Question

he sum of all possible values of n, where n is a non-zero integer when n2 + 8n + 6 is divided by n gives an integer, is

Answer 1

Answer exactly i think it is not perfect answer but i tried

answer is 11

Step-by-step explanation:

990 can be written as the product: 9×10×11

So, So, the factor with highest value here is 11.

So, here n can take the least value of 11

Answer 2

Answer:

The sum of all possible values of n, when $n^{2} + 8n + 6$ is divided by n is an integer, is zero.

Step-by-step explanation:

Given the expression, $n^{2} + 8n + 6$

Dividing the whole expression with $n$,

$\frac{n^{2} + 8n + 6}{n} =\frac{n^{2}}{n}+\frac{8n}{n}+\frac{ 6}{n}$

$=n+8+\frac{ 6}{n}$

For any value of $n$, $n+8$ will always be an integer.

For $\frac{ 6}{n}$ to be an integer, 6 should be divisible by n.

Then the possible divisor values of 6 are 1, 2, 3 and 6.

If $n > 0$ then the possible values of n are $n=1, 2, 3, 6$

If $n < 0$ then the possible values of n are $n=-1, -2, -3, -6$  

The total possible values of n are 8 where $n=\pm 1, \pm 2, \pm3, \pm 6$

Adding all the possible values gives the sum equals to zero.

Therefore, the sum of all possible values of n, when $n^{2} + 8n + 6$ is divided by n is an integer, is zero.