$$\text{Problem: }{x^3-x^2-x-2 \over x^2 + x - 6}$$ My textbook was able to come up with $(x-2)(x^2+x+1)$ $$\text{Textbook: }{(x-2)(x^2+x+1) \over (x-2)(x+3)}$$
I've tried grouping and using the first and last 3 terms as trinomial but it doesn't work. How do I get that solution? Thank you in advance.
For the cubic polynomial, by checking the value of it at some integer values of $x$ like $x = 2$ then $2^3 - 2^2 - 2 - 2 = 0$. This means $x - 2$ is a factor, and using long division you get the other factor. For the quadratic expression, you also have $2$ is a root and also $-3$ is also a root. Then it can be factored as $(x - 2)(x + 3)$.