Math, asked by goviguru, 1 year ago

factorize
8(p-2q)^2-2pq+4q-1

Answers

Answered by mehak2019
2

=8(p-2q)^2-2pq+4q-1

=8(p^2+4q^2-4pq) -2pq+4q-1

=8p^2+32p^2-32pq-2pq+4q-1

=8p^2+32p^2-34pq+4q-1

Answered by rajeev8567
1

Step-by-step explanation:

8(p-2q)^2-2pq+4q-1

= 8{p^2+(2q)^2-2×p×2q} -2pq+4q-1

= 8(p^2+4q^2-4pq) - 2pq+4q-1

= 8p^2+32q^2-32pq-2pq+4q-1

= 8p^2+32q^2-3pq+4q-1

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