It would be very useful if someone can give me an answer to this question with a proper explanation.
One of the factors of the equation $x^2 +9x + c$ is $(x+11)$, where $c$ is a constant. Which of the following is one possible value of $x$?
$-5$; $-2$; $2$; $5$; $11$
On
The question as posed is incomplete; it appears that it should be asking for a possible value of $x$ $\mathbf{making\;\, x^2+9x+c=0}$. Integrator has given one approach. You can also use polynomial long division to divide $x^2+9x+c$ by $x+11$. If you adopt this approach, you’ll find that you get a quotient of $x-2$ and a remainder of $c+22$:
$$\frac{x^2+9x+c}{x+11}=x-2+\frac{c+22}{x+11}\;.$$
Now $x+11$ can be a factor of $x^2+9x+c$ if and only if the remainder is $0$, i.e., $c+22=0$, or $c=-22$. In that case the quotient is $x-2$, and $x^2+9x+c=(x+11)(x-2)$. At this point it’s clear that the two values of $x$ that make $x^2+9x+c=0$ are $-11$, which is not in the list of answers, and ... ?
Hint:
$$(x-a)(x-b)=x^2+-(a+b)x+ab$$
$$-(a+b)=9$$
Now put $a=-11$ and I hope you can take it from there!