How do you factor the expression $2{{x}^{2}}-50$ ?

Hint: Since $2$ is the common factor in both terms, take one $2$ common. Then write them together as the product of sums form. 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 which must be factorized is $2{{x}^{2}}-50$
Now let us take $2$ as a common factor out of both the terms as a factor.
After taking the common factor out we get,
$\Rightarrow 2\left( {{x}^{2}}-25 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow 2\left( {{x}^{2}}-25 \right)$
Now we can see that there is still another polynomial left that can be factored further.
We can write $\left( {{x}^{2}}-25 \right)$ as ${{\left( x \right)}^{2}}-{{\left( 5 \right)}^{2}}$
And since ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can factorize it as,
$\Rightarrow \left( x+5 \right)\left( x-5 \right)$
Now writing it all together we get,
$\Rightarrow 2\left( x+5 \right)\left( x-5 \right)$
Hence the factors for the polynomial $2{{x}^{2}}-50$ are $2\left( x+5 \right)\left( x-5 \right)$
This can also be said in a different way.
The values of x which satisfy the equation are, $5;-5$
When we put these values, we get LHS=RHS and this proves that our answer is correct.

Note: The process of factorization means taking out the common terms or the factors from the expression. Whenever there is a polynomial that is to be solved, the solution contains the roots of the expression which is found out by the factors. Since they are the roots of the expression, when put back into the polynomial, it satisfies the equation. The number of roots is decided by the degree of the polynomial. The process of factorization is reverse multiplication.