Question

How do you long divide `(4x^2 + 8x + 9)/(2x + 1)`?

Answer 1

`color(blue)(2x+3` and remainder of `color(blue)6`

Explanation:

`color(white)(..............)color(blue)(2x+3`
`color(white)(a)2x+1` `|` `overline(4x^2+8x+9)`
`color(white)(..............)ul(4x^2+2x)`
`color(white)(........................)6x+9`
`color(white)(........................)ul(6x+3)`
`color(white)(...............................)6`

`color(blue)(2x+3` and remainder of `color(blue)6`

Answer 2

`2x+3+6/(2x+1)`

Explanation:

`"one way is to use the divisor as a factor in the numerator"`

`"consider the numerator"`

`color(red)(2x)(2x+1)color(magenta)(-2x)+8x+9`

`=color(red)(2x)(2x+1)color(red)(+3)(2x+1)color(magenta)(-3)+9`

`=color(red)(2x)(2x+1)color(red)(+3)(2x+1)+6`

`"quotient "=color(red)(2x+3)" remainder "=6`

`rArr(4x^2+8x+9)/(2x+1)=2x+3+6/(2x+1)`