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.
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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.
The expression follows the pattern of a square + 2ab + b square. Since 9x square is (3x) square and 16y square is (4y) square, we can compress it directly into (3x + 4y) square.
We express 205 as 200 + 5 and use our identity where a = 200 and b = 5. Calculating the individual parts and adding them up gives the final numerical answer.
We transform 105 into 100 + 5 and apply the identity with a = 100 and b = 5. Adding the squared values to the middle cross-product gives the correct total.