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• ## A closer look at generalization of difference of squares factoring.

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• I discovered this simpler proof of the generalization that also gives added insight into its limitations. z = t^s - s^2 z + s^2 = t^2 z + s^2 - (t - A)^2 =
Message 1 of 1 , Jan 11, 2008
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I discovered this simpler proof of the generalization that also gives
added insight
into its limitations.

z = t^s - s^2

z + s^2 = t^2

z + s^2 - (t - A)^2 = t^2 - (t - A)^2

z + (s + A - t) * (s + t - A) = A * ( 2 * t - A )

Z + ( A - (t - s) ) * (s + t - A) = A * (2 * t - A)

Z - ( (t - s) - A) * (( t + s) - A) = A * (2 * t - A)

Z - ( x - A) * ( y - A) = A * ( x + y - A)

where x y = z

which is the simple algebraic identity

xy - (x - A) * (y - A) = A * ( x + y - A)

For every value of A, there is a related factoring, but the identity
gives no clue how
to find any of them.

At the moment, it looks as if finding any of them will be exactly as
difficult as finding the
t or s.

Kermit Rose < kermit@... >
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