AP DSC School Assistant Math Model Test 3

Concept:The problem uses the binomial theorem to expand $(1 + x^{2} - x^{3})^{8}$ and find the coefficient of $x^{10}$. By factoring $x^{2}(1-x)$ from the last two terms, we simplify the expansion and identify only those terms that produce $x^{10}$.Explanation:We rewrite the expression as $[1 + x^{2}(1-x)]^{8}$. Using the binomial theorem:$$[1 + x^{2}(1-x)]^{8} = \sum_{r=0}^{8} \binom{8}{r} \left(x^{2}(1-x)\right)^{r} = \sum_{r=0}^{8} \binom{8}{r} x^{2r} (1-x)^{r}.$$The power of $x$ in the term $x^{2r}(1-x)^{r}$ is $2r$ plus the power contributed by $(1-x)^{r}$. We need total exponent $10$. Check possible $r$ values:For $r=4$: $x^{8}(1-x)^{4}$ contributes a term $x^{8} \cdot \binom{4}{2}(-x)^{2} = x^{8} \cdot 6x^{2} = 6x^{10}$ (coefficient $6$ from $x^{2}$ term). For $r=5$: $x^{10}(1-x)^{5}$ gives $x^{10} \cdot 1$ from the constant term, coefficient $1$. For $r=3$: $x^{6}(1-x)^{3}$ maximum power is $x^{6+3}=x^{9}$, too low. For $r=6$: $x^{12}(1-x)^{6}$ minimum power is $x^{12}$, too high. So only $r=4$ and $r=5$ matter.Thus the coefficient of $x^{10}$ is:$$\binom{8}{4} \cdot [\text{coefficient of }x^{2}\text{ in }(1-x)^{4}] + \binom{8}{5} \cdot 1.$$Now, $(1-x)^{4} = 1 - 4x + 6x^{2} - 4x^{3} + x^{4}$, so the coefficient of $x^{2}$ is $6$.Compute:$\binom{8}{4} = \frac{8!}{4!4!} = 70$, and $\binom{8}{5} = \frac{8!}{3!5!} = 56$.Therefore, coefficient = $70 \times 6 + 56 = 420 + 56 = 476$.Answer:The coefficient of $x^{10}$ is $476$, which corresponds to option A.