Answer:

The factorized expression is [tex]\((3(x-2y) - 4)(x - 2y + 2)\).[/tex]

Step-by-step explanation:

Certainly! Let's factorize the expression \(3(x-2y)^2 - 2(x-2y) - 8\) using the substitution \(k = (x-2y)\).

1.[tex]Substitute (\(k = (x-2y)\):[/tex]

 [tex]\[3k^2 - 2k - 8\][/tex]

2. Rearrange in Standard Form:

  [tex]\[3k^2 - 2k - 8\][/tex]

3. Factorize:

  To factorize the quadratic expression, find two numbers whose product is the product of the coefficient of [tex]\(k^2\) term (3)[/tex] and the constant term (-8), and whose sum is the coefficient of the linear term (-2).

  The numbers are -4 and +2 because[tex]\((-4) \times 2 = -8\) and \((-4) + 2 = -2\).[/tex]

  Rewrite the middle term using these numbers:

  \[tex][3k^2 - 4k + 2k - 8\][/tex]

4. Group and factor by grouping

  [tex]\[ (3k^2 - 4k) + (2k - 8) \][/tex]

  Factor out the common factor from each group:

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

5.Factorize the Common Factor:

  Now, notice that [tex]\((3k - 4)\) is a common factor in both terms. Factor it out:[/tex]

  [tex]\[ (3k - 4)(k + 2) \][/tex]

6. Substitute Back:

  Since [tex]\(k = (x-2y)\)[/tex], substitute [tex]\(k\)[/tex] back in:

  [tex]\[ (3(x-2y) - 4)(x - 2y + 2) \][/tex]

So, the factorized expression is [tex]\((3(x-2y) - 4)(x - 2y + 2)\).[/tex]

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]