Question

How do you simplify and divide `(3c^5+5c^4+c+5)/(c+2)`?

Answer 1

`(3c^5+5c^2+c+5)/(c+2)=3c^4-6c^3+12c^2-19c+39-73/(c+2)`

Explanation:

We use long division method to divide `3c^5+5c^2+c+5` by `(c+2)`

`color(white)(xxxxxxx)3c^4-6c^3+12c^2-19c+39`
`color(white)(xx) c+2| bar(3c^5+0c^4+0c^3+5c^2+c+5)`
`color(white)(xxxxxxx)ul(3c^5+6c^4)color(white)(xx)color(red)(darr)` - subtracting
`color(white)(xxxxxxxxx)-6c^4+0c^3`
`color(white)(xxxxxxx**x)ul(-6c^4-12c^3)`
`color(white)(xxxxxxxxxxxxxxx)12c^3+5c^2`
`color(white)(xxxxxxxxxxxxxxx)ul(12c^3+24c^2`
`color(white)(xxxxxxxxxxxxxxxxxx)-19c^2color(white)(xx)+c`
`color(white)(xxxxxxxxxxxxxxxxxxx)ul(-19c^2-38c)`
`color(white)(xxxxxxxxxxxxxxxxxxxxxxxxx)39c+5`
`color(white)(xxxxxxxxxxxxxxxxxxxxxXXX) ul(39c+78)`
`color(white)(xxxxxxxxxxxxxxxxxxxxxxxxxxx)-73`

WE can check it using Remainder theorem, as `3c^5+5c^2+c+5` is divided by `c+2`,

we should get a remainder `3(-2)^5+5(-2)^2-2+5=3xx(-32)+5xx4-2+5=-96+20-2+5=-73`

and hence `(3c^5+5c^2+c+5)/(c+2)=3c^4-6c^3+12c^2-19c+39-73/(c+2)`