Question

What is the sum of all integers n such that n2+2n+2 divides n3+4n2+4n−14 ?

Answer 1

Using the long division method for polynomials, we get,

$n^3+4n^2+4n-14 = (n^2+2n+2) * (n+ 2) +(-2n-18)$

For $n^2+2n+2$ to divide $n^3+4n^2+4n-14$, it must divide the remainder as well;

$(n^2+2n+2) | (-2n-18)$

To render the remainder divisible, one of the following conditions must be fulfilled;

$|-2n-18|$ ≥ $|n^2+2n+2|$

OR

$(-2n-18) = 0$

In the first scenario, this inequality is valid only if -4 ≤ n ≤ n

On checking the possible values of n within this range, we find that,

n = -4, -2, -1, 0, 1, 4

In the second scenario, we find an additional value; n = -9

∴ the sum of all values = (-9) + (-4) + (-2) + (-1) + 0 + 1 + 4 = (-11)

Ans) The sum of all integers = (-11)

Answer 2

Answer:

Step-by-step explanation: