Question

How do you divide `(2x^3+ 3 x^2-4x-2)/(x+2)`?

Answer 1

`=(2x^2-x-2)+2/(x+2)`

Explanation:

We know that,
Dividend`-:` Divisor = quotient and remainder.In short,
Dividend = Divisor `xx` quotient + remainder.
Now,divisor is (x+2),and
dividend`=2x^3+3x^2-4x-2` `=2x^3+4x^2-x^2-2x-2x-4+2`
`=2x^2(x+2)-x(x+2)-2(x+2)+color(red)(2)`
`=(x+2)(2x^2-x-2)+2`
i.e.quotient=`(2x^2-x-2)`,and remainder=2.
So,
`2x^3+3x^2-4x-2=(x+2)(2x^2-x-2)+2`
Hence,
`(2x^3+3x^2-4x-2)/(x+2)=[(x+2)(2x^2-x-2)+2]/(x+2)`
`=((x+2)(2x^2-x-2))/(x+2)+2/(x+2)`
`=(2x^2-x-2)+color(red)(2/(x+2)`