This expression matches the identity template a square – 2ab + b square. The middle variable terms cancel out completely when multiplying p/4 and 4/p, leaving only the constant integer minus 2.
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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.
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 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 78 as 80 – 2 to use the subtraction identity with a = 80 and b = 2. Combining their squared values and subtracting the double product gives the final result.