How do you factor this equation

x^2 + 16x + 64 - 100y^2

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

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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!