Question

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

Answer 1

`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`