Question

factorise: 4a(a+b) - 6(a+b)2

Answer 1

Given : 4a(a + b) - 6(a + b)^2

= > 4a^2 + 4ab - 6(a^2 + b^2 + 2ab)

= > 4a^2 + 4ab - 6a^2 - 6b^2 - 12ab

= > -2a^2 - 8ab - 6b^2

= > -2(a^2 + 4ab + 3b^2)

= > -2(a^2 + 3ab + ab + 3b^2)

= > -2(a(a + 3b) + b(a + 3b))

= > -2(a + b)(a + 3b).


Hope this helps!

Answer 2

$= > 4a(a + b) - 6 {(a + b)}^{2} \\ \\ = > 4 {a}^{2} + 4ab - 6( {a}^{2} + {b}^{2} + 2ab) \\ \\ = > 4 {a}^{2} + 4ab - 6 {a}^{2} - 6 {b}^{2} - 12ab \\ \\ = > - 2 {a}^{2} - 6 {b}^{2} - 8ab \\ \\ = > - 2( {a}^{2} + 3 {b}^{2} + 4ab) \\ \\ = > - 2( {a}^{2} + ab + 3 {b}^{2} + 3ab) \\ \\ = > - 2(a(a + b) + 3b(a + b)) \\ \\ = > - 2(a + b)(a + 3b)$