1. **Problem statement:** Find the coefficient of $x^{3}$ in the product $(2 - 3x)^{5} (1 + x)^{4}$ using the binomial theorem. 2. **Recall the binomial theorem:** For any integer $n \geq 0$, $$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ where $\binom{n}{k}$ is the binomial coefficient. 3. **Expand each term using the binomial theorem:** - For $(2 - 3x)^5$, let $a=2$ and $b=-3x$: $$ (2 - 3x)^5 = \sum_{i=0}^5 \binom{5}{i} 2^{5-i} (-3x)^i = \sum_{i=0}^5 \binom{5}{i} 2^{5-i} (-3)^i x^i $$ - For $(1 + x)^4$, let $a=1$ and $b=x$: $$ (1 + x)^4 = \sum_{j=0}^4 \binom{4}{j} 1^{4-j} x^j = \sum_{j=0}^4 \binom{4}{j} x^j $$ 4. **The product is:** $$ (2 - 3x)^5 (1 + x)^4 = \left( \sum_{i=0}^5 \binom{5}{i} 2^{5-i} (-3)^i x^i \right) \left( \sum_{j=0}^4 \binom{4}{j} x^j \right) $$ 5. **We want the coefficient of $x^3$ in the product.** This comes from terms where the powers of $x$ add to 3, i.e., $i + j = 3$. 6. **Sum over all pairs $(i,j)$ with $i+j=3$: $i=0,1,2,3$ and $j=3,2,1,0$ respectively.** 7. **Calculate each term's coefficient:** - For $i=0, j=3$: $$ \binom{5}{0} 2^{5-0} (-3)^0 \times \binom{4}{3} = 1 \times 2^5 \times 1 \times 4 = 32 \times 4 = 128 $$ - For $i=1, j=2$: $$ \binom{5}{1} 2^{4} (-3)^1 \times \binom{4}{2} = 5 \times 16 \times (-3) \times 6 = 5 \times 16 \times (-3) \times 6 $$ Calculate stepwise: $$ 5 \times 16 = 80 $$ $$ 80 \times (-3) = -240 $$ $$ -240 \times 6 = -1440 $$ - For $i=2, j=1$: $$ \binom{5}{2} 2^{3} (-3)^2 \times \binom{4}{1} = 10 \times 8 \times 9 \times 4 $$ Calculate stepwise: $$ 10 \times 8 = 80 $$ $$ 80 \times 9 = 720 $$ $$ 720 \times 4 = 2880 $$ - For $i=3, j=0$: $$ \binom{5}{3} 2^{2} (-3)^3 \times \binom{4}{0} = 10 \times 4 \times (-27) \times 1 $$ Calculate stepwise: $$ 10 \times 4 = 40 $$ $$ 40 \times (-27) = -1080 $$ 8. **Sum all contributions:** $$ 128 + (-1440) + 2880 + (-1080) = 128 - 1440 + 2880 - 1080 $$ Calculate stepwise: $$ 128 - 1440 = -1312 $$ $$ -1312 + 2880 = 1568 $$ $$ 1568 - 1080 = 488 $$ 9. **Final answer:** The coefficient of $x^3$ in the product is **488**.