We apply the difference of squares identity, x square – y square = (x + y)(x – y). Rewriting 36s square as (6s) square and 49t square as (7t) square directly provides the two unique linear dimensions.
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The area expression fits the identity template of x square – 2xy + y square. Since 25a square is (5a) square and 9b square is (3b) square, it factors into the product of two identical binomial dimensions.
This matches the three-term identity where the term containing variable b must be negative because both cross-product terms that include b have a minus sign in the expression.
This six-term polynomial matches the three-term square identity. Looking at the negative signs, the bases for both v and w must be negative to yield the correct cross-product signs.
This fits the standard identity a square + 2ab + b square. The first term is the square of 7g and the last is the square of h, which yields the grouped factor (7g + h) square.