How do you factor \[2{{x}^{6}}-3{{x}^{4}}\]?

Hint: Find the greatest common factor of both the terms. Then take that common factor out first. Convert the remaining factor into ${{a}^{2}}-{{b}^{2}}$ form. Then use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ for further factorization.

Complete step by step solution:
Factorization: A polynomial can be written as a product of two or more polynomials of degree less than or equal to that of it. Each polynomial involved in the product will be a factor of it. The process involved in breaking a polynomial into the product of its factors is known as the factorization of polynomials.
The expression we have \[2{{x}^{6}}-3{{x}^{4}}\],
As we know, the greatest common factor of \[2{{x}^{6}}\] and \[3{{x}^{4}}\] is \[{{x}^{4}}\]
So taking common out ‘\[{{x}^{4}}\]’ from both the terms, we get
$\Rightarrow {{x}^{4}}\left( 2{{x}^{2}}-3 \right)$
For $\left( 2{{x}^{2}}-3 \right)$ part,
2 and 3 can be written as ${{\left( \sqrt{2} \right)}^{2}}$ and ${{\left( \sqrt{3} \right)}^{2}}$ respectively
So, $\left( 2{{x}^{2}}-3 \right)$ can be written as ${{\left( \sqrt{2}x \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}$
As we know, ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
So, ${{\left( \sqrt{2}x \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=\left( \sqrt{2}x+\sqrt{3} \right)\left( \sqrt{2}x-\sqrt{3} \right)$
Hence, our expression becomes
$\Rightarrow {{x}^{4}}\left( \sqrt{2}x+\sqrt{3} \right)\left( \sqrt{2}x-\sqrt{3} \right)$
This is the required factorization of the given question.

Note: The common factor should be taken out first as it is a factor itself. The solution should be in maximum simplified form. For example, the factor ${{x}^{4}}\left( 2{{x}^{2}}-3 \right)$ we obtained, during the calculation can be the solution. But for an appropriate solution it should be factored using the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.