Question

$16a {}^{2} + 4b ^{2} + 9c ^{2} - 16ab + 12bc - 24ac$
pls tell answer tomorrow exam plspls​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Factorise : \: 16a {}^{2} + 4b ^{2} + 9c ^{2} - 16ab + 12bc - 24ac$

$\large\underline{\bold{Solution-}}$

Identity Used :-

$\boxed{ \bf{ {(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx}}$

Let's solve the problem now!!

Given that

$\sf \: 16a {}^{2} + 4b ^{2} + 9c ^{2} - 16ab + 12bc - 24ac$

$\sf \: = {(4a)}^{2} + {(2b)}^{2} + {(3c)}^{2} - 2(4a)(2b) + 2(2b)(3c) - 2(4a)(3c)$

$\sf \: = {( - 4a)}^{2} + {(2b)}^{2} + {(3c)}^{2} + 2( - 4a)(2b) + 2(2b)(3c) + 2( - 4a)(3c)$

$\sf \: = {( - 4a + 2b + 3c)}^{2}$

$\sf \: = ( - 4a + 2b + 3c)( - 4a + 2b + 3c)$

Additional Information :-

Useful Identities :-

$(1). \: \boxed{ \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}}$

$(2). \: \boxed{ \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }}$

$(3). \: \boxed{ \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y) }}$

$(4). \: \boxed{ \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$(5). \: \boxed{ \bf{ {(x + y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)}}$

$(6). \: \boxed{ \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2}}}$

$(7). \: \boxed{ \bf{ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2}}}$