How do you long divide #(6x^3 + 4x^2 + 4x + 3) div (2x^2+1)#?

1 Answer
Jan 5, 2016

#(6x^3+4x^2+4x+3) -: (2x^2+1)=3x+2 +(x+1)/(2x^2+1)#

Explanation:

Starting Point:#color(red)( 2x^2)+1|bar(color(blue)(6x^3)+4x^2+4x+3)#

#color(blue)("Step 1") color(white)(....)color(blue)(6x^3)-: color(red)(2x^2) = color(green)(3x)# write as:

#color(white)(xxxxxxx) color(green)(3x)#
# 2x^2+1|bar(6x^3+4x^2+4x+3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2") color(white)(..)color(green)(3x xx (2x^2+1)) =color(red)(6x^3+3x)# write as:

#color(white)(xxxxxxx) color(green)(3x)#
#color(green)( 2x^2+1)|bar(6x^3+4x^2+4x+3)#
#color(white)(xxxxx..) underline(color(red)(6x^3color(white)(xxx.x)+3x)color(white)(x....) )-#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")# Subtract giving:

#color(white)(xxxxxxx) 3x#
#color(green)( 2x^2)+1|bar(6x^3+4x^2+4x+3)#
#color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-#
#color(white)(xxxxxxx) 0 color(red)(+4x^2)+x+3#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")color(white)(...)color(red)( 4x^2) -:color(green)( 2x^2)=color(blue)(+2)# write as:

#color(white)(xxxxxxx) 3xcolor(blue)( + 2)#
# 2x^2+1|bar(6x^3+4x^2+4x+3)#
#color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-#
#color(white)(xxxxxxx) 0 +4x^2+x+3#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")color(white)(..)color(blue)( 2)xxcolor(red)((2x^2+1))=color(green)( 4x^2+2)# write as:

#color(white)(xxxxxxx) 3xcolor(blue)( + 2)#
# color(red)(2x^2+1)|bar(6x^3+4x^2+4x+3)#
#color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-#
#color(white)(xxxxxxx) 0 +4x^2+x+3#
#color(white)(.xxxxxxx....)underline(color(green)(4x^2color(white)(.....)+2) -)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 5")color(white)(..) #Subtract

#color(white)(xxxxxxx) 3xcolor(blue)( + 2)#
# 2color(green)(x^2)+1|bar(6x^3+4x^2+4x+3)#
#color(white)(xxxxx..) underline(6x^3color(white)(xxx.x)+3xcolor(white)(......) )-#
#color(white)(xxxxxxx) 0 +4x^2+x+3#
#color(white)(.xxxxxxx....)underline(4x^2color(white)(.....)+2) -#
#color(white)(xxxxxxxxxx)0color(white)(.)color(red)(+x)+1 #

#"The "color(red)(+x) " from the remaindor "(x+1) " is less than the " color(green)(x^2)" from the devisor"( 2x^2+1)" so we stop."#

Giving #3x+2 +(x+1)/((2x^2+1)#