Question

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

Answer 1

`x^4 + 2x^3 + 3 x^2 + 7x + 12` with the remainder `19`

Explanation:

I know that there are different notations for the polynomial long division in different countries. I will use the one that I'm familiar with and hope that it will be easy for you to adapt it to your notation if needed. :-)

`color(white)(xii) (x^5 color(white)(xxxxx) - x^3 + x^2 - 2x - 5) -: (x-2) = x^4 + 2x^3 + 3 x^2 + 7x + 12`
`- (x^5 - 2x^4)`
`color(white)(xx) color(white)(xxxxxx)/color(white)(x)`
`color(white)(xxxxxx) 2x^4 - x^3`
`color(white)(xxx )- (2x^4 - 4x^3)`
`color(white)(xxxxx) color(white)(xxxxxxxx)/color(white)(x)`
`color(white)(xxxxxxxxxx) 3 x^3 + x^2`
`color(white)(xxxxxxx )- (3x^3 - 6x^2)`
`color(white)(xxxxxxxxx) color(white)(xxxxxxxx)/color(white)(x)`
`color(white)(xxxxxxxxxxxxxx) 7x^2 - 2x`
`color(white)(xxxxxxxxxxx )- (7x^2 - 14x )`
`color(white)(xxxxxxxxxxxxx) color(white)(xxxxxxxx)/color(white)(x)`
`color(white)(xxxxxxxxxxxxxxxxxx) 12x - 5`
`color(white)(xxxxxxxxxxxxxxxii )- (12x - 24)`
`color(white)(xxxxxxxxxxxxxxxxx) color(white)(xxxxxxxx)/color(white)(x)`
`color(white)(xxxxxxxxxxxxxxxxxxxxxxx) 19`

Thus,

`(x^5 - x^3 + x^2 - 2x - 5) / (x-2) = x^4 + 2x^3 + 3 x^2 + 7x + 12`

with the remainder `19`, or:

`(x^5 - x^3 + x^2 - 2x - 5) / (x-2) = x^4 + 2x^3 + 3 x^2 + 7x + 12 + 19/(x-2)`