题目内容
把下列各式因式分
(1)(a+3)(a-7)+25;
(2)(2a+b)2-(4a+3b)2.
(1)(a+3)(a-7)+25;
(2)(2a+b)2-(4a+3b)2.
(1)(a+3)(a-7)+25
=a2-4a-21+25
=a2-4a+4
=(a-2)2;
(2)(2a+b)2-(4a+3b)2
=[(2a+b)+(4a+3b)[(2a+b)-(4a+3b)]
=(2a+b+4a+3b)(2a+b-4a-3b)
=(6a+4b)(-2a-2b)
=-4(3a+2b)(a+b).
=a2-4a-21+25
=a2-4a+4
=(a-2)2;
(2)(2a+b)2-(4a+3b)2
=[(2a+b)+(4a+3b)[(2a+b)-(4a+3b)]
=(2a+b+4a+3b)(2a+b-4a-3b)
=(6a+4b)(-2a-2b)
=-4(3a+2b)(a+b).
练习册系列答案
相关题目