问题
计算题
分解因式
①a2﹣2a(b+c)+(b+c)2;
②(x2﹣5)2+8(x2﹣5)+16
答案
解:
①a2﹣2a(b+c)+(b+c)2
=(a﹣b﹣c)2;
②(x2﹣5)2+8(x2﹣5)+16,
=(x2﹣5+4)2,
=(x2﹣1)2,
=(x+1)2(x﹣1)2.
分解因式
①a2﹣2a(b+c)+(b+c)2;
②(x2﹣5)2+8(x2﹣5)+16
解:
①a2﹣2a(b+c)+(b+c)2
=(a﹣b﹣c)2;
②(x2﹣5)2+8(x2﹣5)+16,
=(x2﹣5+4)2,
=(x2﹣1)2,
=(x+1)2(x﹣1)2.