Verified answer

Answer:

[tex]\boxed{\bf\:3 {(x - 2y)}^{2} - 2(x - 2y) - 8 =\left(x - 2y - 2\right)\left(3x - 6y + 4\right) \: } \\ [/tex]

Step-by-step explanation:

Given algebraic expression is

[tex] \sf \: 3 {(x - 2y)}^{2} - 2(x - 2y) - 8 \\ [/tex]

Let assume that x - 2y = k

So, above expression can be rewritten as

[tex] \sf \: = \: 3 {k}^{2} - 2k - 8 \\ [/tex]

[tex] \sf \: = \: 3 {k}^{2} - 6k + 4k - 8 \\ [/tex]

[tex] \sf \: = \: 3k(k - 2) + 4(k - 2) \\ [/tex]

[tex] \sf \: = \: (k - 2)(3k + 4) \\ [/tex]

On substituting the value of k, we get

[tex] \sf \: = \: \left[x - 2y - 2\right]\left[3(x - 2y) + 4\right] \\ [/tex]

[tex] \sf \: = \: \left(x - 2y - 2\right)\left(3x - 6y + 4\right)\\ [/tex]

Hence,

[tex]\implies\boxed{\bf\:3 {(x - 2y)}^{2} - 2(x - 2y) - 8 =\left(x - 2y - 2\right)\left(3x - 6y + 4\right) \: }\\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information:

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]