Solve $ \dfrac{{3{x^2} - 2x}}{x} $

Hint: First we have to define what the terms we need to solve the problem are.
In order to solve this problem, we must apply the Highest common factor method and take the common terms outside the given problem.

Complete step by step answer:
Highest Common Factor (HCF): The largest or greatest factor common to any two or more given natural numbers is termed as HCF of given numbers and The highest common factor is found by multiplying all the factors which appear in both lists: So the HCF of \[60\] and \[72\] is \[2{\text{ }} \times {\text{ }}2{\text{ }} \times {\text{ }}3\] which is $ 12 $ .
Let us first factorize the given polynomial $ 3{x^2} - 2x $ by finding the Highest common factors of all the terms, which are $ 3{x^2} = 3 \times x \times x $ and $ 2x = 2 \times x $ as we can clear see in both the terms the common factor is $ x $ and which is also the highest common factor too
Thus, highest common factor is $ \Rightarrow x $
Now we factor out the highest common factor from each term of the given polynomial $ 3{x^2} - 2x $ as shown below;
 $ 3{x^2} - 2x $ $ = x(3x - 2) $ [Taking the common values out which is $ x $ is the common value is both terms]
Let us now divide the polynomial $ 3{x^2} - 2x $ by the monomial $ x $ (singleton value in the polynomial function so it is called monomial)
 $ \dfrac{{3{x^2} - 2x}}{x} $ $ = \dfrac{{x(3x - 2)}}{x} $ [Now cancelling $ x $ on the numerator and denominator]
Thus $ \dfrac{{3{x^2} - 2x}}{x} $ = $ 3x - 2 $

Note: If the greatest common divisor is $ 1 $ then it is called GCD of the polynomial of relatively prime elements, there also exists a smallest positive integer that is a multiple of each of the numbers, called their least common multiple (LCM).
While solving these types of problems we must take HCF and take that common term out and try to cancel in it on the numerator and denominator.