We have
[tex]\dfrac{x^4+2x^3-7x^2-8x+12}{x-2}=x^3+4x^2+x-6[/tex]
The rational root theorem suggests that other possible roots may be -6, 6, -3, 3, -2, 2, -1, and 1. It turns out that [tex]x=-2[/tex] is a root, since [tex](-2)^3+4(-2)^2+(-2)-6=0[/tex], so [tex]x+2[/tex] is also a factor and we have
[tex]\dfrac{x^4+2x^3-7x^2-8x+12}{(x-2)(x+2)}=x^2+2x-3[/tex]
Finally, we can factorize the remaining quotient easily:
[tex]x^2+2x-3=(x+3)(x-1)[/tex]
so the other factors are [tex]x+2[/tex], [tex]x+3[/tex], and [tex]x-1[/tex].
