How do you use synthetic division to divide #(4x^2 - 2x + 6) (2x - 3) ^ -1#?

1 Answer
Oct 24, 2015

Answer:

#(2x+2)# with remainder #= 12#

Explanation:

#(4x^2-2x+6)*(2x-3)^(-1)#
is equivalent to #(4x^2-2x+6) div (2x-3)#

Using synthetic division (note that there may be slight variations in the structure depending upon how this is being taught):

Dividend: #color(red)(4)x^2color(red)(-2)xcolor(red)(+6)#
Divisor: #color(blue)(2)xcolor(blue)(-3)#

#{: (,"|",color(red)(4),color(red)(-2),color(red)(+6)), (,"|",,6,6), (,,"-----","-----","-----"), (/color(blue)(2),"|",4,4,12), (color(white)("XX")color(cyan)(+3),"|",2,2,) :}#
Note the change in the sign of the constant term of the divisor.

"Bring down" the #4#
Divide by the #/2# to get #2#
Multiply by #+3# to get #6#

#{: (,"|",4,-2,+6), (,"|",,6,), (,,"-----","-----","-----"), (/2,"|",4,,), (color(white)("XX")+3,"|",2,,) :}#

Add the #-2# and #6# to get #4#
Divide by the #/2# to get #2#
Miultiply by #+3# to get #6#

#{: (,"|",4,-2,+6), (,"|",,6,6), (,,"-----","-----","-----"), (/2,"|",4,4,), (color(white)("XX")+3,"|",2,2,) :}#

Add the #+6# and #6# to get the final remainder #12#

#{: (,,x^2,x^1,x^0), (,"|",4,-2,+6), (,"|",,6,6), (,,"-----","-----","-----"), (/2,"|",4,4,color(green)((12))), (color(white)("XX")+3,"|",2,2,), (,,x^1,x^0,) :}#
Note: the additional rows showing corresponding powers of #x# are not really part of the synthetic division (but hopefully add some clarity).