If n is a positive integer, what is the value of n?

Yes, if x + 4 were a factor of some quadratic, then plugging x = -4 into that quadratic will give you a result of zero. So for example, x + 4 is a factor of this quadratic:

x^2 + 7x + 12 = (x + 4)(x + 3)

and if you plug x = -4 into this quadratic, the result will be zero.

But notice also: if x is a positive integer, then x^2 + 7x + 12 will always be divisible by x+4, because you can write the number x^2 + 7x + 12 as a product of two smaller integers, using the factorization above. In the question Bunuel posted above, n+4 is definitely not a factor of the quadratic n^2 + 6n + 16, because n^2 + 6n + 16 is not divisible by n+4 for every single value of n. It's only divisible by n+4 for certain values of n.

If that's confusing at all, it might help to think of much simpler situations: Say k is a positive integer:

* Algebraically speaking, k is a factor of 2k. So 2k is always divisible by k, no matter what k equals
* Algebraically speaking, k is not a factor of k + 2. But k could still sometimes be a divisor of k+2 for certain values of k. It turns out here that it is, only when k = 1 and k = 2.

Bunuel chetan2u

if n+4 is a factor of any quadartic.... which means when i substitute n = -4 in the quadartic,, the euqation shud give me 0

in this case,,it does not... plz explain

hi ,,,i just want to clarify,,, in a quadratic equation,,
if (n+4) is a factor then n= -4 shud give zero or not?

Statement A:
(n^2 + 6n + 16) is perfectly divisible by (n + 4)
(n^2 + 6n + 16) = (n^2 + 8n + 16) - 2n
= (n+4)^2 - 2n
Therefore 2n s a multiple of (n+4)

2n = n + 4; => n = 4
2n = 2(n + 4); => 0 = 4 Cant be true
2n = 3(n + 4); => n = -12 n is positive
The only way this is true is when n = 4
A is Suff


B clearly does not suffice

Ans: A

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