How do you divide #(4x^3-5x^2-4x-12)/(3x-4) #?

2 Answers
Nov 24, 2017

#4/3 x^2 + 1/9x - 32/27 " Remainder: " (-196)/(27(3x-4))#

Explanation:

This is best solved using long division. The question is, what do I have to multiply by #3x-4# to get #4x^3-5x^2-4x-12#?

#" "4/3 x^2 + 1/9x - 32/27 " Remainder: " (-196)/(27(3x-4))#
#3x-4 |bar(4x^3-5x^2-4x-12)#
#" "-ul((4x^3-16/3 x^2)) downarrow " " downarrow#
#" "1/3x^2-4x#
#" "-ul((1/3x^2-4/9x))#
#" "-32/9x-12#
#" "-ul((-32/9x+128/27))#
#" "-196/27#

Nov 24, 2017

#color(magenta)(1.3x^2+0.1x-1.2# and remainder of #color(magenta)(-6#

Explanation:

#(4x^3-5x^2-4x-12)/(3x-4)#

# color(white)(................)color(magenta)(1.3x^2+0.1x-1.2#
#color(white)(a)3x-4##|##overline(4x^3-5x^2-4x-12)#
#color(white)(..............)ul(4x^3-5.3x^2)#
#color(white)(........................)0.3x^2-4x#
#color(white)(........................)ul(0.3x^2-0.4x)#
#color(white)(...............................)-3.6x-12#
#color(white)(................................)ul(-3.6x+4.8)#
#color(white)(...........................................)color(magenta)(-6#

#color(magenta)((4x^3-5x^2-4x-12)/(3x-4)=1.3x^2+0.1x-1.2#and remainder of #color(magenta)(-6#