Factorize the polynomial ${{x}^{2}}+3\sqrt{3x}+6$ by splitting the middle term
\[\begin{align}
  & A.\left( x+\sqrt{3} \right)\left( x-2\sqrt{3} \right) \\
 & B.\left( x-\sqrt{3} \right)\left( x+2\sqrt{3} \right) \\
 & C.\left( x+\sqrt{3} \right)\left( x+2\sqrt{3} \right) \\
 & D.\left( x-\sqrt{3} \right)\left( x-2\sqrt{3} \right) \\
\end{align}\]

Hint: Generally we have seen a quadratic equation, which does not include any specific method for solving the equation. It is dependent on the student, whether he/she uses the splitting of middle term concept or some other. But here, in our question, it is clearly mentioned that we have to apply the splitting the middle term concept. The approach is quite simple once we get the concept. Basically, we have two ways to solve any quadratic equation. First and most common method is "splitting the middle term" and second and little bit complicated method is using $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ if $a{{x}^{2}}+bx+c=0$ is given.

Complete step-by-step solution:
We have been given equation as
\[{{x}^{2}}+3\sqrt{3x}+6=0\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\left( i \right)\]
The above quadratic equation is of the form ${{x}^{2}}+bx+c=0$
Hence, in order to factorize ${{x}^{2}}+bx+c=0$ we have to find numbers p and q such that $p+q=b\text{ and }pq=c$
After finding p and q, we split the middle term in the quadratic as $px+qx$ and get desired factors by grouping the terms.
Now, \[{{x}^{2}}+3\sqrt{3x}+6=0\]
So here, $p+q=3\sqrt{3},pq=6$
Hence, we get:
\[\begin{align}
  & p=\sqrt{3},q=2\sqrt{3} \\
 & \text{as }\sqrt{3}+2\sqrt{3}=3\sqrt{3} \\
 & \text{and }\left( \sqrt{3} \right)\left( 2\sqrt{3} \right)=2\times 3=6 \\
\end{align}\]
So, we split as \[px+qx=\sqrt{3}x+2\sqrt{3}x\]
Now,
\[\begin{align}
  & {{x}^{2}}+3\sqrt{3x}+6=0 \\
 & {{x}^{2}}+\sqrt{3}x+2\sqrt{3}x+6=0 \\
 & x\left( x+\sqrt{3} \right)+2\sqrt{3}\left( x+\sqrt{3} \right)=0 \\
 & \left( x+\sqrt{3} \right)\left( x+2\sqrt{3} \right)=0 \\
\end{align}\]
Now, according to our question, we have: \[{{x}^{2}}+3\sqrt{3x}+6=\left( x+\sqrt{3} \right)\left( x+2\sqrt{3} \right)\]. Therefore, option C is correct.

Note: Here, there is very less chances of getting an error in the splitting method. There is very little doubt may raise in multiplication of two surds (when we can't simplify a number to remove a square root (or cube root, etc.) then it is a surd e.g. $\sqrt{2},\sqrt{3}$)
Like,
\[\begin{align}
  & \sqrt{3}\times \sqrt{3}={{\left( 3 \right)}^{\dfrac{1}{2}}}\times {{\left( 3 \right)}^{\dfrac{1}{2}}} \\
 & \Rightarrow {{\left( 3 \right)}^{\dfrac{1}{2}\times \dfrac{1}{2}}} \\
 & \Rightarrow {{\left( 3 \right)}^{1}} \\
 & \Rightarrow \left( 3 \right) \\
\end{align}\]
Hence, by multiplying two same surds (in case of square root) we will get the number inside the square root. Here the solution does not relate p and q as the zeros of polynomials.