How do you divide #(3x^3 - 12x^2 - 11x - 20)/(x+5)#?

2 Answers
Dec 24, 2017

#(3x^3−12x^2−11x−20)/(x+5)=3x^2-27x+124-640/(x+5)#

Explanation:

#(3x^3−12x^2−11x−20)-:(x+5)=3x^2-27x+124#
#(3x^3+15x^2)/#
#color(white)(..)0-27x^2-11x-20#
#color(white)(....)(-27x^2-135x)/#
#color(white)(...........)0+124x-20#
#color(white)(..............)(+124x+620)/#
#color(white)(......................)0-640#

#(3x^3−12x^2−11x−20)/(x+5)=3x^2-27x+124+(-640)/(x+5)#

Dec 24, 2017

#3x^2-27x+124-640/(x+5)#

Explanation:

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

#"consider the numerator"#

#color(red)(3x^2)(x+5)color(magenta)(-15x^2)-12x^2-11x-20#

#=color(red)(3x^2)(x+5)color(red)(-27x)(x+5)color(magenta)(+135x)-11x-20#

#=color(red)(3x^2)(x+5)color(red)(-27x)(x+5)color(red)(+124)(x+5)color(magenta)(-620)-20#

#=color(red)(3x^2)(x+5)color(red)(-27x)(x+5)color(red)(+124)(x+5)-640#

#"quotient "=color(red)(3x^2-27x+124)," remainder "=-640#

#rArr(3x^3-12x^2-11x-20)/(x+5)#

#=3x^2-27x+124-640/(x+5)#