How do you factor $3x{{y}^{2}}-12x$ ?

Hint: The process of factorization means taking out the common terms or the factors from the expression. The given expression is a polynomial in two variables. On observing the two terms we can see that there is a common term that can be taken out and set aside as a factor. Now among the factorized terms, find the expression which can be further factorized and then write that also in the product of sums form. Now write all of them together and represent them as the factors of the given expression.

Complete step by step solution:
The given polynomial is $3x{{y}^{2}}-12x$
It is a polynomial in two variables $x,y\;$
Now let us first take the common terms from both terms.
As we can look through, we see that the term $x$ is common in both terms.
After taking the common factor out we get,
$\Rightarrow x\left( 3{{y}^{2}}-12 \right)$
Now let us consider the coefficients in front of the variables.
All the coefficients have a common factor $3$ .
Let us take that also out.
After taking the common factor out we get,
$\Rightarrow 3x\left( {{y}^{2}}-4 \right)$
Now we can see that there is still another polynomial left that can be factored further.
We can write ${{y}^{2}}-4$ as ${{\left( y \right)}^{2}}-{{\left( 2 \right)}^{2}}$
And since ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can factorize it as,
$\Rightarrow \left( y+2 \right)\left( y-2 \right)$
Now writing it all together we get,
$\Rightarrow 3x\left( y+2 \right)\left( y-2 \right)$
Hence the factors for the polynomial $3x{{y}^{2}}-12x$ are $3x\left( y+2 \right)\left( y-2 \right)$.

Note: The process of factorization is reverse multiplication. In the above question, we have multiplied two linear line equations to get a quadratic equation (a polynomial of degree $2$ ) expression using the distributive law. The name quadratic comes from “quad” meaning square because the variable gets squared. Also known as a polynomial of degree $2$ .