How do you simplify \[\dfrac{{8x}}{{{x^2} - 5x - 36}} + \dfrac{{2x}}{{x + 4}}\]

Hint: In order to solve this question we will first make the factor of the quadratic equation and take the common from the whole term then we will take the L.C.M of the remaining part then further we will solve the fractional equation and solve it then if something common present we will cancel it and get the final answer.

Complete step-by-step solution:
For solving this equation we will first make the factors of the quadratic equation:
\[\Rightarrow \dfrac{{8x}}{{{x^2} - 5x - 36}} + \dfrac{{2x}}{{x + 4}}\]
So the quadratic equation will be:
${x^2} - 5x - 36$
For solving this quadratic equation we will apply the dharacharya formula in order to solve this:
According to dharacharya formula;
\[\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Now putting the value of a=1, b=-5 and c=-36 from above equation:
$\Rightarrow x = \dfrac{{ - \left( { - 5} \right) \pm \sqrt {{{\left( { - 5} \right)}^2} - 4\left( 1 \right)\left( { - 36} \right)} }}{{2\left( 1 \right)}}$
On further solving this we will get:
$\Rightarrow x = \dfrac{{5 \pm \sqrt {169} }}{2}$
Now on finalizing this equation we will get two values of x that will be:
$\Rightarrow x = \dfrac{{18}}{2},\dfrac{{ - 8}}{2}$
So the values of x will be 9 and -4.
So the factors of this quadratic equation will be $\left( {x - 9} \right)$ and $\left( {x + 4} \right)$
Now coming to our main question we will put all these factors in that equation:
\[\Rightarrow \dfrac{{8x}}{{\left( {x + 4} \right)\left( {x - 9} \right)}} + \dfrac{{2x}}{{x + 4}}\]
Now as we can see that the common coming from whole equation is $\dfrac{1}{{x + 4}}$ so taking the common from it will be:
$\Rightarrow \dfrac{1}{{\left( {x + 4} \right)}}\left\{ {\dfrac{{8x}}{{\left( {x - 9} \right)}} - \dfrac{{2x}}{1}} \right\}$
Now taking the L.C.M form this bracket we will get:
$\Rightarrow \dfrac{1}{{\left( {x + 4} \right)}}\left\{ {\dfrac{{8x - 2x\left( {x - 9} \right)}}{{\left( {x - 9} \right)}}} \right\}$
Now again taking common 2x from this we will get:
$\Rightarrow \dfrac{{2x}}{{\left( {x + 4} \right)}}\left\{ {\dfrac{{4 - x + 9}}{{x - 9}}} \right\}$
Now on further solving this we will get:
$\Rightarrow \dfrac{{2x}}{{\left( {x + 4} \right)}}\left\{ {\dfrac{{5 - x}}{{x - 9}}} \right\}$
So after this further simplification we will get:
$\dfrac{{\left( {2x} \right)}}{{\left( {x + 4} \right)}}\dfrac{{\left( {5 - x} \right)}}{{\left( {x - 9} \right)}}$

Thus the final answer is $\dfrac{{\left( {2x} \right)}}{{\left( {x + 4} \right)}}\dfrac{{\left( {5 - x} \right)}}{{\left( {x - 9} \right)}}$

Note: While solving these types of question we should keep in mind that for factorization we can also use the factorization method of this equation and the precaution which should be taken is that the factor made are correct and if there are some factors to be cancelled they it must be done.