How do you factor by grouping $2{{n}^{2}}+5n+2$?

Hint: The method of factoring a given polynomial involves grouping of the pairs of the terms of the polynomials such that one factor is common to each pair. In the given polynomial $2{{n}^{2}}+5n+2$, there are three terms involved. For grouping, we need to divide these into pairs. For this we can use the middle term technique to split the middle term $5n$ into terms such that their product is equal to the product of the first and the last term of the polynomial $2{{n}^{2}}+5n+2$, that is $4{{n}^{2}}$. The only way to do this is $5n=4n+n$ and so our polynomial will be written as $2{{n}^{2}}+4n+n+2$. Then we have to take out the highest factor common from each of the two pairs. Finally, one factor common to each of the pairs will be taken common and the polynomial $2{{n}^{2}}+5n+2$ will be factored.

Complete step by step answer:
Let us write the polynomial given in the question as
$f\left( n \right)=2{{n}^{2}}+5n+2.......(i)$
For grouping the terms of the above polynomial, we have to make the pairs of the terms. But since there are three terms, we need to split the middle term by the middle term technique so as to get four terms and divide them into two pairs.
By the middle term technique, we split the middle term $5n$ as a sum of two terms such that their product is equal to the product of the first and last term of the polynomial $2{{n}^{2}}+5n+2$, that is, equal to $\left( 2{{n}^{2}} \right)\left( 2 \right)=4{{n}^{2}}$. The only way to do so is to write the middle term as $5n=4n+n$. Putting this in (i), we get
\[\begin{align}
  & \Rightarrow f\left( n \right)=2{{n}^{2}}+4n+n+2 \\
 & \Rightarrow f\left( n \right)=\left( 2{{n}^{2}}+4n \right)+\left( n+2 \right) \\
\end{align}\]
So we have formed two pairs of terms, which are \[\left( 2{{n}^{2}}+4n \right)\] and \[\left( n+2 \right)\]. Now, the highest factor common in \[\left( 2{{n}^{2}}+4n \right)\] is $2n$ and that in \[\left( n+2 \right)\] is $1$. Taking these common, we have
$\Rightarrow f\left( n \right)=2n\left( n+2 \right)+1\left( n+2 \right)$
Now, we have the factor \[\left( n+2 \right)\] common to both the pairs. Taking it common, we get\[f\left( n \right)=\left( n+2 \right)\left( 2n+1 \right)\]
Hence, the given polynomial is factored.

Note: If in the last step of the grouping no factor comes out common from each of the pairs, then this means you have made some mistake. Either the middle term is split in the wrong way, or there is some mistake in taking the highest factor common from both the pairs.