How do you factorize the equation ${{x}^{2}}-27=0$?

Hint: We know that the given equation is a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ . Comparing the given equation with the general form we will get the values of a, b and c. Now we know that the roots of any quadratic equation is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . Hence we will substitute the values of a, b and c in the formula to find the roots of the given quadratic equation. Now once we have roots we can easily find the factors as the factors of the equation are $\left( x-\alpha \right)$ and $\left( x-\beta \right)$ where $\alpha $ and $\beta $ are the roots of the equation.

Complete step-by-step solution:
Now consider the given equation ${{x}^{2}}-27=0$
We know that the given equation is a quadratic equation in x.
Comparing the given equation ${{x}^{2}}-27=0$ with the general form of quadratic equation $a{{x}^{2}}+bx+c=0$ we get a = 1 b = 0 and c = 27.
Now we know that the solution to any quadratic equation of the form $a{{x}^{2}}+bx+c=0$ is given by the formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
Hence substituting the values of a, b and c in the formula we get,
$\begin{align}
  & \Rightarrow x=\dfrac{0\pm \sqrt{0-\left( 4 \right)\left( 1 \right)\left( -27 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{\pm 2\sqrt{27}}{2}=\pm \sqrt{27} \\
 & \Rightarrow x=\pm 3\sqrt{3} \\
\end{align}$
Hence the roots of the equation are $x=3\sqrt{3}$ and $x=-3\sqrt{3}$
Now if $\alpha $ and $\beta $ are the roots of the equation then $\left( x-\alpha \right)$ and $\left( x-\beta \right)$ are the factors of the equation.
Hence the factors of the equation are $\left( x-3\sqrt{3} \right)$ and $\left( x+3\sqrt{3} \right)$.

Note: We can also solve this equation easily by just taking the term 27 on RHS and then taking square root on both sides. While taking square roots remember to take both the cases positive and negative. Hence we will get the roots of the equation and hence the factors of the equation.