Question

factorise 25a²— 4b²+28bc —49c²​

Answer 1

$\bf\huge\textbf{\underline{\underline{According\:to\:the\:Question}}}$  

$\bf\huge\bf\huge{\boxed{\bigstar{{5a^2 - 4b^2 + 28bc - 49c^2}}}}$        

$\bf\huge\bf\huge{\boxed{\bigstar{{(5a)^2 - (4b^2 - 28bc + 49c^2)}}}}$        

$\bf\huge\bf\huge{\boxed{\bigstar{{(5a)^2 - (2b)^2 - 2(2b)(7c) + (7c)^2}}}}$        

$\bf\huge\bf\huge{\boxed{\bigstar{{(5a)^2 - (2b - 7c)^2}}}}$          

$\bf\huge\bf\huge{\boxed{\bigstar{{(5a - 2b + 7c)(5a + 2b - 7c)}}}}$        

Answer 2

Answer:

Step-by-step explanation:

I think in ur procedure 4b² should be written as (4b)²

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