I just don't quite fully understand what I'm being asked to do here. here's [one of] the examples I was given in the book.
6(x+1)^2-5(x+1)
=[6(x+1)](x+1)-5(x+1)
=[6(x+1)-5](x+1)
=(6x+6-5)(x+1)
=(6x+1)(x+1)
Let's try Denis' suggestion, and make a substitution for the expression x+1, in the example above.
Let symbol u = x+1
Then, the original expression becomes
6u^2 - 5u
Factoring this gives
u(6u - 5)
That's easy to see, yes?
Now, reverse the earlier substitution by replacing symbol u with x+1, and we get the same result as the book.
(x+1)(6[x+1] - 5)
Apply the distributive property to 6[x+1]
(x+1)(6x + 6 - 5)
Simplify 6-5
(x+1)(6x + 1)
You can always substitute a single symbol for a lengthier expression; sometimes, doing this helps to recognize repeated factors more easily and to complete the algebraic steps. Reverse the substitution after factoring, to simplify.
With enough practice, you will be able to recognize these sorts of repeated factors (eg: binomials) and complete the factorization without making substitutions, but, until then, try substituting and see whether it helps!