Question

How do you long divide `(6x^3 + 4x^2 + 4x + 3) div (2x^2+1)`?

Answer 1

`(6x^3+4x^2+4x+3) -: (2x^2+1)=3x+2 +(x+1)/(2x^2+1)`

Explanation:

Starting Point:`color(red)( 2x^2)+1|bar(color(blue)(6x^3)+4x^2+4x+3)`

`color(blue)("Step 1") color(white)(....)color(blue)(6x^3)-: color(red)(2x^2) = color(green)(3x)` write as:

`color(white)(xxxxxxx) color(green)(3x)`
`2x^2+1|bar(6x^3+4x^2+4x+3)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 2") color(white)(..)color(green)(3x xx (2x^2+1)) =color(red)(6x^3+3x)` write as:

`color(white)(xxxxxxx) color(green)(3x)`
`color(green)( 2x^2+1)|bar(6x^3+4x^2+4x+3)`
`color(white)(xxxxx..) underline(color(red)(6x^3color(white)(xxx.x)+3x)color(white)(x....) )-`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 3")` Subtract giving:

`color(white)(xxxxxxx) 3x`
`color(green)( 2x^2)+1|bar(6x^3+4x^2+4x+3)`
`color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-`
`color(white)(xxxxxxx) 0 color(red)(+4x^2)+x+3`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 4")color(white)(...)color(red)( 4x^2) -:color(green)( 2x^2)=color(blue)(+2)` write as:

`color(white)(xxxxxxx) 3xcolor(blue)( + 2)`
`2x^2+1|bar(6x^3+4x^2+4x+3)`
`color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-`
`color(white)(xxxxxxx) 0 +4x^2+x+3`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 5")color(white)(..)color(blue)( 2)xxcolor(red)((2x^2+1))=color(green)( 4x^2+2)` write as:

`color(white)(xxxxxxx) 3xcolor(blue)( + 2)`
`color(red)(2x^2+1)|bar(6x^3+4x^2+4x+3)`
`color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-`
`color(white)(xxxxxxx) 0 +4x^2+x+3`
`color(white)(.xxxxxxx....)underline(color(green)(4x^2color(white)(.....)+2) -)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 5")color(white)(..)`Subtract

`color(white)(xxxxxxx) 3xcolor(blue)( + 2)`
`2color(green)(x^2)+1|bar(6x^3+4x^2+4x+3)`
`color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-`
`color(white)(xxxxxxx) 0 +4x^2+x+3`
`color(white)(.xxxxxxx....)underline(4x^2color(white)(.....)+2) -`
`color(white)(xxxxxxxxxx)0color(white)(.)color(red)(+x)+1`

`"The "color(red)(+x) " from the remaindor "(x+1) " is less than the " color(green)(x^2)" from the devisor"( 2x^2+1)" so we stop."`

Giving `3x+2 +(x+1)/((2x^2+1)`