How do you long divide #(4x^2 + 8x + 9)/(2x + 1)#?

2 Answers
May 28, 2017

#color(blue)(2x+3# and remainder of #color(blue)6#

Explanation:

# color(white)(..............)color(blue)(2x+3#
#color(white)(a)2x+1##|##overline(4x^2+8x+9)#
#color(white)(..............)ul(4x^2+2x)#
#color(white)(........................)6x+9#
#color(white)(........................)ul(6x+3)#
#color(white)(...............................)6#

#color(blue)(2x+3# and remainder of #color(blue)6#

May 28, 2017

#2x+3+6/(2x+1)#

Explanation:

#"one way is to use the divisor as a factor in the numerator"#

#"consider the numerator"#

#color(red)(2x)(2x+1)color(magenta)(-2x)+8x+9#

#=color(red)(2x)(2x+1)color(red)(+3)(2x+1)color(magenta)(-3)+9#

#=color(red)(2x)(2x+1)color(red)(+3)(2x+1)+6#

#"quotient "=color(red)(2x+3)" remainder "=6#

#rArr(4x^2+8x+9)/(2x+1)=2x+3+6/(2x+1)#