Is there any way to calculate:
1. (a+2b+3c)(a+2b-3c)
THX!
Mathematics (Factorization)
2009-07-31 3:45 am
回答 (3)
2009-07-31 5:00 am
✔ 最佳答案
1. (a+2b+3c)(a+2b-3c)=[(a+2b)+3c][(a+2b)-3c]
=(a+2b)^2-(3c)^2 [ a^2-b^2 = (a+b)(a-b) ]
=a^2+4ab+4b^2-9c^2//
2009-07-31 6:26 am
By Using Factorization :
[(a+2b)+3c][(a+2b)-3c]
=[(a+2b)(a+2b)][(3c)(-3c)]
=(a+2b)^2-9c^2
=a^2+2(a)(2b)+(2b)^2-9c^2
=a^2+4ab+(4b^2)-9c^2
[(a+2b)+3c][(a+2b)-3c]
=[(a+2b)(a+2b)][(3c)(-3c)]
=(a+2b)^2-9c^2
=a^2+2(a)(2b)+(2b)^2-9c^2
=a^2+4ab+(4b^2)-9c^2
2009-07-31 3:54 am
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