Question

Factorise
${(x - 2y)}^{2} - 5(x - 2y) + 6$
Please fast​

Answer 1

Answer:

(x-y-3). (x-y-2)

Step-by-step explanation:

(x-2y)² - 5(x-y) + 6

a² - 5a +6... Let (x-y) = a

=a²-(3+2) a +6

= a²-3a - 2a +6

= a. ( a-3) -2. (a-3)

= (a-3) . (a-2)

= (x-y-3). (x-y-2)... Since, a= (x-y).

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Answer 2

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

Given expression is

$\rm \: {(x - 2y)}^{2} - 5(x - 2y) + 6$

Let assume that

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \rm{ \:x - 2y = z \: }}$

So, above expression can be rewritten as

$\rm \: = \: {z}^{2} - 5z + 6$

On splitting the middle terms, we get

$\rm \: = \: {z}^{2} - 3z - 2z + 6$

$\rm \: = \: ( {z}^{2} - 3z) - (2z - 6)$

$\rm \: = \: z(z - 3) - 2(z - 3)$

$\rm \: = \: (z - 3)(z - 2)$

On substituting the value of z, we get

$\rm \: = \: (x - 2y - 3)(x - 2y - 2)$

Hence,

$\rm \: \: \: \: \: \: \: {(x - 2y)}^{2} - 5(x - 2y) + 6 \\ \\ = \rm \: \: (x - 2y - 3)(x - 2y - 2)$

$\rule{190pt}{2pt}$

Basic Concept Used :-

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers m and n such that m + n = b and mn = ac.

After finding m and n, we split the middle term i.e bx in the quadratic equation as mx + nx and get the required factors by grouping the terms.

$\rule{190pt}{2pt}$

Additional Information :-

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