Question

What is the sum of all positive integers x such that $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$ is an integer.
(a) 2
(b) 4
(c) 6
(d) 8

Answer 1

We start solving the problem by finding the values of x at which the given function is not defined. We then simplify the given function $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$ as $x+\dfrac{15}{{{x}^{2}}-1}$ by making necessary calculations. We then find the positive integer values of x such that $\dfrac{15}{{{x}^{2}}-1}$ will become an integer. We then add the obtained positive integer values of x to get the required result.

Complete step-by-step solution
According to the problem, we need to find the sum of all positive integers x such that $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$ is an integer.
We can see that the function $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$ is not defined at $x=-1$ and $x=1$.
So, the values of x cannot be equal to 1 as x should be positive integers.
Now, let us simplify the given function $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$.
So, we have $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}=\dfrac{x\left( {{x}^{2}}-1 \right)+15}{{{x}^{2}}-1}$.
$\Rightarrow \dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}=x+\dfrac{15}{{{x}^{2}}-1}$.
We need to find the values of x (positive integer values) in which $\dfrac{15}{{{x}^{2}}-1}$ is an integer.
We know that the factors of 15 are 1, 3, 5, and 15.
So, let us find the values of x for which ${{x}^{2}}-1=1,3,5,15$.
Now, we have ${{x}^{2}}-1=1$.
$\Rightarrow {{x}^{2}}=2$.
$\Rightarrow x=\sqrt{2}$, which is not an integer.
Now, we have ${{x}^{2}}-1=3$.
$\Rightarrow {{x}^{2}}=4$.
$\Rightarrow x=2$, which is an integer.
Now, we have ${{x}^{2}}-1=5$.
$\Rightarrow {{x}^{2}}=6$.
$\Rightarrow x=\sqrt{6}$, which is not an integer.
Now, we have ${{x}^{2}}-1=15$.
$\Rightarrow {{x}^{2}}=16$.
$\Rightarrow x=4$, which is an integer.
So, we have found that the positive integer values of x are 2, 4 in order to get $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$ as an integer.
Let us find the sum of the obtained positive integer values of x.
So, the sum is $2+4=6$.
$\therefore$The correct option for the given problem is (c).

Note: Whenever we get this type of problem, we should first resolve it to a sum of smaller fractions to make calculations easier. We should only consider positive integers of x while solving this problem as many students will neglect this condition. We can also find the product of the obtained values of x. Similarly, we can expect problems to find the domain and range of the given function $\dfrac{{{x}^{3}}-x+15}{\left( x+1 \right)\left( x-1 \right)}$.