Question

Using the factorised trinomial (n-2)(4n-7)" align="absmiddle" class="latex-formula">, prove that there are only two values of n for which $4 n^{2} - 15n + 14$ is a prime number.

Answer 1

4n² - 15n + 14 is always the product of two numbers, for it to be prime number, one of these factors must be either 1 or -1.

Case n - 2 = 1
That would be n = 3
Then 4n² - 15n + 14 = 5 , which is prime.

Case n - 2 = -1
That would be n = 1
Then 4n² - 15n + 14 = 3, which is also prime.

Case 4n - 7 = 1
That would be n = 2 and that makes other factor (n-2) zero so it's not prime

Case 4n-7 = -1
That would be n = 3/2 which is not integer, so 4n² - 15n + 14 will not be interger.

For any other n values, 4n² - 15n + 14 will be composite number since it is product of two factors.

Therefore we are left with n = 1 and n = 3; only two values of n.