How do you factor the trinomial $4{{x}^{2}}-9x-9$?

Hint: We have been given a quadratic equation of $x$ as $4{{x}^{2}}-9x-9$. 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 solve the value of the x. We use the process of grouping. We put the values and find the solution.

Complete step by step solution:
We need to form the square part in $4{{x}^{2}}-9x-9$.
The square form of subtraction of two numbers be ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$.
We have $4{{x}^{2}}-9x-9={{\left( 2x \right)}^{2}}-2\times \left( 2x \right)\times \dfrac{9}{4}+{{\left( \dfrac{9}{4} \right)}^{2}}-9-{{\left( \dfrac{9}{4} \right)}^{2}}$.
Forming the square, we get \[4{{x}^{2}}-9x-9={{\left( 2x-\dfrac{9}{4} \right)}^{2}}-\dfrac{225}{16}={{\left( 2x-\dfrac{9}{4} \right)}^{2}}-{{\left( \dfrac{15}{4} \right)}^{2}}\].
We know the identity of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We get \[{{\left( 2x-\dfrac{9}{4} \right)}^{2}}-{{\left( \dfrac{15}{4} \right)}^{2}}\]. Taking solution,
we get ${{\left( 2x-\dfrac{9}{4} \right)}^{2}}-{{\left( \dfrac{15}{4} \right)}^{2}}=\left( 2x-\dfrac{9}{4}+\dfrac{15}{4} \right)\left( 2x-\dfrac{9}{4}-\dfrac{15}{4} \right)=\left( 2x+\dfrac{3}{2} \right)\left( 2x-6 \right)$.
Simplifying we get $4{{x}^{2}}-9x-9=\left( 2x+\dfrac{3}{2} \right)\left( 2x-6 \right)=\left( 4x+3 \right)\left( x-3 \right)$
Thus, the factorisation of the equation $4{{x}^{2}}-9x-9$ is $\left( 4x+3 \right)\left( x-3 \right)$.

Note: We apply the middle-term factoring or grouping to factorise the polynomial.
In the case of $4{{x}^{2}}-9x-9$, we break the middle term $-9x$ into two parts of $-12x$ and $3x$.
So, $4{{x}^{2}}-9x-9=4{{x}^{2}}-12x+3x-9$. We have one condition to check if the grouping is possible or not. If we order the individual elements of the polynomial according to their power of variables, then the multiple of end terms will be equal to the multiple of middle terms.
Here multiplication for both cases give $-36{{x}^{2}}$. The grouping will be done for $4{{x}^{2}}-12x$ and $3x-9$. We try to take the common numbers out.
For $4{{x}^{2}}-12x$, we take $4x$ and get $4x\left( x-3 \right)$.
For $3x-9$, we take 3 and get $3\left( x-3 \right)$.
The equation becomes $4{{x}^{2}}-9x-9=4{{x}^{2}}-12x+3x-9=4x\left( x-3 \right)+3\left( x-3 \right)$.
Both the terms have $\left( x-3 \right)$ in common. We take that term again and get
$\begin{align}
  & 4{{x}^{2}}-9x-9 \\
 & =4x\left( x-3 \right)+3\left( x-3 \right) \\
 & =\left( 4x+3 \right)\left( x-3 \right) \\
\end{align}$
Therefore, the factorisation of $4{{x}^{2}}-9x-9$ is $\left( 4x+3 \right)\left( x-3 \right)$.