Question

初二数学：两道因式分解题！

把下列各式因式分解：
（1） （a^2+a+1）(a^2-6a+1)+12a^2

(2) (x+y)^4+(x^2-y^2)^2+(x-y)^4

重点要详细的过程！！！！！！！！！！

Answer 1

1、(a^2+a+1)(a^2-6a+1)+12a^2
=[(a^2+1)+a][(a^2+1)-6a]+12a^2
=(a^2+1)^2-5a*(a^2+1)+6a^2
=(a^2-3a+1)(a^2-2a+1)
=[a-(3+根号5)/2][a-(3-根号5)/2](a-1)^2；
2、(x+y)^4+(x^2-y^2)^2+(x-y)^4
=[(x+y)^4+2(x^2-y^2)^2+(x-y)^4]-(x^2-y^2)^2
=[(x+y)^2+(x-y)^2]^2-(x^2-y^2)^2
=[(x+y)^2+(x-y)^2+(x^2-y^2)][(x+y)^2+(x-y)^2-(x^2-y^2)]
=(3x^2+y^2)(x^2+3y^2)

Answer 2

1：原式=a^4-5a^3+8a^2-5a+1
令原式=0 则a=1容易看出来为一解。
原式=(a-1)*t=a^4-5a^3+8a^2-5a+1
t=a^3-4a^2+4a-1=(a-1)(a^2-3a+1)
即（a^2+a+1）(a^2-6a+1)+12a^2=(a-1)^2*(a^2-3a+1)

2：(x+y)^4+(x^2-y^2)^2+(x-y)^4
=((x+y)^2)^3-((x-y)^2)^3)/((x+y)^2-(x-y)^2)
=((x+y)^3+(x-y)^3)*((x+y)^3-(x-y)^3)/(x+y+x-y)(x+y-(x-y))
=((x+y)^2+(x-y)^2+(x+y)(x-y))((x+y)^2+(x-y)^2-(x+y)(x-y))
=(3x^2+y^2)(x^2+3y^2)

Answer 3

1 、(a2+a+1)(a2-6a+1)+12a2
=[(a2+1)+a][(a2+1)-6a]+12a2
=(a2+1)2-5(a2+1)+6a2
=[(a2+1)-2a][(a2+1)-3a]
=(a2-2a+1)(a2-3a+1)
=(a-1)2(a2-3a+1)
2、(x+y)4+(x2-y2)2+(x-y)4
=(x+y)4+(x+y)2(x-y)2+(x-y)4
=(x+y)4+2(x+y)2(x-y)2+(x-y)4-(x+y)2(x-y)2
=[(x+y)2+(x-y)2]2-(x+y)2(x-y)2
=[(x+y)2+(x-y)2+(x+y)(x-y)][(x+y)2+(x-y)2-(x+y)(x-y)]
=(3x2+y2)(x2+3y2)