This fits the identity a square + 2ab + b square. The first term is the square of 3/2 s and the last is the square of 2t, which yields the factor (3/2 s + 2t) square.
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The polynomial matches the identity template a square – 2ab + b square. Since 16y square is (4y) square and 9 is (3) square, it condenses directly into (4y – 3) square.
We treat 1120 as 1100 + 20 and apply the standard two-term addition square identity. Squaring both parts and adding their double product provides the final answer.
We separate 1104 into 1000 + 100 + 4 and expand it using the three-term identity. Adding the squares and all two-term product pairings gives the total value.
We rewrite 214 as 200 + 10 + 4 and apply the identity for the square of a three-term sum. Summing the three individual squares and their cross-products gives 45796.