How do you factor the trinomial ${{c}^{2}}+9c+18$?

Hint: We have been given a quadratic equation of $c$ as ${{c}^{2}}+9c+18$. We first try to form the square form of the given equation and find its root value from the square. We also use the quadratic formula to factor the equation. We have the solution in the form of $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for general equation of $a{{x}^{2}}+bx+c=0$. We put the values and find the solution.

Complete step by step solution:
We need to form the square part in ${{c}^{2}}+9c+18$.
The square form of subtraction of two numbers be ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$.
We have ${{c}^{2}}+9c+18={{c}^{2}}+2\times c\times \dfrac{9}{2}+{{\left( \dfrac{9}{2} \right)}^{2}}+18-{{\left( \dfrac{9}{2} \right)}^{2}}$.
Forming the square, we get ${{c}^{2}}+9c+18={{\left( c+\dfrac{9}{2} \right)}^{2}}-\dfrac{9}{4}={{\left( c+\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}$.
We know the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We get ${{\left( c+\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}$. Taking solution, we get
${{\left( c+\dfrac{9}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}=\left( c+\dfrac{9}{2}+\dfrac{3}{2} \right)\left( c+\dfrac{9}{2}-\dfrac{3}{2} \right)=\left( c+6 \right)\left( c+3 \right)$.
Thus, the factorisation of the equation ${{c}^{2}}+9c+18$ is $\left( c+6 \right)\left( c+3 \right)$.

Note: We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This is the quadratic equation solving method.
In the given equation we have ${{c}^{2}}+9c+18$. The values of a, b, c is $1,9,18$ respectively.
We put the values and get x as \[x=\dfrac{-9\pm \sqrt{{{9}^{2}}-4\times 18\times 1}}{2\times 1}=\dfrac{-9\pm \sqrt{9}}{2}=\dfrac{-9\pm 3}{2}=-3,-6\].
The roots of the equation are rational numbers.
The discriminant value being square, we get the rational numbers a root value.
In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is a full square. ${{b}^{2}}-4ac={{9}^{2}}-4\times 18\times 1=9$.
This is a square value. That’s why the roots are rational.