1. **Stating the problem:** We have an area model with tiles arranged in 3 rows and 5 columns representing algebraic terms. The first two rows have 3 blue squares labeled $x^2$ and 2 green rectangles labeled $x$. The last row has 3 green rectangles labeled $x$ and 2 yellow squares labeled 1. We want to express the total area algebraically and verify the factorizations given: $(2x + 2)(x + 1)$, $(3x + 2)(2x + 1)$, $(2x + 2)(2x + 1)$, and $(3x + 2)(x + 1)$. 2. **Understanding the model:** - Each $x^2$ tile represents an area of $x^2$. - Each $x$ tile represents an area of $x$. - Each 1 tile represents an area of 1. 3. **Calculate total area by summing tiles:** - Number of $x^2$ tiles: $3$ columns $\times$ $2$ rows $= 6$ tiles, so total $6x^2$. - Number of $x$ tiles: $2$ columns $\times$ $2$ rows $= 4$ tiles in first two rows plus $3$ tiles in last row, total $4 + 3 = 7x$. - Number of 1 tiles: $2$ tiles in last row, total $2$. So total area is: $$6x^2 + 7x + 2$$ 4. **Check factorizations:** We want to factor $6x^2 + 7x + 2$. Try $(2x + 1)(3x + 2)$: $$ (2x + 1)(3x + 2) = 2x \times 3x + 2x \times 2 + 1 \times 3x + 1 \times 2 = 6x^2 + 4x + 3x + 2 = 6x^2 + 7x + 2 $$ This matches the total area. 5. **Verify other given factorizations:** - $(2x + 2)(x + 1) = 2(x + 1)(x + 1) = 2(x + 1)^2 = 2(x^2 + 2x + 1) = 2x^2 + 4x + 2$ which is not equal to $6x^2 + 7x + 2$. - $(3x + 2)(2x + 1)$ is the same as $(2x + 1)(3x + 2)$ above, so it matches. - $(2x + 2)(2x + 1) = 2(x + 1)(2x + 1) = 2(2x^2 + x + 2x + 1) = 2(2x^2 + 3x + 1) = 4x^2 + 6x + 2$ which is not equal to $6x^2 + 7x + 2$. - $(3x + 2)(x + 1) = 3x^2 + 3x + 2x + 2 = 3x^2 + 5x + 2$ which is not equal to $6x^2 + 7x + 2$. 6. **Conclusion:** The correct factorization of the total area $6x^2 + 7x + 2$ is: $$ (2x + 1)(3x + 2) $$ **Final answer:** $(2x + 1)(3x + 2)$