How do you simplify $\dfrac{8{{a}^{2}}-6a-9}{6{{a}^{2}}-5a-6}\div \dfrac{4{{a}^{2}}+11a+6}{9{{a}^{2}}+12a+4}$ ?

Hint: In this question we have been asked to simplify the given expression $\dfrac{8{{a}^{2}}-6a-9}{6{{a}^{2}}-5a-6}\div \dfrac{4{{a}^{2}}+11a+6}{9{{a}^{2}}+12a+4}$. For doing that we will factorise the quadratic expressions present in the numerator and denominator. For factorising we will use traditional factorization methods.

Complete step by step solution:
Now considering from the question we have been asked to simplify the given expression $\dfrac{8{{a}^{2}}-6a-9}{6{{a}^{2}}-5a-6}\div \dfrac{4{{a}^{2}}+11a+6}{9{{a}^{2}}+12a+4}$ .
Firstly we will factorise the quadratic expressions present in the numerator and denominator of the expression.
The quadratic expression $8{{a}^{2}}-6a-9$ can be factored as
$\begin{align}
  & \Rightarrow 8{{a}^{2}}-6a-9=8{{a}^{2}}-12a+6a-9 \\
 & \Rightarrow 4a\left( 2a-3 \right)+3\left( 2a-3 \right)=\left( 4a+3 \right)\left( 2a-3 \right) \\
\end{align}$
Similarly we will factorise the expression $6{{a}^{2}}-5a-6$ as
$\begin{align}
  & \Rightarrow 6{{a}^{2}}-5a-6=6{{a}^{2}}-9a+4a-6 \\
 & \Rightarrow 3a\left( 2a-3 \right)+2\left( 2a-3 \right)=\left( 3a+2 \right)\left( 2a-3 \right) \\
\end{align}$
Now we will factorise the expression $4{{a}^{2}}+11a+6$ and write it as
$\begin{align}
  & \Rightarrow 4{{a}^{2}}+11a+6=4{{a}^{2}}+8a+3a+6 \\
 & \Rightarrow 4a\left( a+2 \right)+3\left( a+2 \right)=\left( 4a+3 \right)\left( a+2 \right) \\
\end{align}$
Now we will factorise the last quadratic expression $9{{a}^{2}}+12a+4$ after that we will have $\begin{align}
  & \Rightarrow 9{{a}^{2}}+12a+4=9{{a}^{2}}+6a+6a+4 \\
 & \Rightarrow 3a\left( 3a+2 \right)+2\left( 3a+2 \right)=\left( 3a+2 \right)\left( 3a+2 \right) \\
 & \Rightarrow {{\left( 3a+2 \right)}^{2}} \\
\end{align}$
Now we will replace all the four quadratic expressions present in the numerators and denominators with its factors. After doing that we will have
$\Rightarrow \dfrac{8{{a}^{2}}-6a-9}{6{{a}^{2}}-5a-6}\div \dfrac{4{{a}^{2}}+11a+6}{9{{a}^{2}}+12a+4}=\dfrac{\left( 4a+3 \right)\left( 2a-3 \right)}{\left( 3a+2 \right)\left( 2a-3 \right)}\div \dfrac{\left( 4a+3 \right)\left( a+2 \right)}{{{\left( 3a+2 \right)}^{2}}}$
Now we will cancel all the common terms in the numerator and denominator after doing that we will have $\Rightarrow \dfrac{\left( 4a+3 \right)\left( 2a-3 \right)}{\left( 3a+2 \right)\left( 2a-3 \right)}\div \dfrac{\left( 4a+3 \right)\left( a+2 \right)}{{{\left( 3a+2 \right)}^{2}}}=\dfrac{4a+3}{3a+2}\div \dfrac{\left( 4a+3 \right)\left( a+2 \right)}{{{\left( 3a+2 \right)}^{2}}}$
Now we will shift the expressions from denominator to numerator then we will have $\Rightarrow \dfrac{\left( 4a+3 \right)\left( 2a-3 \right)}{\left( 3a+2 \right)\left( 2a-3 \right)}\div \dfrac{\left( 4a+3 \right)\left( a+2 \right)}{{{\left( 3a+2 \right)}^{2}}}=\dfrac{4a+3}{3a+2}\times \dfrac{{{\left( 3a+2 \right)}^{2}}}{\left( 4a+3 \right)\left( a+2 \right)}$
Now we will again cancel the common terms in the numerator and denominator then we will have $\Rightarrow \dfrac{4a+3}{3a+2}\times \dfrac{{{\left( 3a+2 \right)}^{2}}}{\left( 4a+3 \right)\left( a+2 \right)}=\dfrac{3a+2}{a+2}$
Now we will simplify this further and write it as
$\begin{align}
  & \Rightarrow \dfrac{3a+2}{a+2}=\dfrac{2a+a+4-2}{a+2} \\
 & \Rightarrow \dfrac{2\left( a+2 \right)+a-2}{a+2}=2+\dfrac{a-2}{a+2} \\
\end{align}$
Therefore we can conclude that the simplified form of the given expression is expressed as $\dfrac{8{{a}^{2}}-6a-9}{6{{a}^{2}}-5a-6}\div \dfrac{4{{a}^{2}}+11a+6}{9{{a}^{2}}+12a+4}=\dfrac{3a+2}{a+2}=2+\dfrac{a-2}{a+2}$

Note: While answering questions of this type we should be sure with the steps we take. The quadratic expressions can also be factored using the formulae for finding the roots of any quadratic expression in the form of $a{{x}^{2}}+bx+c$ where roots are given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . If we assume ${{x}_{1}},{{x}_{2}}$ as roots then the factored form will be $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$ . For example we can factorize the expression $8{{a}^{2}}-6a-9$ as the roots are $\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 8 \right)\left( -9 \right)}}{2\left( 8 \right)}=\dfrac{6\pm \sqrt{36+288}}{16}=\dfrac{6\pm \sqrt{324}}{16}=\dfrac{6\pm \sqrt{18}}{16}=\dfrac{-3}{4},\dfrac{3}{2}$ the factored form is $\left( 4a+3 \right)\left( 2a-3 \right)$ . Similarly we can simplify all the other quadratic expressions.