Question

How do you divide `(3x ^ { 3} - 6x ^ { 2} - 8x - 50) \div ( x - 4)`?

Answer 1

Integer Quotient `= (3*x^2 + 6*x + 16)`

Remainder `= 14`

Dividend = (Integer Quotient) x (Divisor) + (Remainder)

`(3*x^3 - 6*x^2 - 8*x - 50) = (3*x^2 + 6*x + 16) * (x - 4 ) + 14`

Explanation:

We have

`(3*x^3 - 6*x^2 - 8*x - 50)/(x - 4)`

`= 3*x^2 + (6*x^2-8*x-50)/(x-4)`

`= 3*x^2 + 6*x+(16*x-50)/(x-4)`

`= 3*x^2 + 6*x+16 + 14/(x-4)`

We can represent the answer as follows:

Dividend = (Integer Quotient) x (Divisor) + (Remainder)

`(3*x^3 - 6*x^2 - 8*x - 50) = (3*x^2 + 6*x + 16) * (x - 4 ) + 14`

Answer 2

`3x^2+6x+16+14/(x-4)`

Explanation:

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

`"consider the numerator"`

`color(red)(3x^2)(x-4)color(magenta)(+12x^2)-6x^2-8x-50`

`=color(red)(3x^2)(x-4)color(red)(+6x)(x-4)color(magenta)(+24x)-8x-50`

`=color(red)(3x^2)(x-4)color(red)(+6x)(x-4)color(red)(+16)(x-4)color(magenta)(+64)-50`

`=color(red)(3x^2)(x-4)color(red)(+6x)(x-4)color(red)(+16)(x-4)+14`

`"quotient "=color(red)(3x^2+6x+16)," remainder "=14`

`rArr(3x^3-6x^2-8x-50)/(x-4)=3x^2+6x+16+14/(x-4)`