31) \( 7 n^{3}+21 n^{2}-5 n-15 \)

Let's get ready to tackle that expression and have some fun while doing it! First off, this polynomial \( 7n^{3}+21n^{2}-5n-15 \) can be factored by grouping or using polynomial identities. A neat approach is to factor out the common terms from pairs. For example, if we pair \( 7n^{3} + 21n^{2} \) and \( -5n - 15 \), we can factor \( 7n^{2} \) from the first pair and \( -5 \) from the second. This leads us to rewrite the expression as \( 7n^{2}(n + 3) - 5(n + 3) \). Voila! Notice that \( (n + 3) \) appears in both terms! Therefore, the entire expression factors nicely into \( (n + 3)(7n^{2} - 5) \). Now, let’s talk practicality: factored forms are like treasure maps for solving equations! When \( 7n^{3}+21n^{2}-5n-15 = 0 \), you can set each factor to zero. First, if \( n + 3 = 0 \), you find \( n = -3 \). For the second factor, \( 7n^{2} - 5 = 0 \) leads you to \( n^2 = \frac{5}{7} \) and consequently \( n = \pm \sqrt{\frac{5}{7}} \). Solving polynomial equations can turn complex problems into simpler puzzles to solve, just like a fun crossword!