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# How do you factor $3{{x}^{2}}-2$ ?

Last updated date: 20th Jun 2024
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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 $\left( x-a \right)\left( x-b \right)$ .

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 $a{{x}^{2}}+bx+c=\left( x-\alpha \right)\left( x-\beta \right)$ .