Question #1c2d4

1 Answer
Feb 4, 2018

#(3x^2+11x-16)/(x+5)=3x-4# Remainder #4#

Explanation:

#(x+5)|bar(3x^2+11x-16)#

Start by dividing the first term by #x#, and put the quotient above it:

#color(white)(888888888)3x#
#(x+5)|bar(3x^2+11x-16)#

Next multiply #bb((x+5))# by #bb(3x)# and subtract this from the dividend:

#color(white)(888888888)3x#
#(x+5)|bar(3x^2+11x-16)#
#color(white)(8888888//)3x^2+15x#
#color(white)(8888888//)0color(white)(88)-4xcolor(white)(8)-16#

Notice that we brought down the #bb(-16)# and we kept everything in its respective column. This is good practice if you are new to this, and stops you getting confused as to were you are.

#bb(0-4x-16)# is our new dividend and we shall be dividing this by #bb((x+5))# next.

Divide #bb(-4x)# by #bb(x)# and put the result above the bar:

#color(white)(8888.8888)3x-4#
#(x+5)|bar(3x^2+11x-16)#
#color(white)(8888888//)3x^2color(white)(8)+15x#
#color(white)(8888888//)0color(white)(88)-4xcolor(white)(8)-16#

Subtract #bb(-4(x+5))# from #bb(0-4x-16)#

#color(white)(888888888)3x-4#
#(x+5)|bar(3x^2+11x-16)#
#color(white)(8888888//)3x^2+15x#
#color(white)(8888888//)0color(white)(88)-4xcolor(white)(8)-16#
#color(white)(8888888//)0color(white)(88)-4xcolor(white)(8)-20#
#color(white)(8888888//)0color(white)(88)+0color(white)(88)+4#

Notice we cannot divide #bb(4)# by #bb(x)#. This is our remainder. So:

#(3x^2+11x-16)/(x+5)=3x-4# Remainder #4#

Hopefully this is helpful. This process is actually much easier to do than it is to explain. With a bit of practice you will even be able to do this in your head.