How do you factor $6{{x}^{2}}+10$ ?

Hint: For solving these types of problems, we need to keep some basic axioms in mind which are an arithmetic term always has an arithmetic term as its factor but, an algebraic expression can have an arithmetic term or an algebraic term or both as its factors. At first, we take $2$ common from the two terms and the final answer becomes $2\left( 3{{x}^{2}}+5 \right)$ .

Complete step by step solution:
The given expression that we need to factor in this problem is,
$6{{x}^{2}}+10$
The term factor means a number (not necessarily) or an expression which can completely or exactly divide another number or expression. By completely or exactly dividing, we mean that after division, there is no remainder left or that the remainder is zero. For example, $12$ has factors $1,2.3,4,6,12$ which exactly divide $12$ . Also, the algebraic expression ${{x}^{2}}+2x+1$ has the factor $x+1$ in its name. Now, there are some basic axioms regarding factors. Some of them are that an arithmetic term always has an arithmetic term as its factor but, an algebraic expression can have an arithmetic term or an algebraic term or both as its factors. For example, $2{{x}^{2}}+4x+2$ has both $2$ and $x+1$ as its factors.
Now, coming to the problem, we need to factorize $6{{x}^{2}}+10$ . We see that we have two types of terms in the expression, one being the algebraic and the other being the arithmetic. Keeping the aforesaid axioms in mind, both the terms can have only one type of factor in common, which is arithmetic. That arithmetic term clearly is $2$ . So, taking $2$ common from the two terms, we get,
$\Rightarrow 2\left( 3{{x}^{2}}+5 \right)$
$3{{x}^{2}}+5$ cannot further be factorized.

Therefore, we can conclude that $6{{x}^{2}}+10$ upon factorization gives $2\left( 3{{x}^{2}}+5 \right)$.

Note: These problems are pretty easy and thus, should be errorless. We should always remember the axioms as those help the problem get easier. At last, we should cross-check if the final expression can be further factorized or not.