Question

How do you find the quotient of `(2y^2-3y+1)div(y-2)` using long division?

Answer 1

The quotient is `=(2y+1)` and the remainder is `=3`

Explanation:

Let's perform the long division

`color(white)(aaaa)` `y-2` `|` `color(white)(aaaa)` `2y^2-3y+1` `|` `color(white)(aa)` `2y+1`

`color(white)(aaaaaaaaaaaaaa)` `2y^2-4y`

`color(white)(aaaaaaaaaaaaaaaa)` `0+y+1`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `y-2`

`color(white)(aaaaaaaaaaaaaaaaaaa)` `0+3`

So,

`(2y^2-3y+1)/(y-2)=(2y+1)+3/(y-2)`

The quotient is `=(2y+1)` and the remainder is `=3`

Answer 2

Quotient `= (2y+1)`

Explanation:

I know the question asked for long division, but there is an easier way to do it.

We want `(2y^2-3y+1)/(y-2)`

First, let `f(y)=2y^2-3y+1`

Then take `f(2)=2(2^2)-3(2)+1=3`

So `(2y^2-3y+1)/(y-2)` has a remainder of `3`.

Therefore `2y^2-3y+1=(ay+b)(y-2)+3`

`2y^2-3y-2=(ay+b)(y-2)`

`y^2` has a coefficient of `2`, so `ay^2=2y^2 rArr a=2`

The constant is `-2`, so `-2b=-2rArrb=1`

Thus the quotient, `(ay+b)`, equals `(2y+1)`