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**Hint**: Here in this question, given an algebraic equation which resembles the quadratic equation \[a{x^2} + bx + c = 0\]. We have to find factoring or finding the factor by using the method of factorisation and taking out the greatest common division and by pairing the terms we get the required solution.

**:**

__Complete step-by-step answer__Factorization means the process of creating a list of factors otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.

The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\].

Consider the given expression:

\[ \Rightarrow {x^2} - xy + y - x\].

The equation has two variables \[x\] and \[y\].

Here, we have to factorise or find the factors of the given algebraic equation.

To solve this, firstly making pairs of terms in the above algebraic expression

\[ \Rightarrow \left( {{x^2} - xy} \right) + \left( {y - x} \right)\]

Take out greatest common divisor GCD from the both pairs, then

\[ \Rightarrow x\left( {x - y} \right) - 1\left( {x - y} \right)\]

In the second pair of terms we take -1 as common because to make the first and second pair equal.

Take \[\left( {x - y} \right)\] common

\[ \Rightarrow \left( {x - y} \right)\left( {x - 1} \right)\]

Or

\[ \Rightarrow \left( {x - 1} \right)\left( {x - y} \right)\]

Therefore, the factors of the equation \[{x^2} - xy + y - x\] is \[\left( {x - 1} \right)\left( {x - y} \right)\].

**So, the correct answer is “\[\left( {x - 1} \right)\left( {x - y} \right)\].”.**

**Note**: The equation is a quadratic equation. This problem can be solved by using the sum product rule. This defines as for the general quadratic equation \[a{x^2} + bx + c\], the product of \[a{x^2}\] and c is equal to the sum of bx of the equation. Hence we obtain the factors. The factors for the equation depend on the degree of the equation.

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