How do you divide #(3x ^ { 3} - 35x ^ { 2} + 128x - 140) \div ( x - 5)#?

1 Answer
Jim G.
Jul 1, 2017

#3x^2-20x+28#

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)-35x^2+128x-140#

#=color(red)(3x^2)(x-5)color(red)(-20x)(x-5)color(magenta)(-100x)+128x-140#

#=color(red)(3x^2)(x-5)color(red)(-20x)(x-5)color(red)(+28)(x-5)color(magenta)(+140)-140#

#=color(red)(3x^2)(x-5)color(red)(-20x)(x-5)color(red)(+28)(x-5)+0#

#"quotient "=color(red)(3x^2-20x+28)," remainder " =0#

#rArr(3x^3-35x^2+128x-140)/(x-5)#

#=(cancel((x-5))(3x^2-20x+28))/cancel((x-5))#

#=3x^2-20x+28#

Related questions
Impact of this question
824 views around the world
You can reuse this answer
Creative Commons License