两到初一数学题。急需过程！！！！

来源：百度知道 编辑：UC知道 时间：2024/05/24 16:42:38
分解因式
【1】（m^2+3m)^2-8(m^2+3m）-20
【2】（y^2+3y)-(2y+6)^2
过程一定要详细！！

【1】（m^2+3m)^2-8(m^2+3m）-20
令a=m^2+3m
则原式=a^2-8a-20
=(a-10)(a+2)
=(m^2+3m-10)(m^2+3m+2)
=(m+5)(m-2)(m+1)(m+2)

【2】（y^2+3y)-(2y+6)^2
=y(y+3)-[2(y+3)]^2
=y(y+3)-4(y+3)^2
=(y+3)[y-4(y+3)]
=(y+3)(-3y-12)
=-3(y+3)(y+4)

（m^2+3m)^2-8(m^2+3m）-20
=(m^2+3m-10)(m^2+3m+2)
=(m+5)(m-2)(m+1)(m+2)

（y^2+3y)^2-(2y+6)^2
=（y^2+3y)^2-4(y+3)^2
=(y^2+3y-2y-6)(y^2+3y+2y+6)
=(y^2+y-6)(y^2+5y+6)
=(y+3)(y-2)(y+3)(y+2)
=(y+3)^2(y-2)(y+2)

（y^2+3y)-(2y+6)^2
=y^2+3y-4y^2-24Y-36
=-3y^2-21y-36
=-3(y^2+7y+12)
=-3(y+3)(y+4)

(m^2+3m)^2-8(m^2+3m）-20
=(m^2+3m-10)(m^2+3m+2)
=(m+5)(m-2)(m+1)(m+2)

2)
（y^2+3y)-(2y+6)^2
=y^2+3y-4y^2-24Y-36
=-3y^2-21y-36
=-3(y^2+7y+12)
=-3(y+3)(y+4)

1=(m^2+3m-10)(m^2+3m+2)
=(m+5)(m-2)(m+1)(m+2)
2=y(y+3)-4(y+3)^2
=(y+3)(y-4y-12)
=-3(y+3)(y+4)

1=(m^2+3m-10)(m^2+3m+2)=(m+5