How do you divide `x^2/(2x + 4)`?

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`