Question

初中数学因式分解

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Answer 1

（a+c)(2ab-2bc)+（c-a)(ab+b^2+ac+bc)
=2b(a+c)(a-c)+（c-a)(ab+b^2+ac+bc)
=2b(a+c)(a-c)-(a-c)(ab+b^2+ac+bc)
=(a-c)(2ab+2bc-ab-b^2-ac-bc)
=(a-c)(ab+bc-ac-b^2)
=(a-c)[(ab-ac)+(bc-b^2)]
=(a-c)[a(b-c)+b(c-b)]
=(a-c)[(b-c)(a-b)]
=(a-c)(b-c)(a-b)

Answer 2

(a+c)(2ab-2bc)+(c-a)(ab+b2+ac+bc)
=(a+c)(2b(a-c))-(a-c)(ab+b2+ac+bc)
=(a-c)(2ab+2bc-ab-b2-ac-bc)
=(a-c)(ab+bc-ac-b2)
=(a-c)(a(b-c)-b(b-c))
=(a-c)(b-c)(a-b)
=-(a-b)(b-c)(c-a)

Answer 3

(a+c)(2ab-2bc)+(c-a)(ab+b^2+ac+bc)........................................................1
=2(a+c)(ab-bc)+(c-a)(b+a)(b+c)
=2b(a+c)(a-c)+(c-a)(b+a)(b+c)
=2b(a+c)(a-c)-(a-c)(b+a)(b+c)
=(a-c)( 2b(a+c)-(b+a)(b+c) )
=(a-c)( 2bc+2ba-ab-b^2-ac-bc)...................................................................2
=(a-c)(ab+bc-ac-b^2)
=(a-c)[(ab-ac)+(bc-b^2)]
=(a-c)[a(b-c)+b(c-b)]
=(a-c)(b-c)(a-b)