How do you factor $2{x^2} - 6xy + 2x$ completely?

Hint: In this question, we want to factor given algebraic expression. The process of finding two or more expressions whose product is the given expression is called the factorization of algebraic expressions. Thus, the factorization of algebraic expressions is the reverse process of multiplication.
Factorization using identities can be solved using three methods.
Taking out common factors.
Grouping.
The difference of two squares.

Complete step-by-step solution:
In this question, we want to factor the algebraic expression:
 $ \Rightarrow 2{x^2} - 6xy + 2x$
To solve this expression we will use the method of taking out the common factors because, in this question, the different terms of the given polynomial have common factors.
The first step is to find the H.C.F of all the terms of the given polynomial.
The H.C.F is the product of each common literal raised to the lowest power.
The factors of $2{x^2} = 2 \times x \times x$.
The factors of $6xy = 6 \times x \times y$.
The factors of $2x = 2 \times x$.
Here, the common literal raised to the lowest power is 2 and x. So, their product will be 2x.
Therefore, the H.C.F. of $\left( {2{x^2},6xy,2x} \right) = 2x$
The second step is to divide each term of the polynomial by H.C.F. the quotient will be enclosed within the brackets and the common factor will be kept outside the bracket.
Here, the H.C.F is 2x and the quotient will be $x - 3y + 1$.
Therefore,
$ \Rightarrow 2{x^2} - 6xy + 2x = 2x\left( {x - 3y + 1} \right)$

Hence, the factors of the given expression is $2x\left( {x - 3y + 1} \right)$.

Note: The factorization of algebraic expressions is the reverse process of multiplication. So, by multiplying the brackets we can get our algebraic expression back.
$ \Rightarrow 2x\left( {x - 3y + 1} \right)$
Let us remove the bracket. Multiply 2x with each term in the brackets.
$ \Rightarrow 2x\left( x \right) - 2x\left( {3y} \right) + 2x\left( 1 \right)$
That is equal to,
$ \Rightarrow 2{x^2} - 6xy + 2x$