factorise 25a²-4b²+28bc-49c²
Answers
Answered by
131
(5a²)-(2b)²-28bc-(7c)²
(5a²)-{(2b)²+28bc+(7c)²}
(5a²)-{(2b)²+2×2b×7c+(7c)²}
(5a²)-(2b+7c)² [(x+y)²=x²+2xy+y²]
{(5a²)-(2b+7c)²} {(5a²)+(2b+7c)²} [x²-y²=(x-y) (x+y)
= (5a-2b-7c) (5a+2b+7c)
hence proved
(5a²)-{(2b)²+28bc+(7c)²}
(5a²)-{(2b)²+2×2b×7c+(7c)²}
(5a²)-(2b+7c)² [(x+y)²=x²+2xy+y²]
{(5a²)-(2b+7c)²} {(5a²)+(2b+7c)²} [x²-y²=(x-y) (x+y)
= (5a-2b-7c) (5a+2b+7c)
hence proved
Answered by
36
done see its a little bit tricky but after 1 once you make it it's easy
Attachments:
Similar questions