How do you divide #( -14x^2 + 4x^2 + 19 +4x)/(2x-5)#?

1 Answer
Barney V.
Jan 4, 2018

#color(magenta)( (-14x^3+4x^2+4x+19)/(2x-5) = -7x^2-15.5x-36.75# and remainder #color(magenta)(-164.75/(2x-5)#

Explanation:

#(-14x^2+4x^2+19+4x)/(2x-5)#

I think# -14x^2# should be #-14x^3#

#:.= (-14x^3+4x^2+4x+19)/(2x-5)#

#color(white)(......)color(white)(......)-7x^2-15.5x-36.75#
#2x-5|overline(-14x^3+4x^2+4x+19)#
#color(white)(.............)ul(-14x^3+35x^2)#
#color(white)(......................)-31x^2+4x#
#color(white)(.......................)ul(-31x^2+77.5x)#
#color(white)(..................................)-73.5x+19#
#color(white)(....................................)ul(-73.5x+183.75)#
#color(white)(................................................)-164.75#

#color(magenta)( (-14x^3+4x^2+4x+19)/(2x-5) = -7x^2-15.5x-36.75# and remainder #color(magenta)(-164.75/(2x-5)#

Related questions
See all questions in Division of Polynomials
Impact of this question
1537 views around the world
You can reuse this answer
Creative Commons License