Answer: $3x^2 + 14x + 8$

Set up the multiplication: Area = $(3x + 2)(x + 4)$. Distribute each term: $3x \cdot x = 3x^2$, $3x \cdot 4 = 12x$, $2 \cdot x = 2x$, $2 \cdot 4 = 8$. Combine like terms: $3x^2 + 12x + 2x + 8 = 3x^2 + 14x + 8$. Why distractors fail: Option B ($3x^2 + 6x + 8$) omits the $12x$ term, only recording $2 \cdot x + 2 \cdot 4$. Option C ($3x^2 + 12x + 8$) omits the $2x$ cross term. Option D ($3x^2 + 14x + 6$) miscalculates the constant $2 \times 4$ as $6$.

Answer: $2n^2 + 5n - 12$

Translate descriptions to expressions: Width = $2n - 3$ ("3 less than twice a number"). Length = $n + 4$ ("4 more than the number"). Multiply for area: $(2n - 3)(n + 4) = 2n^2 + 8n - 3n - 12 = 2n^2 + 5n - 12$. Why distractors fail: Option A ($2n^2 - 5n - 12$) gives $-5n$, likely from $-8n + 3n$ with reversed signs. Option C ($2n^2 + 11n - 12$) adds $8n + 3n$ instead of subtracting. Option D ($2n^2 + 5n + 12$) makes the constant positive.

Answer: Valid, because $(x + 4)(x + 6) = x^2 + 10x + 24$

Verify by expansion: $(x + 4)(x + 6) = x^2 + 6x + 4x + 24 = x^2 + 10x + 24$. This matches the given area. Why the correct answer works: Since area = length × width, if the area is $x^2 + 10x + 24$ and one side is $(x + 4)$, dividing gives the other side as $(x + 6)$. Priya's reasoning is valid. Why distractors fail: Option A is wrong because the expression factors as $(x+4)(x+6)$. Option B gives $(x+8)$, but $(x+4)(x+8) = x^2 + 12x + 32 \neq x^2 + 10x + 24$. Option D adds an unnecessary restriction; the algebraic identity holds for all values of $x$.

Answer: $2x^2 - 7x - 4$

Translate words to algebra: Length = $2x + 1$ (twice a number, increased by 1). Width = $x - 4$ (the number decreased by 4). Compute the area: $(2x + 1)(x - 4) = 2x^2 - 8x + x - 4 = 2x^2 - 7x - 4$. Why distractors fail: Option A ($2x^2 - 8x + 4$) omits the $+x$ cross term and flips the constant sign. Option B ($2x^2 + 7x - 4$) gets the wrong sign on the middle term. Option D ($2x^2 - 9x + 4$) incorrectly combines $-8x + x$ as $-9x$ and flips the constant sign.

Answer: Distributing $(x+a)(x+b)$ yields $x^2 + bx + ax + ab$, and combining like terms gives $x^2 + (a+b)x + ab$ for all values of $a$ and $b$

Verify the general expansion: $(x+a)(x+b) = x^2 + bx + ax + ab = x^2 + (a+b)x + ab$. This uses the distributive property, which is valid for all real numbers $a$ and $b$. Why the correct answer works: Option B correctly identifies the algebraic justification: distribution and combining like terms, with no restriction on $a$ or $b$. Why distractors fail: Option A adds a false restriction to positive integers. Option C confuses sum and product — $a + b \neq ab$ in general. Option D restricts unnecessarily to squares.

Answer: $x^2 + 5x$

Translate the description: Width = $x$. Length = $x + 5$ ("5 more than the number"). Compute the area: Area = $x(x + 5) = x^2 + 5x$. Why distractors fail: Option A ($x^2 + 5$) fails to distribute $x$ across $+5$. Option C ($5x^2$) incorrectly multiplies the coefficients. Option D ($x + 5x = 6x$) adds the expressions instead of multiplying.

Answer: $11x + 18$

Compute the room area: Room area = $(x + 9)(x + 2) = x^2 + 2x + 9x + 18 = x^2 + 11x + 18$. Subtract the column area: Column area = $x^2$. Usable area = $(x^2 + 11x + 18) - x^2 = 11x + 18$. Why distractors fail: Option B adds $x^2$ to the room area. Option C is the room area before subtracting the column. Option D incorrectly makes the constant negative.

Answer: $x^2 + 3x + 24$

Find the total garden area: $(x + 8)(x + 5) = x^2 + 5x + 8x + 40 = x^2 + 13x + 40$. Subtract the path area: Remaining area = $(x^2 + 13x + 40) - (10x + 16) = x^2 + 3x + 24$. Why distractors fail: Option A ($x^2 + 13x + 56$) adds the path area instead of subtracting. Option C ($x^2 + 3x + 56$) subtracts the linear term but adds the constant. Option D ($x^2 + 23x + 56$) adds both terms of the path area.

Answer: $8x + 48$

Determine the inner dimensions: A uniform border of $2$ on each side reduces each dimension by $2 \times 2 = 4$. Inner length = $(x+10) - 4 = x + 6$. Inner width = $(x+6) - 4 = x + 2$. Compute outer and inner areas: Outer area = $(x+10)(x+6) = x^2 + 16x + 60$. Inner area = $(x+6)(x+2) = x^2 + 8x + 12$. Subtract to find border area: Border area = $(x^2 + 16x + 60) - (x^2 + 8x + 12) = 8x + 48$. Why distractors fail: Option A ($28x + 48$) likely results from an incorrect calculation of the inner dimensions. Option C ($8x + 32$) subtracts the wrong constant: the correct computation is $60 - 12 = 48$, not $32$. Option D ($x^2 + 8x + 48$) forgets that the $x^2$ terms cancel when subtracting.

Answer: Widths of $(x + 2)$ and $5$

Factor the target area: $x^2 + 12x + 35 = (x+5)(x+7)$. Interpret the shared side: Both rectangles share a side of $(x+5)$, so combined area = $(x+5)(w_1 + w_2)$. For this to equal $(x+5)(x+7)$, we need $w_1 + w_2 = x + 7$. Check each option: Option A: $3 + 4 = 7$, giving area $(x+5)(7) = 7x + 35$, not quadratic. Option B: $(x+2) + 5 = x + 7$, giving area $(x+5)(x+7) = x^2 + 12x + 35$. ✓ Option C: $6 + 1 = 7$, giving $7x + 35$ again. Option D: $(x+3) + 3 = x + 6$, giving $(x+5)(x+6) = x^2 + 11x + 30$. Why distractors fail: Options A and C use constant-only widths whose sum ($7$) has no $x$ term, producing linear expressions, not quadratic. Option D's widths sum to $x + 6$ instead of $x + 7$.

Answer: $x^2 + 6x + 8$

Set up the area formula: Area of a rectangle = length × width = $(x + 4)(x + 2)$. Expand using distribution: $(x + 4)(x + 2) = x^2 + 2x + 4x + 8 = x^2 + 6x + 8$. Why distractors fail: Option B ($x^2 + 8x + 6$) swaps the middle-term coefficient and constant. Option C ($x^2 + 6x + 6$) miscalculates the constant as $4 + 2$ instead of $4 \times 2$. Option D ($2x + 6$) adds the expressions instead of multiplying them.

Answer: Added the constants ($6 + 3 = 9$) instead of multiplying them ($6 \times 3 = 18$) for the last term

Identify the correct expansion: $(x + 6)(x + 3) = x^2 + 3x + 6x + 18 = x^2 + 9x + 18$. Locate the student's error: The student wrote $+9$ instead of $+18$ as the constant. The middle term $9x$ is correct ($6 + 3 = 9$), but the student also used $6 + 3 = 9$ for the constant instead of $6 \times 3 = 18$. Why distractors fail: Option B is wrong because the student's answer includes $x^2$. Option C is wrong because the middle coefficient $9 = 6 + 3$ is actually correct. Option D is wrong because all signs in the student's answer are positive.

Answer: Multiply the two expressions

Recall the area formula: The area of a rectangle is length × width. When both dimensions are algebraic expressions, we multiply them. Why the correct answer works: Multiplying $(2x + 1)(x + 3)$ produces the quadratic area expression $2x^2 + 7x + 3$. Why distractors fail: Adding gives $3x + 4$, which relates to perimeter, not area. Subtracting and dividing have no standard geometric interpretation for computing a rectangle's area.

Answer: $x^2 + 8x + 7$

Apply the area formula: Area = $(x + 7)(x + 1)$. Expand the product: $(x + 7)(x + 1) = x^2 + x + 7x + 7 = x^2 + 8x + 7$. Why distractors fail: Option A gets the middle coefficient wrong as $7$ (only one cross product) instead of $7 + 1 = 8$. Option C swaps the middle coefficient and constant. Option D incorrectly computes the constant as $8$ instead of $7 \times 1 = 7$.

Answer: No; the remaining area should be $2x^2 + 18x + 13$

Expand the large rectangle: $(4x+3)(x+5) = 4x^2 + 20x + 3x + 15 = 4x^2 + 23x + 15$. Expand the small rectangle: $(2x+1)(x+2) = 2x^2 + 4x + x + 2 = 2x^2 + 5x + 2$. Subtract to find remaining area: $(4x^2 + 23x + 15) - (2x^2 + 5x + 2) = 2x^2 + 18x + 13$. Evaluate the colleague's answer: The correct remaining area is $2x^2 + 18x + 13$. The colleague wrote $2x^2 + 20x + 13$, which has $20x$ instead of $18x$ — likely a subtraction error on the linear terms ($23x - 5x = 18x$, not $20x$). Why distractors fail: Option A incorrectly validates the colleague's wrong answer. Option C ($2x^2 + 20x + 17$) has errors in both the middle and constant terms. Option D ($6x^2 + 28x + 17$) adds the areas instead of subtracting.

Answer: $2x^2 + 9x - 18$

Multiply the binomials: $(2x - 3)(x + 6) = 2x \cdot x + 2x \cdot 6 + (-3) \cdot x + (-3) \cdot 6 = 2x^2 + 12x - 3x - 18$. Combine like terms: $12x - 3x = 9x$, so the area is $2x^2 + 9x - 18$. Why distractors fail: Option B ($2x^2 + 15x - 18$) adds $12x + 3x$ instead of subtracting. Option C ($2x^2 + 9x + 18$) incorrectly makes the constant positive. Option D ($2x^2 - 9x - 18$) reverses the sign of the middle term.

Answer: $x^2 + 10x + 25$

Area of a square: Area = side² = $(x + 5)^2 = (x + 5)(x + 5)$. Expand the expression: $(x + 5)(x + 5) = x^2 + 5x + 5x + 25 = x^2 + 10x + 25$. Why distractors fail: Option A uses $10$ as the constant instead of $5^2 = 25$. Option B only counts one of the two $5x$ terms. Option D omits the middle term entirely, a common error when squaring binomials.

Answer: $(3x + 1)(2x + 5)$

Expand the correct option: $(3x+1)(2x+5) = 6x^2 + 15x + 2x + 5 = 6x^2 + 17x + 5$. This matches the target. Check the distractors: $(6x+1)(x+5) = 6x^2 + 30x + x + 5 = 6x^2 + 31x + 5$. $(3x+5)(2x+1) = 6x^2 + 3x + 10x + 5 = 6x^2 + 13x + 5$. $(6x+5)(x+1) = 6x^2 + 6x + 5x + 5 = 6x^2 + 11x + 5$. Key insight: All options produce $6x^2$ and $+5$ as leading and constant terms, but only $(3x+1)(2x+5)$ yields the correct middle term of $17x$.

Answer: Because multiplying two linear expressions in $x$ produces a highest power of $x^2$

Define quadratic: A quadratic expression has degree 2, meaning the highest power of the variable is $2$. Why multiplying two linear expressions gives degree 2: Each factor, $(x + 3)$ and $(x - 1)$, is degree 1 (linear). When multiplied, the highest-degree terms multiply: $x \cdot x = x^2$, making the result degree 2. Why distractors fail: Option A describes the sign of a constant, not the degree. Option B confuses the number of terms (trinomial) with the degree — not all trinomials are quadratic. Option D is too vague; a variable alone does not determine the degree.

Answer: $5x^2 + 20x + 18$

Compute courtyard area: $(3x+4)(2x+5) = 6x^2 + 15x + 8x + 20 = 6x^2 + 23x + 20$. Compute fountain area: $(x+1)(x+2) = x^2 + 2x + x + 2 = x^2 + 3x + 2$. Subtract to find remaining area: $(6x^2 + 23x + 20) - (x^2 + 3x + 2) = 5x^2 + 20x + 18$. Why distractors fail: Option B ($5x^2 + 26x + 22$) adds the fountain area instead of subtracting. Option C ($5x^2 + 20x + 22$) subtracts the linear terms correctly but adds the constants. Option D ($7x^2 + 26x + 22$) adds everything.