How do you factor the trinomial \[{{x}^{2}}+6x+27\]?

Hint: Assume the obtained expression as y and compare it with the general form given as: - \[y=a{{x}^{2}}+bx+c\]. Find the respective values of a, b and c. Now, find the discriminant of the given quadratic equation by using the formula: - \[D={{b}^{2}}-4ac\], where D = discriminant. Now, write the expression as: - \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\]. If the value of D is negative then the expression will be of the form ${{m}^{2}}+{{n}^{2}}$ whose factored form will be $\left( m+in \right)\left( m-in \right)$ where $i=\sqrt{-1}$. If the value of D is positive then use the identity ${{m}^{2}}-{{n}^{2}}=\left( m+n \right)\left( m-n \right)$ to get the factored form.

Complete step by step solution:
Here, we have been provided with the quadratic equation: - \[{{x}^{2}}+6x+27\] and we are asked to factorize it. That means we have to write it as a product of two linear terms.
Now, we know that any quadratic equation of the form \[y=a{{x}^{2}}+bx+c\] can be simplified as \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\], using completing the square method. Here, ‘D’ denotes the discriminant. So, on assuming \[{{x}^{2}}+6x+27=y\] and comparing it with the general quadratic equation, we get,
\[\Rightarrow \] a = 1, b = 6, c = 27
Applying the formula for discriminant of a quadratic equation given as, \[D={{b}^{2}}-4ac\], we get,
\[\begin{align}
  & \Rightarrow D={{\left( 6 \right)}^{2}}-4\left( 1 \right)\left( 27 \right) \\
 & \Rightarrow D=36-108 \\
 & \Rightarrow D=-72 \\
\end{align}\]
As we can see that the value of D is negative it means that we will not get any real factors but we need to find the complex factors. Therefore, substituting the values in the simplified form of y, we get,
\[\begin{align}
  & \Rightarrow y=1\left[ {{\left( x+\left( \dfrac{6}{2\times 1} \right) \right)}^{2}}-\dfrac{\left( -72 \right)}{4\times {{\left( 1 \right)}^{2}}} \right] \\
 & \Rightarrow y=\left[ {{\left( x+3 \right)}^{2}}+18 \right] \\
 & \Rightarrow y=\left[ {{\left( x+3 \right)}^{2}}+{{\left( 3\sqrt{2} \right)}^{2}} \right] \\
\end{align}\]
Now, the above equation is of the form ${{m}^{2}}+{{n}^{2}}$ so the factored form can be given as $\left( m+in \right)\left( m-in \right)$ where $i=\sqrt{-1}$, so we have,
\[\Rightarrow y=\left( \left( x+3 \right)+3\sqrt{2}i \right)\left( \left( x+3 \right)-3\sqrt{2}i \right)\]
Hence, the above relation is the factored form of the given quadratic expression.

Note: One may note that here it may be difficult to use the middle term split method to get the factors because first of all the factors are not real and second it will be difficult to think of the factors like \[\left( \left( x+3 \right)+3\sqrt{2}i \right)\] and \[\left( \left( x+3 \right)-3\sqrt{2}i \right)\]. You must remember the expression: \[y=a\left[ {{\left( x+\dfrac{b}{2a} \right)}^{2}}-\dfrac{D}{4{{a}^{2}}} \right]\] otherwise you have to solve the quadratic expression by substituting it equal to 0 and get the factors from there.