How do you use synthetic division to divide #(4x^2 - 2x + 6)# by #2x-3#?

1 Answer
Oct 20, 2015

Answer:

#(4x^2-2x+6) div (2x-3) = (2x+2) R:12#

Explanation:

#(color(brown)(4)x^2color(brown)(-2)xcolor(brown)(+6))div (color(cyan)(2)xcolor(blue)(-3))#

Remember to reverse the sign on #color(blue)((-3))#
and since we have a non-monic divisor, we need to divide each column sum by (in this case) #color(cyan)(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)#