1. **State the problem:** Find the value of $k$ given that in the expansion of $(1 + kx)^{15}$, the coefficient of $x^3$ is $-29120$. 2. **Recall the binomial expansion formula:** $$(a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r$$ 3. **Identify the term containing $x^3$:** The general term is $$\binom{15}{r} (1)^{15-r} (kx)^r = \binom{15}{r} k^r x^r$$ We want the term where $r=3$: $$\binom{15}{3} k^3 x^3$$ 4. **Calculate the binomial coefficient:** $$\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 455$$ 5. **Set up the equation for the coefficient of $x^3$:** $$455 k^3 = -29120$$ 6. **Solve for $k^3$:** $$k^3 = \frac{-29120}{455} = -64$$ 7. **Find $k$ by taking the cube root:** $$k = \sqrt[3]{-64} = -4$$ **Final answer:** $$k = -4$$