Question

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

Answer 1

`(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)`

Answer 2

`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)`