Factor completely. \( 3 j^{2}+8 j+5 \)

To factor the quadratic expression \( 3j^{2}+8j+5 \), we look for two numbers that multiply to \( 3 \cdot 5 = 15 \) (the product of the leading coefficient and the constant term) and add up to \( 8 \) (the coefficient of the middle term). The numbers \( 3 \) and \( 5 \) work perfectly! We can rewrite the quadratic as \( 3j^{2}+3j+5j+5 \) and then factor by grouping: \( 3j(j+1) + 5(j+1) \). Now, we can factor out the common binomial factor: \( (3j+5)(j+1) \). So, the complete factorization is \( (3j+5)(j+1) \). Let me know if you need further assistance!