Question

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

Answer 1

Answer:

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