We take out the common factor 1/5 from the entire expression. The remaining polynomial inside is 9s square + 30sv + 25v square, which condenses smoothly into (3s + 5v) square.
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Following the hint in the book, we factor out 1/3 as a common factor. This leaves the expression 9a square + 12ab + 4b square inside brackets, which factors into (3a + 2b) square.
We match this with the identity a square + 2ab + b square. Here, 64p square is (8p) square and 4/9 q square is (2/3 q) square, which simplifies cleanly to the factor (8p + 2/3 q) square.
We apply the standard square identity to this expression. By recognizing that 49x square is (7x) square and 4y square is (2y) square, we can easily write the final factors as (7x + 2y) square.
This polynomial fits the standard identity a square + 2ab + b square. We rewrite 4s square as (2s) square and 25t square as (5t) square, giving the factored result of (2s + 5t) square.