2007-03-04 02:33:09 · 3 answers · asked by Crystal 3 in Science & Mathematics ➔ Mathematics
That factors into |(n-2)(n-5)|. So the only way it can be prime if one of them has an absolute value of 1.
Case I: n-2=1
So n=3 and |(n-2)(n-5)|=|(1)(-2)|=2, a prime.
Case II: n-2=-1
So n=1 and |(n-2)(n-5)|=|(-1)(-4)|=4, which is composite.
Case III: n-5=1
So n=6 and |(n-2)(n-5)|=|(4)(1)|=4, which is composite.
Case IV: n-5=-1
So n=4 and |(n-2)(n-5)|=|(2)(-1)|=2, a prime.
So the only answers are 3 and 4.
2007-03-04 02:42:08 · answer #1 · answered by Adam 2 · 2⤊ 0⤋
This is same as |(n-5)(n-2)|
This is satisfied by n = 3, 4 only
2007-03-04 10:41:42 · answer #2 · answered by FedUp 3 · 1⤊ 0⤋
Uh, well, there should be none.
|n^2 - 7n + 10| = |(n - 2)(n - 5)|
Thus, whatever n you think up of, it will always be factorable.
2007-03-04 10:38:10 · answer #3 · answered by Moja1981 5 · 1⤊ 2⤋