1+15c-34c squared, two common factors of 34c squared ?

You don't have "common factors" of a single term like 34c². There's nothing else for 34c² to have factors in common with.

The factors are 2, 17, and c (twice).

To factor that trinomial, it's clear that the form wll be (ac + 1)(bc + 1), where a and b are constants to be determined. Foil it out and:

abc² + (a+b)c + 1 = -34c² + 15c + 1

...implying that ab = -34 and a+b = 15. What two numbers multiply to -34 and add to 15? I get (17)(-2) = -34 and 17+(-2) = 15. So:

(15c + 1)(-2c + 1) = -34c² + 15c + 1

(1+17c)(1-2c)