How do you divide #(-2x^3-23x^2-4x+11)/(x-5) #?

1 Answer
Jim G.
Nov 12, 2017

#-2x^2-33x-169-834/(x-5)#

Explanation:

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

#"consider the numerator"#

#color(red)(-2x^2)(x-5)color(magenta)(-10x^2)-23x^2-4x+11#

#=color(red)(-2x^2)(x-5)color(red)(-33x)(x-5)color(magenta)(-165x)-4x+11#

#=color(red)(-2x^2)(x-5)color(red)(-33x)(x-5)color(red)(-169)(x-5)color(magenta)(-845)+11#

#=color(red)(-2x^2)(x-5)color(red)(-33x)(x-5)color(red)(-169)(x-5)-834#

#"quotient "=color(red)(-2x^2-33x-169)," remainder "=-834#

#rArr(-2x^3-23x^2-4x+11)/(x-5)#

#=-2x^3-33x-169-834/(x-5)#

Related questions
See all questions in Division of Polynomials
Impact of this question
1312 views around the world
You can reuse this answer
Creative Commons License