Question 1177390
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            The solution presented by  @MathLover1 is  INCORRECT  and may confuse you.


            I came to bring a correct solution.



<pre>
You consider this rational function

    f(x) = {{{(2x^3+9x^2-4x-18)/(2x+9)}}}.


Notice that the domain where the function is defined, is the set of all real numbers excluding  x = {{{-9/2}}}.


Next, factor the numerator  

    {{{2x^3+9x^2-4x-18}}} = {{{(2x+9)*(x^2-2)}}}.



Now, considering the given rational function f(x), you can cancel the factor (2x+9) everywhere except of x= {{{-9/2}}}.


It means that the given rational function f(x) is equal to  {{{x^2-2}}} everywhere, except x= {{{-9/2}}}.


Now, the zeroes of the function f(x) are the zeroes of this quadratic binomial   {{{x^2 - 2}}} =0,

i.e.  x= +/- {{{sqrt(2)}}}.


<U>ANSWER</U>.  The given function f(x) has two zeroes in its domain. They are the values  x= {{{sqrt(2)}}}  and  x= - {{{sqrt(2)}}}.
</pre>

Solved, answered and explained.  &nbsp;&nbsp;&nbsp;&nbsp;// &nbsp;&nbsp;&nbsp;&nbsp;And completed.



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For your better understanding, &nbsp;the function &nbsp;f(x) &nbsp;has a &nbsp;"hole" &nbsp;at the point &nbsp;&nbsp;x= - {{{9/2}}}.


This rational function &nbsp;IS &nbsp;NOT &nbsp;DEFINED &nbsp;at this point, &nbsp;although it has the limits at this point from the left side 

and from  the right side, &nbsp;and these limits are equal.


NETHERTHELESS, &nbsp;the function &nbsp;f(x) &nbsp;IS &nbsp;NOT &nbsp;DEFINED &nbsp;at x= - {{{9/2}}}:  &nbsp;this point is the &nbsp;"hole" &nbsp;point for the function.


You should clearly understand it, &nbsp;and the entire problem is designed and intended to teach you to it &nbsp;(!)



Happy learning (!)