Question

factorise this question plz fast urgent​

Answer 1

25a2 – 4b2 + 28bc – 49c2

= (5a)2 – (4b2 – 28bc + 49c2)

= (5a)2 – {(2b)2 – 2(2b)(7c) + (7c)2}

= (5a)2 – (2b – 7c)2

= (5a – 2b + 7c)(5a + 2b – 7c)

Answer 2

$\red{ \huge{ \underline{ \overline{ \mid{ \bold{ \: \: QUESTION \: \: \mid}}}}}}$

$\huge{\underline{ \bold{To \: \: Factorise \: \: : }}}$

$\boxed{ \pink{ \bold{ \: \: 25a {}^{2} - 4b {}^{2} + 28bc - 49c {}^{2} \: \: }}}$

$\red{ \huge{ \underline{ \overline{ \mid{ \bold{ \: \: SOLUTION\: \: \mid}}}}}}$

$\bold{ \blue{25a {}^{2} - 4b {}^{2} + 28bc - 49c {}^{2} }} \\ \\ \bold{ \blue{ = 25a {}^{2} - (4b {}^{2} - 28bc + 49c {}^{2}) }} \\ \\ \\ \\ \underline{\bold{ \green{Using \: \: \boxed{(x - y) {}^{2} = x {}^{2} - 2xy + y {}^{2} } \: \: this \: \: identity \: \: }}}\\ \\ \\ \bold {\blue{25a {}^{2} - (4b {}^{2} - 28bc + 49c {}^{2}) }} \\ \\ \bold{ \blue{ = 25a {}^{2} - [(2b) {}^{2} - 2 \times 2b \times 7c + (7c) {}^{2} ] }} \\ \\ \bold{ \blue{ = (5a) {}^{2} -( 2b - 7c) {}^{2} }} \: \: \\ \\ \\ \\ \underline{\bold{ \green{Using \: \: \boxed{x {}^{2} - y {}^{2} = (x + y)(x - y) } \: \: this \: \: identity}}} \\ \\ \\ \bold{ \blue{(5a) {}^{2} - (2b - 7c) {}^{2} }} \\ \\ \bold{ \blue{ = [5a + (2b - 7c)][5a - (2b - 7c)]}} \\ \\ \bold{ \blue{ = (5a + 2b - 7c)(5a - 2 b+ 7c)}}$


$\red{ \huge{ \underline{ \overline{ \mid{ \bold{ \: \: ANSWER\: \: \mid}}}}}}$

$\green{ \boxed{ \boxed{ \bold{ \pink{25a {}^{2} - 4b {}^{2} + 28bc - 49c {}^{2} = (5a + 2b - 7c)(5a - 2b + 7c)}}}}}$