Factoring

9 = (r-2)(2r+sqrt5)(2r-sqrt5)

10 = (7k+6)(3k^2+8)

11 = (6x+7)(sqrt7x+sqrt6)(sqrt7x-sqrt6)

12 = (p+6)(p+sqrt3)((p-sqrt3)

Given a rough inspection I think they are all the same sort of factorization.

So if I do the first one and you understand then you are capable of doing the other three yourself.

9)     

 $4r^3-8r^2-5r+10\\ =4r^3-8r^2\qquad\qquad -5r+10\\ \text{now factorize the pairs}\\ =4r^2(r-2)\qquad\qquad -5(r-2)\\$

now you can think of (r-2) as just one thing.

So you have  $4r^2$  lots of (r-2)   and then you take away  $5$ lots of (r-2)

That will give you   $(4r^2-5)$    lots of   (r-2)

which is

$(4r^2-5)(r-2)$

you could take this one step further by treating $4r^2-5$ as the difference of 2 squares

$4r^2 = (2r)^2$     and      $5=(\sqrt5)^2$    so

$(4r^2-5) = (2r-\sqrt5)(2r+\sqrt5)$

$(4r^2-5)(r-2)=(2r-\sqrt5)(2r+\sqrt5)(r-2)$