Eight problems of factoring in Mathematics （m+n）²-n² 49（a-b）²-16（a+b）² （2x+y）²-（x+2y）² 1/3m²-3n² （a+b）²-（a-b）² （9a+4b）²-（3a-b）²

Eight problems of factoring in Mathematics （m+n）²-n² 49（a-b）²-16（a+b）² （2x+y）²-（x+2y）² 1/3m²-3n² （a+b）²-（a-b）² （9a+4b）²-（3a-b）²

（m+n）²-n²
=(m+n+n)(m+n-n)
=m(m+2n)
49（a-b）²-16（a+b）²
[7(a-b)+4(a+b)][7(a-b)-4(a+b)]
=(11a-3b)(3a-11b)
（2x+y）²-（x+2y）²
=[(2x+y)+(x+2y)][(2x+y)-(x+2y)]
=(3x+3y)(x-y)
=3(x+y)(x-y)
1/3m²-3n²
=1/3(m²-9n²)
=1/3(m+3n)(m-3n)
（a+b）²-（a-b）²
=[(a+b)+(a-b)][(a+b)-(a-b)]
=2a×2b
=4ab
（9a+4b）²-（3a-b）²
=[(9a+4b)+(3a-b)][9a+4b)-(3a-b)]
=(12a+3b)(6a+5b)
=3(4a+b)(6a+5b)

Factorization of Mathematics
(2x-5) ^ 2 = 9 this type of I don't seem to ask my friends who will tell me the detailed process. There is another (2x-1) (3x + 4) = 2x-1

First question
9 is the square of 3. To the left, take 2x-5 as a number, then (2x-5) ^ 2-3 ^ 2 can use the square difference method to decompose the factor (2x-5 + 3) (2x-5-3) = 0
Second question
Invert 2x-1 to the left to get (2x-1) (3x + 4) - (2x-1) = 0
We can extract the common factor and get (2x-1) (3x + 4-1) = 0

Using factorization
（1）202²+198²
（2）1998²-1997²×1999
（3）、（2+1）（2²+1）（2⁴+1）（2^8+1）····（2^128+1）

（1）202²+198²
=(200+2)²+(200-2)²
=40000+800+4+40000-800+4
=80000+8
=80008
（2）1998²-1997²×1999
=1998²-(1998-1)(1998+1)
=1998²-(1998²-1)
=1
（3）、（2+1）（2²+1）（2⁴+1）（2^8+1）····（2^128+1）
=(2-1)(2+1)(2²+1)(2⁴+1)(2^8+1）····（2^128+1）
=(2^2-1)(2²+1)(2⁴+1)(2^8+1）····（2^128+1）
=(2^4-1)(2⁴+1)(2^8+1）····（2^128+1）
=……
=2^256-1