underneath:
#color(white)( (2x+4)/color(black)(2x+4))(1/2x-1color(white)(+0))/(")" color(white)(x)x^2+0x + 0)#
#color(white)(".............")ul(-x^2-2x)#
#color(white)(".....................")-2x+0#
#color(white)("........................")ul(2x+4)#
#color(white)("................................")4" " larr# This is the remainder.
You can write the remainder as #4/(2x+4)# or #2/(x+2)#
The results of the division is:
#x^2/(2x + 4)= 1/2x-1+2/(x+2#
