Question

Factorise : x2 + 9y2 + 422 + 6xy – 12yz – 4xz

Answer 1

Step-by-step explanation:

$\large{ \green{ \underline{ \sf GIVEN \: \: \: QUESTION : - }}}$

$\sf \: Factorise : {x}^{2} + 9 {y}^{2} + 4 {z}^{2} + 6xy – 12yz – 4xz$

$\large{ \green{ \underline{ \sf SOLUTION : - }}}$

$\sf \: First \: \: arrange \: \: the \: \: que.$

$\sf We \: \: get$

$\sf {x}^{2} + (3 {y})^{2} + ( - 2 {z})^{2} + 2(x)(3y) - 2(3y)(2z) - 2( 2z)(x)$

$\sf \: Using \: \: the \: \: formula : -$

$\tt \: {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$

$\tt \: = ( {a + b + c)}^{2}$

$\sf \: Here \: \: compare \: \: with \: the \: formula : -$

$\sf We \: \: get$

$\sf a = x \: \: \\ \sf b =3 y$

$\sf c = - 2z$

$\sf \: Hence \: we \: get : -$

$\sf {x}^{2} + (3 {y})^{2} + ( - 2 {z})^{2} + 2(x)(3y) - 2(3y)(2z) - 2( 2z)(x)$

$\sf = (x + 3y - 2z )^{2}$

$\sf \: Or \: \: we \: \: can \: \: write \: it \: \: as \: \: factor \: \: form$

$\sf \: = (x + 3y - 2z)(x + 3y - 2z)$

$\large{ \green{ \underline{ \tt Additional \: \: \: Information : - }}}$

$\green{ \star}\rm(a + b) {}^{2} = {a}^{2} + 2ab + {b}^{2}$

$\green{ \star}\rm(a - b) {}^{2} = {a}^{2} - 2ab + {b}^{2}$

$\green{ \star}\rm(a + b) {}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

$\green{ \star}\rm(a - b) {}^{3} = {a}^{3} - {b}^{3} + 3ab(a - b)$

$\green{ \star}\rm ( {a + b + c)}^{2} = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: ( {a}^{2} + {b}^{2} + c {}^{2} + 2ab + 2bc + 2ca)$

$\green{ \star}\rm \: {a}^{3} + {b}^{3} + {c}^{3} - 3abc = \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \rm \: (a + b + c)( {a}^{2} + b {}^{2} + {c}^{2} + ab + bc + ca)$