因式分解：1.2（x²+6x+1）²+5（x²+1）（x²+6x+1）+2（x²+1）²2.（x+1）（x+3）（x+5）（x+7）+15看答案看了半天也没搞懂,唉,我初高中脱节了

1. 2（x²+6x+1）²+5（x²+1）（x²+6x+1）+2（x²+1）²
=[（x²+6x+1）+2（x²+1）]*[2（x²+6x+1）+（x²+1）]
=(3x²+6x+3)*(3x²+12x+3)
=3(x²+2x+1)*3(x²+4x+1)
=3(x+1)²3*(x²+4x+1)
=9(x+1)²(x²+4x+1)
2. （x+1）（x+3）（x+5）（x+7）+15
=[(x+1)(x+7)]*[(x+3)(x+5)]+15
=[(x²+8x)+7]*[(x²+8x)+15]+15
=(x²+8x)²+22(x²+8x)+120
=(x²+8x+10)*(x²+8x+12)
=(x²+8x+10)*(x+2)*(x+6)
其实，这两道题就是障眼法，你只要把第一题的（x²+6x+1）看成是a，（x²+1）看成是b，那题目就变成了2a²+5ab+2b²，十字相乘一下子就出来了。第二题也是类似的想法，就不赘述了。

太简单了，先合并脱括号再分解就行了

第一题（2（x²+6x+1）+（x²+1））*（（x²+6x+1）+2（x²+1））
第二题 (（x+1）(x+7))(（x+3）（x+5）)+15
=(x²+8x+7)(x²+8x+15)+15
=(x²+8x+7)(x²+8x+15)+8(x²+8x+15)-8(x²+8x+15)+15
=(x²+8x+15)²-8(x²+8x+15)+15
=(x²+8x+15-3)(x²+8x+15-5)

1
2（x²+6x+1）²+5（x²+1）（x²+6x+1）+2（x²+1）²
=(2（x²+6x+1）+（x²+1）)(（x²+6x+1）+2（x²+1）)
2
(x+1)(x+3)(x+5)(x+7)+15
=[(x+1)(x+7)][(x+3)(x+5)]+15
=(x^2+8x+7)(x^2+8x+15)+15
=(x^2+8x+7)[(x^2+8x+7)+8]+15
=(x^2+8x+7)^2+8(x^2+8x+7)+15
=[(x^2+8x+7)+3][(x^2+8x+7)+5]
=(x^2+8x+10)(x^2+8x+12)
=(x^2+8x+10)(x+2)(x+6)