Decompose $5^{1985}-1$ into factors 
Decompose the number $5^{1985}-1$ into a product of three integers, each of which is larger than $5^{100}$.

We first notice the factorization $x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$. Now to factorize $x^4+x^3+x^2+x+1$ we get $$(x^2+ax+1)(x^2+bx+1) = x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1 = x^4+x^3+x^2+x+1$$ implies $a+b = 1,ab+2 = 1$. Thus, $$x^4+x^3+x^2+x+1 = (x^2+\left(\frac{1+\sqrt{5}}{2}\right)x+1)(x^2+\left(\frac{1-\sqrt{5}}{2}\right)x+1).$$ Is it possible to continue from this approach because now the factors I have aren't integers or is there a better way?

I would try Aurifeuillian factorization here.
Let $f(x)=x^4+x^3+x^2+x+1$ be the fifth cyclotomic polynomial. The Aurifeuillian stuff says that $f(5x^2)$ factors into a product as follows
$$
f(5x^2)=\left(25 x^4-25 x^3+15 x^2-5 x+1\right) \left(25 x^4+25 x^3+15 x^2+5
   x+1\right).
$$
Call those two factors on the RHS $g_1(x)$ and $g_2(x)$.
Your number is
$$
\begin{aligned}
5^{1985}-1&=(5^{397}-1)f(5^{397})\\
&=(5^{397}-1)g_1(5^{198})g_2(5^{198}),
\end{aligned}
$$
and this fits the bill.
