Question

Hey i just need some help with this question thanks :)

Answer 1

There are no positive integer solutions for n.

Explanation:

A prime number is a positive integer greater than 1 that has no positive divisors other than 1 and itself. Therefore, the factors of a prime number are limited to 1 and the number itself.

To find all positive integers n such that 2n² + 7n + 5 is a prime number, first factor the quadratic.

`2n^2+7n+5`

`= 2n^2+2n+5n+5`

`= 2n(n+1)+5(n+1)`

`= (2n+5)(n+1)`

As the factors of a prime number are limited to 1 and the number itself, set each factor to 1 and solve for n:

`\begin{aligned}2n+5&=1\\2n&=-4\\n&=-2\end{aligned}`
`\begin{aligned}n+1&=1\\n&=1-1\\n&=0\end{aligned}`

Since -2 and 0 are not positive integers, it is therefore not possible for 2n² + 7n + 5 to be a prime number for any positive integer value of n.

Hence, there are no positive integer solutions for n in this case.

Answer 2

Let's factor by grouping.

2n^2 + 7n + 5

2n^2 + 2n + 5n + 5

(2n^2 + 2n) + (5n + 5)

2n(n + 1) + 5(n + 1)

(2n+5)(n+1)

If 2n^2+7n+5 is prime, then one of the factors 2n+5 or n+1 must be 1. This is because a prime number can only be factored into 1 times itself. Example: 7 = 7*1.

Let's set each factor equal to 1 and see what happens.

• 2n+5 = 1 leads to n = -2. But it was stated that n > 0. So we rule out this case.
• n+1 = 1 leads to n = 0. This is ruled out as well since n > 0.

It is impossible for (2n+5)(n+1) to be prime since none of the factors is 1. Therefore, (2n+5)(n+1) = 2n^2+7n+5 is composite for positive integers n = 1, 2, 3, ...

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Another way to think about it:

2n^2 + 7n + 5 = (2n+5)(n+1)

Since n > 0 both 2n+5 and n+1 are larger than 1.

This means we have two different positive factors. These factors may be prime or composite. They multiply to some composite value.

Example: If n = 4 then

(2n+5)(n+1) = (2*4+5)(4+1) = 13*5 = 65