This six-term polynomial matches the expanded three-term identity. The squared bases are m/3, k/2, and 3n, which merge into the single grouped factor (m/3 + k/2 + 3n) square.
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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.
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 express 198 as 200 – 2 and use the subtraction square identity with a = 200 and b = 2. Working out the arithmetic components gives the final plain text answer.
We rewrite 79 as 80 – 1 to use the subtraction square identity with a = 80 and b = 1. Combining their squared values and subtracting the double product gives 6241.