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**Hint:**

We will use the concept of the factorization of the quadratic equation to solve the above polynomial. At first, we will make the coefficient of the $ {{x}^{2}} $ as 1 by taking 3 as common and then we will find the root of the equation $ {{x}^{2}}-\dfrac{2}{3}=0 $ . Let us say that root is a, b then the factor of $ {{x}^{2}}-\dfrac{2}{3} $ is $ \left( x-a \right)\left( x-b \right) $ .

**Complete step by step answer:**

We know from the question that we have to factorize $ 3{{x}^{2}}-2 $ . So, we will use the concept of the factorization of the quadratic polynomial to solve the above polynomial.

At first, we will make the coefficient of the $ {{x}^{2}} $ as 1 of the polynomial $ 3{{x}^{2}}-2 $ . We can do this by taking 3 commons from it. Then we will get:

$ \Rightarrow 3{{x}^{2}}-2=3\left( {{x}^{2}}-\dfrac{2}{3} \right) $

Now, we will find the root of the quadratic polynomial $ {{x}^{2}}-\dfrac{2}{3} $ by equating it with zero.

$ \Rightarrow {{x}^{2}}-\dfrac{2}{3}=0 $

If we consider the above in the form of $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ so we can apply this and we will get

$ \Rightarrow \left( x-\sqrt{\dfrac{2}{3}} \right)\left( x+\sqrt{\dfrac{2}{3}} \right)=0 $

Equating each term to 0, we will get the values of x as

$ \therefore x=\sqrt{\dfrac{2}{3}},-\sqrt{\dfrac{2}{3}} $

So, root of the equation $ {{x}^{2}}-\dfrac{2}{3}=0 $ is $ \sqrt{\dfrac{2}{3}},-\sqrt{\dfrac{2}{3}} $ .

So, we can express $ {{x}^{2}}-\dfrac{2}{3} $ as $ \left( x-\sqrt{\dfrac{2}{3}} \right)\left( x+\sqrt{\dfrac{2}{3}} \right) $ .

$ \Rightarrow 3{{x}^{2}}-2=3\left( {{x}^{2}}-\dfrac{2}{3} \right)=3\left( x-\sqrt{\dfrac{2}{3}} \right)\left( x+\sqrt{\dfrac{2}{3}} \right) $

Hence, the factorization of the polynomial $ 3{{x}^{2}}-2 $ is:

\[=3\left( x-\sqrt{\dfrac{2}{3}} \right)\left( x+\sqrt{\dfrac{2}{3}} \right)\]

This is our required solution.

**Note:**

Students are required to note that to find the root of the quadratic equation $ a{{x}^{2}}+bx+c=0 $ we will use middle term splitting method or the formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ to find the root. And , And, suppose the root is $ \alpha ,\beta $ then the factor of $ a{{x}^{2}}+bx+c=\left( x-\alpha \right)\left( x-\beta \right) $ .

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