How do you divide #(x^2+4x+12)div(x-4)#?

2 Answers
Dec 22, 2016

Answer:

#x+8" Remainder: "44#
or (which means the same thing)
#x+8 +(44)/(x-4)#

Explanation:

Using Synthetic division:
#(color(red)1x^2+color(red)4x+color(red)(12)) div (x-color(blue)(4))#

#{: (,,underline(color(red)(""(x^2))),,underline(color(red)(""(x))),,underline(color(red)(" "))), (,,color(red)1,,color(red)4,,color(red)12), (underline(" + ")," | ",underline(color(white)("XX")),,underline(color(blue)4xxcolor(green)1),,underline(color(blue)4xxcolor(orange)8)), (color(blue)(4)," | ",color(green)1,,color(orange)8,,color(magenta)(44)), (,,color(green)(""(x)),,,,color(magenta)("Remainder")) :}#

Dec 23, 2016

Answer:

The following is an attempt to expand on the Synthetic Division process and to compare it to Long Division.

Explanation:

#color(white)("XXXXXXXX")#Set Up

Long Division
#color(white)("XXXx")underline(color(white)("xxxxxxxxxxxxx"))#
#x-4 ) x^2+4x+12#

#color(white)("XXXXXXXXXXXX")#Synthetic Division
#color(white)("XXXXXXXXXXXX")##color(white)("XX")"|"color(white)("X")1color(white)("X")4color(white)("x")12#
#color(white)("XXXXXXXXXXXX")##underline("+ |"color(white)("XXXXXXXX"))#
#color(white)("XXXXXXXXXXXX")##4color(white)("xx")"|"#
#color(white)("XXXXXXXXXXXX")#First row are the coefficients of the dividend.
#color(white)("XXXXXXXXXXXX")#4 in the third row is the constant subtracted
#color(white)("XXXXXXXXXXXX")#from #x# in the divisor

#color(white)("XXXXXXXX")#Division of First Term

Long Division
#color(white)("XXXx")underline(1xcolor(white)("x")color(white)("XXxxxxx"))#
#x-4 ) x^2+4x+12#
Divide the first term of the divisor into the first term
of the dividend and write the result as the first term
of the quotient (above the line)

#color(white)("XXXXXXXXXXXX")#Synthetic Division
#color(white)("XXXXXXXXXXXX")##color(white)("XX")"|"color(white)("X")1color(white)("X")4color(white)("x")12#
#color(white)("XXXXXXXXXXXX")##underline(+" |"color(white)("XXx")color(white)(4)color(white)("x")color(white)(32))#
#color(white)("XXXXXXXXXXXX")##4color(white)("xx")"|"color(white)("X")1color(white)("X")color(white)(8)color(white)("X")color(white)(44)#
#color(white)("XXXXXXXXXXXX")#Add first dividend coefficient column
#color(white)("XXXXXXXXXXXX")#(1 + "nothing") and write sum in
#color(white)("XXXXXXXXXXXX")#first quotient column

#color(white)("XXXXXXXX")#Generate New Dividend Terms

Long Division
#color(white)("XXXx")underline(1xcolor(white)("x")color(white)(+8)color(white)("xxxxx"))#
#x-4 ) x^2+4x+12#
#color(white)("XXXx")underline(x^2color(white)("x")-4xcolor(white)("xxxx"))#
#color(white)("XXXXXXX")8x+12#
Multiply most recent divisor term (#1x#)
by the divisor (#x-4#) and subtract
the product from the (previous) dividend

#color(white)("XXXXXXXXXXXX")#Synthetic Division
#color(white)("XXXXXXXXXXXX")##color(white)("XX")"|"color(white)("X")1color(white)("X")4color(white)("x")12#
#color(white)("XXXXXXXXXXXX")##underline(+" |"color(white)("XXx")4color(white)("x"color(white)(32))#
#color(white)("XXXXXXXXXXXX")##4color(white)("xx")"|"color(white)("X")1color(white)("X")8color(white)("X")color(white)(44)#
#color(white)("XXXXXXXXXXXX")#Multiply the divisor constant (#4#)
#color(white)("XXXXXXXXXXXX")#by the most recent quotient coefficient (#1#)
#color(white)("XXXXXXXXXXXX")#and write the result on row 2
#color(white)("XXXXXXXXXXXX")#under the second dividend coefficient;# color(white)("XXXXXXXXXXXX")#then add the second column to get the next#color(white)("XXXXXXXXXXXX")#quotient term coefficient.

#color(white)("XXXXXXXX")#Repeat Process Until All Terms Used

Long Division
#color(white)("XXXx")underline(1xcolor(white)("x")+8color(white)("xxxxx"))#
#x-4 ) x^2+4x+12#
#color(white)("XXXx")underline(x^2color(white)("x")-4xcolor(white)("xxxx"))#
#color(white)("XXXXXXX")8x+12#
#color(white)("XXXXXXX")underline(8x-32)#
#color(white)("XXXXXXXXXX")44#

#color(white)("XXXXXXXXXXXX")#Synthetic Division
#color(white)("XXXXXXXXXXXX")##color(white)("XX")"|"color(white)("X")1color(white)("X")4color(white)("x")12#
#color(white)("XXXXXXXXXXXX")##underline(+" |"color(white)("XXx")4color(white)("x")32)#
#color(white)("XXXXXXXXXXXX")##4color(white)("xx")"|"color(white)("X")1color(white)("X")8color(white)("X")44#
#color(white)("XXXXXXXXXXXX")#Note that the final sum on the 3rd line (#44#)
#color(white)("XXXXXXXXXXXX")#is the Remainder;
#color(white)("XXXXXXXXXXXX")#the other sums are the coefficients of #x#
#color(white)("XXXXXXXXXXXX")#and the quotient constant, respectively.