Math, asked by pradhansaditya2356, 6 hours ago

100-[121p^2-88pq+16q^2]
factorise using identites

Answers

Answered by bhumi4653
0

Answer:

100 - ( {21p}^{2}  - 88pq +  {16q}^{2} \\ 100( -  {21p}^{2} + 88pq -  {16q}^{2}   ) \\ (100 \times  -  {21p}^{2}) + (100 \times 88pq) + (100 \times  -  {16q}^{2}  ) \\  - 2100 {p}^{2}  + 8800pq - 1600{q}^{2}

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