Question

怎样分解因式（1）（a+1)(a+2)(a+4)(a-1)-27

(2)(a+b﹚²+﹙a+c﹚²-(c+d)²-(b+d)² 求解 并说明具体方法 谢谢

Answer 1

解：

（1）

(a+1)(a+2)(a+4)(a-1)-27
=[(a+1)(a+2)][(a+4)(a-1)]-27
=(a²+3a+2)(a²+3a-4)-27  

=[(a²+3a)+2][(a²+3a)-4]-27   a²+3a看着一个整体

=(a²+3a)²-2(a²+3a)-8-27
=(a²+3a)²-2(a²+3a)-35   
=[(a²+3a)+5][(a²+3a)-7]   十字相乘法
=(a²+3a+5)(a²+3a-7)

(2)

(a+b﹚²+﹙a+c﹚²-(c+d)²-(b+d)²

=(a+b﹚²-(c+d)²+﹙a+c﹚²-(b+d)²

=(a+b+c+d)(a+b-c-d)+(a+c+b+d)(a+c-b-d)   平方差公式

=(a+b+c+d)(a+b-c-d+a+c-b-d)   提取公因式
=2(a+b+c+d)(a-d)

Answer 2

（1）（a+1)(a+2)(a+4)(a-1)-27
=（a²+3a+2)(a²+3a-4)-27
=[(a²+3a)+2][(a²+3a)-4]-27
=(a²+3a)²-2(a²+3a)-8-27
=(a²+3a)²-2(a²+3a)-35
=[(a²+3a)-7][(a²+3a)+5]
=(a²+3a-7)(a²+3a+5)

(2)(a+b﹚²+﹙a+c﹚²-(c+d)²-(b+d)²

=(a+b)²-(b+d)²+(a+c)²-(c+d)²
=[(a+b)+(b+d)][(a+b)-(b+d)]+[(a+c)+(c+d)][(a+c)-(c+d)]
=(a+2b+d)(a-d)+(a+2c+d)(a-d)
=(a-d)[(a+2b+d)+(a+2c+d)]
=(a-d)(2a+2b+2c+2d)
=2(a-d)(a+b+c+d)

Answer 3

(a+1)(a+2)(a+4)(a-1)-27
=[(a+1)(a+2)][(a+4)(a-1)]-27
=(a^2+3a+2)(a^2+3a-4)-27
=[(a^2+3a)+2][(a^2+3a)-4]-27
=(a^2+3a)^2-2(a^2+3a)-8-27
=(a^2+3a)^2-2(a^2+3a)-35
=[(a^2+3a)+5][(a^2+3a)-7]
=(a^2+3a+5)(a^2+3a-7)

Answer 4

（1）(a+1)(a+2)(a+4)(a-1)-27=(a^2+3a+2)(a^2+3a-4)-27
=(a^2+3a)^2-2(a^2+3a)-8-27
=(a^2+3a)^2-2(a^2+3a)-35
=(a^2+3a+3)(a^2+3a-5)
（2）(a+b)^2+(a+c)^2-(c+d)^2-(b+d)^2
=(a+b)^2-(c+d)^2+(a+c)^2-(b+d)^2
=(a+b+c+d)(a+b-c-d)+(a+c+b+d)(a+c-b-d)
=(a+b+c+d)(a+b-c-d+a+c-b-d)
=2(a+b+c+d)(a-d)

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