#{:
(,,2x,+4,),
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12),
(,,2x^2,+6x,),
(,,"------","------","------"),
(,,,4x,+12),
(,,,4x,+12),
(,,,"------","------"),
(,,,,0)
:}#
Detailed explanation (ignore if the above made adequate sense)
Starting from:
#{:
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12)
:}#
considering only the first term of the divisor (#x#) and of the dividend (#2x^2#)
we see that #2x^2 div x = 2x#; so we write #2x# above #2x^2#
#{:
(,,2x,,),
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12)
:}#
now we multiply the divisor (#x+3#) by the term we just wrote above the line (i.e. #2x#) to get #2x^2+6x# which we write under the dividend
#{:
(,,2x,+4,),
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12),
(,,2x^2,+6x,)
:}#
Subtract from the dividend
#{:
(,,2x,+4,),
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12),
(,,2x^2,+6x,),
(,,"------","------","------"),
(,,,4x,+12)
:}#
Again using only the first term of the divisor (#x#) and the first term of our remainder (#4x#)
we see that (#4x div x = 4#; so we write #4# as the next term above the top line.
#{:
(,,2x,+4,),
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12),
(,,2x^2,+6x,),
(,,"------","------","------"),
(,,,4x,+12)
:}#
Multiply the divisor (#x+3#) by the new quotient term (#+4#) to get #4x+12#; then subtract to get a final remainder of #0#
#{:
(,,2x,+4,),
(,,"------","------","------"),
(x+3,")",2x^2,+10x,+12),
(,,2x^2,+6x,),
(,,"------","------","------"),
(,,,4x,+12),
(,,,4x,+12),
(,,,"------","------"),
(,,,,0)
:}#
