Factorise:4 (2a+b)^2-(a-b)^2
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Answered by
12
(8a + 4b - a - b ) ^2
(8a - a + 4b - b ) ^2
(7a + 3b ) ^2
As we know ( A^2 - B ^2 )
=(7a + 3b) (7a - 3b)
(8a - a + 4b - b ) ^2
(7a + 3b ) ^2
As we know ( A^2 - B ^2 )
=(7a + 3b) (7a - 3b)
siddhartharao77:
kindly correct ur answer
Answered by
64
Your answer is --
we have ,
4(2a+b)^2 -(a-b)^2
= 4(4a^2+b^2+4ab) - (a^2+b^2 - 2ab)
= 16a^2 + 4b^2 + 16ab - a^2 - b^2 + 2ab
= 15a^2 + 3b^2 + 18ab
= 3(5a^2 + b^2 + 6ab)
= 3(5a^2 + 6ab + b^2)
= 3(5a^2 + 5ab + ab + b^2)
= 3{5a (a+b) + b(a+b)}
= 3 (5a+b) ( a+b)
【 Hope it helps you 】
we have ,
4(2a+b)^2 -(a-b)^2
= 4(4a^2+b^2+4ab) - (a^2+b^2 - 2ab)
= 16a^2 + 4b^2 + 16ab - a^2 - b^2 + 2ab
= 15a^2 + 3b^2 + 18ab
= 3(5a^2 + b^2 + 6ab)
= 3(5a^2 + 6ab + b^2)
= 3(5a^2 + 5ab + ab + b^2)
= 3{5a (a+b) + b(a+b)}
= 3 (5a+b) ( a+b)
【 Hope it helps you 】
Kindly correct ur answer.
shi hai
yes but answer 3 (5a+b)(a+b)
Thats why i told to correct answer.
so correct it please
see again
The answer will be - 3 (5a+b)(a+b)
15a^2 + 3b^2 + 18ab. Dont equate this equation to 0. Let it be and take ca common factor i.e 3
now, right
Yes
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