1. **State the problem:** We have a large rectangle divided into six smaller rectangles arranged in two columns and three rows. The height is divided into three parts: $c^3$, $6c^2$, and $2c$. The width is divided into two parts: $c^2$ and $2$. 2. **Find the area of each smaller rectangle:** Each smaller rectangle's area is the product of its height and width. - Top-left rectangle area: $c^3 \times c^2 = c^{3+2} = c^5$ - Top-right rectangle area: $c^3 \times 2 = 2c^3$ - Middle-left rectangle area: $6c^2 \times c^2 = 6c^{2+2} = 6c^4$ - Middle-right rectangle area: $6c^2 \times 2 = 12c^2$ - Bottom-left rectangle area: $2c \times c^2 = 2c^{1+2} = 2c^3$ - Bottom-right rectangle area: $2c \times 2 = 4c$ 3. **Express the total area as a sum of these areas:** $$\text{Total area} = c^5 + 2c^3 + 6c^4 + 12c^2 + 2c^3 + 4c$$ 4. **Combine like terms:** $$c^5 + 6c^4 + (2c^3 + 2c^3) + 12c^2 + 4c = c^5 + 6c^4 + 4c^3 + 12c^2 + 4c$$ 5. **Express the total area as a product of polynomials:** The total height is $c^3 + 6c^2 + 2c$ and the total width is $c^2 + 2$. So the total area is: $$ (c^3 + 6c^2 + 2c)(c^2 + 2) $$ 6. **Verify by expanding:** $$\begin{aligned} (c^3 + 6c^2 + 2c)(c^2 + 2) &= c^3 \times c^2 + c^3 \times 2 + 6c^2 \times c^2 + 6c^2 \times 2 + 2c \times c^2 + 2c \times 2 \\ &= c^5 + 2c^3 + 6c^4 + 12c^2 + 2c^3 + 4c \\ &= c^5 + 6c^4 + 4c^3 + 12c^2 + 4c \end{aligned}$$ **Final answer:** - Product form: $$(c^3 + 6c^2 + 2c)(c^2 + 2)$$ - Simplified standard form: $$c^5 + 6c^4 + 4c^3 + 12c^2 + 4c$$