3(a平方+4a+4)-a(a+3)(3a+4)
问题描述:
3(a平方+4a+4)-a(a+3)(3a+4)
答
-3a的立方-10a的平方+12
答
=3a²+12a+12-(3a³+13a²+12)
= -3a³-10a²+12a
= -a(3a²+10a-12)
答
3(a²+4a+4)-a(a+3)(3a+4)
=3a²+12a+12-a(3a²+13a+12)
=3a²+12a+12-3a³-13a²-12a
=-3a³-10a²+12