9a²(b-2c)+6(b+2c)² factorise
Answers
Answered by
0
Answer:
Hi there!
9a² - b² + 4b - 4
= 9a² - (b² - 4b + 4). [Taking '-' common from the three terms]
= 9a² - [(b)² - (2 x b x 2) + (2)²]
= 9a² - (b - 2)² [As, a² - 2ab + b² = (a - b)²]
= (3a)² - (b - 2)²
= (3a + b - 2)[3a - (b - 2)] [As, (a + b)(a - b) = a² - b²]
= (3a + b - 2)(3a - b + 2)
Cheers!
follow me
Answered by
0
answer is 3(b-2c)(3a^2+2b-4c)
Attachments:
Similar questions