Question

Factorise:9a^2+4b^2+16c^2+12ab-16bc-24ca

Answer 1

$identity \\ ( {a + b + c)}^{2} = ( {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ac) \\ {(3a + 2b - 4c)}^{2} = (9 {a}^{2} + 4{b}^{2} + 16{c}^{2} + 12ab - 16bc - 24ac)$
so according to that identity,try to converts that terms so to fit into the formula
so factors are
$( {3a + 2b - 4c)}^{2}$

Answer 2

Answer:

Step-by-step explanation:

9a² + 4b² + 16c² + 12ab - 16bc - 24ca

=> ( - 3a )² + ( - 2b )² + ( 4 c )² + 2[ ( -3a × -2b ) + ( -2b × 4c ) + ( 4c × - 3a ) ]

We know, a² + b² + c² + 2( ab + bc + ca ) = ( a + b + c )²

Hence,

.

=> [ ( -3a ) + ( -2b ) + ( 4c ) ]²

=> ( -3a - 2b + 4c )²