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关于因式分解的常用方法,中学课本中已作了介绍。本文要探讨的是根据题目的特征,运用比较特殊的方法,进行因式分解的问题。例1 在复域内分解: (x+1)(x+2)(x+3)(x+6)-3x~2 解原式=(x~2+7x+6)(x~2+5x+6)-3x~2推敲上式的特征,可知若令y=x~2+6x+6,原式就化为: (y+x)(y-x)-3x~2 =y~2-4x~2=(y+2x)(y-2x) =(x~2+8x+6)(x~+4x+6) =(x+4-10~(1/2))(x+4+10~(1/2)) (x+2-(2~(1/2))i)(x+2-(2~(1/2))i) 例2分解:(ab+1)(a+1)(b+1)+ab 解原式即(ab+1)[ab+1+a+b]+ab,若令(ab+1)=A,可得: 原式=A(A+a+b)+ab =A~2+(a+b)A+ab=(A+a)(A+b)
The commonly used method of factorization has been introduced in the middle school textbooks. This article discusses the problem of factorization based on the characteristics of the topic and the use of more specific methods. Example 1 Decomposition in complex domain: (x+1)(x+2)(x+3)(x+6)-3x~2 Solution primitive = (x~2+7x+6)(x~2+5x +6)-3x~2 Inferring the features of the above formula, we know that if we let y=x~2+6x+6, the original formula is: (y+x)(yx)-3x~2=y~2-4x ~2=(y+2x)(y-2x) =(x~2+8x+6)(x~+4x+6) =(x+4-10~(1/2))(x+4+ 10~(1/2)) (x+2-(2~(1/2))i)(x+2-(2~(1/2))i) Example 2 Decomposition: (ab+1)( a+1)(b+1)+ab The solution primitive is (ab+1)[ab+1+a+b]+ab. If (ab+1)=A, then it can be: Original = A ( A+a+b)+ab =A~2+(a+b)A+ab=(A+a)(A+b)