a^2+4a+4-b^2


ac+ad+bc+bd

3a+ab+3c+bc

ac+ad+bc+bd

Because this is the simplest variable one:
Look at any common variables
There are 2 "a" in the first two terms
a(c+d) = ac + ad by the distributive property right?
b(c+d) = because + bd by the same property
So if you have a(c+d) + b(c+d)
Therefore (a+b)(c+d) because both a and b distribute through c and d.

Will you factor these?
1. 10ty-35y-14tx+49x
2. 9s^2+12s+4-t^2
3. 3x^7-9x^5+4x^4-12x^2
4. 3x^2+15x+2x+10
5. 27u^3+125

1. 10ty-35y-14tx+49x
Common factor between the first two terms is 5
Common factor between the second two is 7
Solve in the same format.

2. 9s^2+12s+4-t^2
Reorganize to:
9s^2 - t^2 + 12s + 4
(3s+t)(3s-t) + 4 (s+1)

Well, I have to go so I'll leave it to someone else to do the rest, or maybe I'll do it later.