(color(brown)(4)x^2color(brown)(-2)xcolor(brown)(+6))div (color(cyan)(2)xcolor(blue)(-3))(4x2−2x+6)÷(2x−3)
Remember to reverse the sign on color(blue)((-3))(−3)
and since we have a non-monic divisor, we need to divide each column sum by (in this case) color(cyan)(2)2
"Bring down" the first coefficient
Then divide by 2
{:
( ," | ",color(brown)(4),color(brown)(-2),color(brown)(+6)),
(color(blue)(+3)," | ", , , ),
( ," | ","----","----","----"),
(color(cyan)(/2)," | ",4,,),
(," | ",2,color(white)("X")2,)
:}
Multiply the last column quotient (2) by 3 and write in the next column.
Add that column.
{:
( ," | ",4,-2,+6),
(+3," | ", , color(white)("X")6 , ),
( ," | ","----","----","----"),
(/2," | ",4,color(white)("X")4,),
(," | ",2,,)
:}
Repeat this process until done
{:
( ," | ",4,-2,+6),
(+3," | ", , color(white)("X")6 , color(white)("X")6),
( ," | ","----","----","----"),
(/2," | ",4,color(white)("X")4,color(white)("X")color(red)(12)),
(," | ",color(green)(2),color(white)("X")color(orange)(2),)
:}
The last sum (undivided), color(red)(12), is the remainder.
The quotients preceding the last column, color(green)(2) and color(orange)(2), are the coefficients of the quotient expression.
That is the solution is
color(green)(2)x+color(orange)(2) with a Remainder of color(red)(12)
