For how many positive integers ‘n’, $$n^3 - 8n^2 + 20n - 13$$ is a prime?
$$n^3- 8n^2+20n -13$$
= $$(n-1)\times (n^2 - 7n + 13 )$$
In order to make (n-1) prime it has to be +/-1 or $$(n^2 - 7n + 13 )$$ has to be +/-1
n-1 = +/-1
= n = 0 or 2
$$n^3- 8n^2+20n -13$$ = -13 or 3
$$(n^2 - 7n + 13 ) = +/-1$$
= n is 3 or 4.
For this we will get $$n^3- 8n^2+20n -13$$ = 2 or 3
We know that prime numbers are positive so the prime numbers will be 2,3 and 4
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