Question

factorisation
25a2-4b2+28bc-49c2​

Answer 1

$\boxed{\large\mathrm{Solution}}$

$\mathrm{25 {a}^{2} - 4 {b}^{2} + 28bc - 49 {c}^{2} }$

$\mathrm{ = (5a)^{2} - (2b)^{2} + 28bc - (7c)^{2} }$

= (5a)² - {(2b)² + 2 × 2b × 7c − (7c)²}

$\mathrm{\because \: [(a + b)² = a² + b² + 2ab]}$

$\mathrm{ = (5a)^{2} - (2b \: + 7c)^{2} }$

$\mathrm{\because \: [ {a}^{2} - {b}^{2} = (a - b)(a + b) ] }$

= { (5a) − (2b + 7c)} {(5a) + (2b + 7c)}

$\mathrm{ = (5a - 2b - 7c)(5a + 2b + 7c)}$