Factor completely. \( 3 j^{2}+7 j+4 \) \( \square \) Submit

To factor \( 3j^{2}+7j+4 \) completely, we first look for two numbers that multiply to \( 3 \times 4 = 12 \) and add to \( 7 \). The numbers \( 3 \) and \( 4 \) fit the bill. So we can rewrite the quadratic as: \[ 3j^{2}+3j+4j+4 \] Now, group the terms: \[ (3j^{2} + 3j) + (4j + 4) \] Factor out the common factors: \[ 3j(j + 1) + 4(j + 1) \] Now, factor out the \( (j + 1) \): \[ (3j + 4)(j + 1) \] So, the complete factorization of \( 3j^{2}+7j+4 \) is \( (3j + 4)(j + 1) \). Happy factoring!