How do you divide #(2x^3+7x^2-5x-4) / (2x+1)# using polynomial long division?

2 Answers
Jun 15, 2018

Answer:

Quotient is #x^2+3x-4# and remainder is #0#

Explanation:

#2x^3+7x^2-5x-4#

=#2x^3+x^2+6x^2+3x-8x-4#

=#x^2*(2x+1)+3x*(2x+1)-4*(2x+1)#

=#(2x+1)*(x^2+3x-4)#

Hence quotient is #x^2+3x-4# and remainder is #0#

Jun 15, 2018

Answer:

#x^2+3x-4 #

Explanation:

Numerator #->color(white)("d")2x^3+7x^2-5x-4#
#color(magenta)(x^2)(2x+1)-> color(white)("d")ul(2x^3+x^2larr" Subtract"#
#color(white)("dddddddddddddd") 0+6x^2-5x-4#
#color(magenta)(3x)(2x+1)-> color(white)("dddddd") ul(6x^2+3xlarr" Subtract")#
#color(white)("ddddddddddddddddddd")0-8x-4#
#color(magenta)(-4)(2x+1) ->color(white)("ddddddd.d")ul(-8x-4 larr" Subtract")#
#"Remainder"-> color(white)("ddddddddddd")0color(white)("d")+0#

#color(white)("dddddddddddddd")color(magenta)( x^2+3x-4 )#