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.
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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.
We split the middle coefficient minus 1 into minus 7 and positive 6 because their sum equals minus 1 and their product equals minus 42, giving the linear binomial factors.
The polynomial matches the identity template a square – 2ab + b square. Since 16s square is (4s) square and 25t square is (5t) square, it condenses directly into (4s – 5t) square.