How do you divide #(2x^3-5x^2+4x+12)/(x-7) #?

2 Answers
Jun 16, 2017

#2x^2+9x+67+481/(x-7)#

Explanation:

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

#"consider the numerator"#

#color(red)(2x^2)(x-7)color(magenta)(+14x^2)-5x^2+4x+12#

#=color(red)(2x^2)(x-7)color(red)(+9x)(x-7)color(magenta)(+63x)+4x+12#

#=color(red)(2x^2)(x-7)color(red)(+9x)(x-7)color(red)(+67)(x-7)color(magenta)(+469)+12#

#=color(red)(2x^2)(x-7)color(red)(+9x)(x-7)color(red)(+67)(x-7)+481#

#"quotient "=color(red)(2x^2+9x+67)", remainder "=481#

#rArr(2x^3-5x^2+4x+12)/(x-7)#

#=2x^2+9x+67+481/(x-7)#

Jun 16, 2017

#color(blue)(2x^2+9x+67# plus remainder of #color(blue)481#

Explanation:

# color(white)(.............)ul(color(blue)(2x^2+9x+67)#
#color(white)(aa)x-7##|##2x^3-5x^2+4x+12#
#color(white)(..............)ul(2x^3-14x^3)#
#color(white)(........................)9x^2+4x#
#color(white)(........................)ul(9x^2-63x)#
#color(white)(..............................)67x+12#
#color(white)(..............................)ul(67x-469)#
#color(white)(........................................)481#