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**Hint:**We use both grouping method and vanishing method to solve the problem. We take common terms out to form the multiplied forms. Factorising a polynomial by grouping is to find the pairs which on taking their common divisor out, give the same remaining number. In the case of vanishing method, we use the value of $x$ which gives the polynomial value 0.

**Complete step by step solution:**

We apply the middle-term factoring or grouping to factorise the polynomial.

In case of $4{{x}^{4}}-5{{x}^{2}}-9$, we break the middle term $-5{{x}^{2}}$ into two parts of $-9{{x}^{2}}$ and $4{{x}^{2}}$.

So, $4{{x}^{4}}-5{{x}^{2}}-9=4{{x}^{4}}-9{{x}^{2}}+4{{x}^{2}}-9$. 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 gives $-36{{x}^{2}}$. The grouping will be done for $4{{x}^{4}}-9{{x}^{2}}$ and $4{{x}^{2}}-9$.

We try to take the common numbers out.

For $4{{x}^{4}}-9{{x}^{2}}$, we take ${{x}^{2}}$ and get ${{x}^{2}}\left( 4{{x}^{2}}-9 \right)$.

For $4{{x}^{2}}-9$, we take 1 and get $\left( 4{{x}^{2}}-9 \right)$.

The equation becomes $4{{x}^{4}}-5{{x}^{2}}-9=4{{x}^{4}}-9{{x}^{2}}+4{{x}^{2}}-9={{x}^{2}}\left( 4{{x}^{2}}-9 \right)+\left( 4{{x}^{2}}-9 \right)$.

Both the terms have $\left( 4{{x}^{2}}-9 \right)$ in common. We take that term again and get

$\begin{align}

& 4{{x}^{4}}-5{{x}^{2}}-9 \\

& ={{x}^{2}}\left( 4{{x}^{2}}-9 \right)+\left( 4{{x}^{2}}-9 \right) \\

& =\left( 4{{x}^{2}}-9 \right)\left( {{x}^{2}}+1 \right) \\

\end{align}$

We still can use the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to factor $\left( 4{{x}^{2}}-9 \right)$ where we take $a=2x,b=3$. So, $\left( 4{{x}^{2}}-9 \right)=\left( 2x+3 \right)\left( 2x-3 \right)$.

The final factorisation is $4{{x}^{4}}-5{{x}^{2}}-9=\left( 2x+3 \right)\left( 2x-3 \right)\left( {{x}^{2}}+1 \right)$.

Therefore, $\left( 4{{x}^{2}}-9 \right)\left( {{x}^{2}}+1 \right)=0$ has multiplication of two polynomial giving value of 0. This means at least one of them has to be 0.

Therefore, either $\left( 2x+3 \right)=0$ or $\left( 2x-3 \right)=0$.

So, values of x are $x=\pm \dfrac{3}{2}$.

The polynomial $\left( {{x}^{2}}+1 \right)>0$. The solution of $\left( {{x}^{2}}+1 \right)=0$ is imaginary where $x=\pm i$.

**The solution for $4{{x}^{4}}-5{{x}^{2}}-9=0$ is $x=\pm \dfrac{3}{2},\pm i$.**

**Note:**We find the value of x for which the function $f\left( x \right)=4{{x}^{4}}-5{{x}^{2}}-9$. We can see $f\left( \dfrac{3}{2} \right)=4{{\left( \dfrac{3}{2} \right)}^{4}}-5{{\left( \dfrac{3}{2} \right)}^{2}}-9=\dfrac{81}{4}-\dfrac{45}{4}-9=0$. So, the root of the $f\left( x \right)=4{{x}^{4}}-5{{x}^{2}}-9$ will be the function $\left( x-\dfrac{3}{2} \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.

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