How do you factor completely: $16{{x}^{2}}+40x+25$?

Hint: To factor the given polynomial $16{{x}^{2}}+40x+25$, we have to use the middle term splitting method to split the middle term as a sum of two terms so as to generate four terms. The splitting will have to be done in such a manner that the product of the two terms must be equal to the product of the first and the third terms. The first and the third terms in the given polynomial are $16{{x}^{2}}$ and $25$ respectively, which means that their product equals $400{{x}^{2}}$. Therefore, the middle term will be split as $40x=20x+20x$. Then finally using the factor by grouping method, we will be able to completely factorize the given polynomial.

Complete step by step solution:
Let us write the polynomial given as
$\Rightarrow p\left( x \right)=16{{x}^{2}}+40x+25$
By the middle term splitting method, we split the middle term of $40x$ as the sum of two terms such that their product is equal to the product of the first term, $16{{x}^{2}}$ and the third term, $25$, that is, equal to $400{{x}^{2}}$. Therefore we split the middle term as $40x=20x+20x$ in the above polynomial to get
$\Rightarrow p\left( x \right)=16{{x}^{2}}+20x+20x+25$
Now, by the factor by grouping method, we group the first two and the last two terms to form two pairs of terms as
$\Rightarrow p\left( x \right)=\left( 16{{x}^{2}}+20x \right)+\left( 20x+25 \right)$
Taking $4x$ and $5$ common from the two pairs, we get
$\Rightarrow p\left( x \right)=4x\left( 4x+5 \right)+5\left( 4x+5 \right)$
Finally we take $\left( 4x+5 \right)$ common to get
$\begin{align}
  & \Rightarrow p\left( x \right)=\left( 4x+5 \right)\left( 4x+5 \right) \\
 & \Rightarrow p\left( x \right)={{\left( 4x+5 \right)}^{2}} \\
\end{align}$
Hence, the given polynomial is completely factored as ${{\left( 4x+5 \right)}^{2}}$.

Note:
We can also use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ to factor the given polynomial. For this, we have to put $16={{4}^{2}}$ and $25={{5}^{2}}$ in the given polynomial to get \[{{\left( 4x \right)}^{2}}+40x+{{5}^{2}}\]. Then the middle term is to be written as $40x=2\left( 4x \right)\left( 5 \right)$ to get \[{{\left( 4x \right)}^{2}}+2\left( 4x \right)\left( 5 \right)+{{5}^{2}}\] which can be compared with the RHS of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ so that we can factor it as ${{\left( 4x+5 \right)}^{2}}$.