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 Subject: algebra, factoring Category: Miscellaneous Asked by: dj2965-ga List Price: \$2.50 Posted: 16 Nov 2003 12:09 PST Expires: 16 Dec 2003 12:09 PST Question ID: 276470
 ```How do you factor this equation x^2 + 16x + 64 - 100y^2```
 Subject: Re: algebra, factoring Answered By: answerguru-ga on 16 Nov 2003 13:27 PST
 ```Hi dj2965-ga, Although the comment by nautico-ga was close, there is no benefit in placing 100y^2 in binomial factored form. The correct factorization of the expression you provided is actually: (x+8)^2 + (10y)^2 OR (x+8)*(x+8) + (10y)^2 Hope that helps :) answerguru-ga``` Clarification of Answer by answerguru-ga on 18 Nov 2003 19:10 PST ```My mistake - that should be: (x+8)*(x+8) - (10y)^2 Thanks for the heads up p2006-ga and thegimp55-ga```
 ```It's been more than 50 years since I took algebra, but thought I would give it a shot: (x+8)(x+8) - (10y+0)(10y+0)```
 ```I believe that answerguru's answer is wrong, but only in a small way. It should be - (minus) (10y)^2, not + (10y)^2 because you had the 100y^2 term subtracted from the rest of the polynomial, ( - 100y^2). I believe that answerguru might have meant (x+8)^2 = (10y)^2 It could have been a typo since + and = are on the same key. Make sense?```
 ```x^2 + 16x + 64 - 100y^2 = (x+8)^2 - (10y)^2 / you reconize now a^2-b^2 (a=x+8 and b=10y) =(x+8+10y).(x+8-10y) /which is equal to (a+b)(a-b) the end. good luck with your studies!```