How to solve this?

Given:

`(ax^3+bx^2-3x+2b)/(x^2-x-2)`

Note that:

`x^2-x-2 = (x-2)(x+1)`

So for the quotient to be a polynomial, the numerator must be divisible by both `(x-2)` and `(x+1)`.

Hence it must be zero when `x=2` and when `x=-1`

So we find:

`0 = a(color(blue)(2))^3+b(color(blue)(2))^2-3(color(blue)(2))+2b`

`color(white)(0) = 8a+4b-6+2b`

`color(white)(0) = 8a+6b-6`

`color(white)(0) = 2(4a+3b-3)`

and:

`0 = a(color(blue)(-1))^3+b(color(blue)(-1))^2-3(color(blue)(-1))+2b`

`color(white)(0) = -a+b+3+2b`

`color(white)(0) = -a+3b+3`

So:

`{ (4a+3b-3=0), (-a+3b+3=0) :}`

Subtracting the second equation from the first (to eliminate the term in `3b`), we find:

`5a-6=0`

and hence:

`a = 6/5`

Putting this value for `a` into the second equation, we get:

`0 = -6/5+3b+3 = 3b+9/5 = 3(b+3/5)`

and hence:

`b = -3/5`

Substituting these values for `a` and `b` in the numerator, we have:

`6/5x^3-3/5x^2-3x+6/5 = 1/5(6x^3-3x^2-15x-6)`

`color(white)(6/5x^3-3/5x^2-3x+6/5) = 3/5(2x^3-x^2-5x-2)`

`color(white)(6/5x^3-3/5x^2-3x+6/5) = 3/5(x-2)(2x^2+3x+1)`

`color(white)(6/5x^3-3/5x^2-3x+6/5) = 3/5(x-2)(x+1)(2x+1)`

So:

`(6/5x^3-3/5x^2-3x+6/5)/(x^2-x-2) = (6/5x^3-3/5x^2-3x+6/5)/((x-2)(x+1)) = 3/5(2x+1)`