Question 156565
None of the coefficients have anything in common besides 1. So, we are going to need a 5 coefficient for a,c in (ax+b)(cx+d) factorization.
So we need (5x+b)(5x+d)=25x^2+20x+4. We can suspect b=d=2 because if b=4, our x coefficient becomes >20. So (5x+2)^2 is our factorization.

Now, we could solve this slightly differently, yet more concisely:


(5x+b)(5x+d)=25x^2+20x+4
25x^2+5xd+5xb+bd=25x^2+20x+4


This means that:
5d+5b=20
bd=4

from the first equation: d=4-b
putting that into the second equation: b(4-b)=4
4b-b^2-4=0
b^2-4b+4=0

We must necessarily have solution b=2 and d=4-2=2 as well. So we place those values in our hybrid form, getting the answer already supplied.