1. The problem is to find the value of $A$ in the vertical multiplication of the polynomials $(4a^3 - 2a + 3a^2 + 1)$ and $(3 - 2a + a^2)$. 2. The vertical multiplication method involves multiplying each term of the first polynomial by each term of the second polynomial and then adding the results. 3. The given partial products are: - $12a^3 + 9a^2 - 6a + 3$ - $-8a^4 - 6a^3 + A - 2a$ - $4a^5 + 3a^4 - 2a^3 + a^2$ 4. Let's multiply the polynomials step-by-step: Multiply $4a^3$ by each term in $(3 - 2a + a^2)$: $$4a^3 \times 3 = 12a^3$$ $$4a^3 \times (-2a) = -8a^4$$ $$4a^3 \times a^2 = 4a^5$$ Multiply $3a^2$ by each term: $$3a^2 \times 3 = 9a^2$$ $$3a^2 \times (-2a) = -6a^3$$ $$3a^2 \times a^2 = 3a^4$$ Multiply $-2a$ by each term: $$-2a \times 3 = -6a$$ $$-2a \times (-2a) = 4a^2$$ $$-2a \times a^2 = -2a^3$$ Multiply $1$ by each term: $$1 \times 3 = 3$$ $$1 \times (-2a) = -2a$$ $$1 \times a^2 = a^2$$ 5. Now, group the terms by powers of $a$: - $a^5$: $4a^5$ - $a^4$: $-8a^4 + 3a^4 = -5a^4$ - $a^3$: $12a^3 - 6a^3 - 2a^3 = 4a^3$ - $a^2$: $9a^2 + 4a^2 + a^2 = 14a^2$ - $a$: $-6a - 2a = -8a$ - Constant: $3$ 6. The term $A$ appears in the $a^3$ terms of the second partial product. From the multiplication, the $a^3$ terms are $-6a^3$ (from $3a^2 \times (-2a)$) and $-2a^3$ (from $-2a \times a^2$), summing to $-8a^3$. The given partial product has $-6a^3 + A - 2a$, so $A$ must be the missing $-2a^3$ term to complete the $a^3$ terms. 7. However, the options for $A$ are in terms of $a$ or $a^2$, so let's check carefully. The $A$ term is in the second line, which corresponds to multiplying $-2a$ by the second polynomial: $-2a \times 3 = -6a$ $-2a \times (-2a) = 4a^2$ $-2a \times a^2 = -2a^3$ The $-2a^3$ term is the one missing and corresponds to $A$. 8. Therefore, $A = -2a^3$. But since the options do not include $a^3$ terms, we must check the problem's given options again. 9. The options are: - $4a$ - $-4a$ - $4a^2$ - $-4a^2$ 10. From the partial product line: $-8a^4 - 6a^3 + A - 2a$ We see that $A$ replaces the term $4a^2$ (from $-2a \times (-2a)$), which is $4a^2$. 11. Hence, $A = 4a^2$. Final answer: $\boxed{4a^2}$