1. **State the problem:** Prove that 6 is a factor of $n(n^2 + 5)$ for all integers $n \geq 1$ using mathematical induction. 2. **Base case:** For $n=1$, evaluate $1(1^2 + 5) = 1 \times 6 = 6$. Since 6 is divisible by 6, the base case holds. 3. **Inductive hypothesis:** Assume for some $k \geq 1$, 6 divides $k(k^2 + 5)$, i.e., $k(k^2 + 5) = 6m$ for some integer $m$. 4. **Inductive step:** Show that 6 divides $(k+1)((k+1)^2 + 5)$. Calculate: $$(k+1)((k+1)^2 + 5) = (k+1)(k^2 + 2k + 1 + 5) = (k+1)(k^2 + 2k + 6).$$ Expand: $$= (k+1)(k^2 + 2k + 6) = (k+1)(k^2 + 2k) + 6(k+1) = (k+1)k^2 + 2(k+1)k + 6(k+1).$$ Rewrite terms: $$(k+1)k^2 + 2(k+1)k = k^2(k+1) + 2k(k+1) = k(k+1)(k + 2).$$ So, $$(k+1)((k+1)^2 + 5) = k(k+1)(k+2) + 6(k+1).$$ 5. **Analyze divisibility:** Note that among three consecutive integers $k$, $k+1$, and $k+2$, one is divisible by 3 and at least one is even, so their product $k(k+1)(k+2)$ is divisible by 6. Since $k(k+1)(k+2)$ is divisible by 6 and $6(k+1)$ is clearly divisible by 6, their sum is divisible by 6. 6. **Conclusion:** By the principle of mathematical induction, 6 divides $n(n^2 + 5)$ for all integers $n \geq 1$. **Final answer:** 6 is a factor of $n(n^2 + 5)$ for all $n \geq 1$.