Factor completely. \[ 3 n^{2}+8 n+5 \]

To factor the quadratic expression \( 3n^2 + 8n + 5 \) completely, we want to find two binomials that multiply to give the original quadratic. We can look for two numbers that multiply to \( 3 \times 5 = 15 \) (the product of the leading coefficient and the constant term) and add to \( 8 \) (the coefficient of the middle term). These two numbers are \( 3 \) and \( 5 \). We can rewrite the middle term using these numbers: \[ 3n^2 + 3n + 5n + 5 \] Next, we can group the terms: \[ (3n^2 + 3n) + (5n + 5) \] Now, factor out the common factors from each group: \[ 3n(n + 1) + 5(n + 1) \] Both groups contain the binomial \( (n + 1) \): \[ (3n + 5)(n + 1) \] Thus, the completely factored form of \( 3n^2 + 8n + 5 \) is: \[ (3n + 5)(n + 1) \]