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

1 Answer
Jim G.
Jun 2, 2017

#3x^2+19x+121+845/(x-7)#

Explanation:

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

#"consider the numerator"#

#color(red)(3x^2)(x-7)color(magenta)(+21x^2)-2x^2-12x-2#

#=color(red)(3x^2)(x-7)color(red)(+19x)(x-7)color(magenta)(+133x)-12x-2#

#=color(red)(3x^2)(x-7)color(red)(+19x)(x-7)color(red)(+121)(x-7)color(magenta)(+847)-2#

#=color(red)(3x^2)(x-7)color(red)(+19x)(x-7)color(red)(+121)(x-7)+845#

#"quotient "=color(red)(3x^2+19x+121),"remainder "=845#

#rArr(3x^3-2x^2-12x-2)/(x-7)=3x^2+19x+121+845/(x-7)#

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