Answer: $(3x + 4)(2x - 5)$

Apply the AC method: $ac = 6 \times (-20) = -120$. Find two numbers that multiply to $-120$ and add to $-7$: $8$ and $-15$. Split and group: $6x^2 + 8x - 15x - 20 = 2x(3x + 4) - 5(3x + 4) = (3x + 4)(2x - 5)$. Why distractors fail: Option B: $(6x+5)(x-4) = 6x^2 - 19x - 20$. Option C: $(3x-4)(2x+5) = 6x^2 + 7x - 20$, sign error on $b$. Option D: $(2x+4)(3x-5) = 6x^2 + 2x - 20$, incorrect.

Answer: No, the two numbers must multiply to $ac$ and add to $b$, not the reverse

Identify the student's error: The student reversed the conditions. The correct requirement is: find two numbers whose product equals $ac$ and whose sum equals $b$. Why the correct answer works: The AC method is named for the product $a \times c$, which is the target product. The middle coefficient $b$ is the target sum. The student swapped these roles. Why distractors fail: Option A accepts the incorrect statement. Option C introduces the irrelevant sum $a + c$. Option D incorrectly limits the correction to $a = 1$.

Answer: $(6x + 5)(2x - 3)$

Apply the AC method: $ac = 12 \times (-15) = -180$. Find two numbers with product $-180$ and sum $-8$: $10$ and $-18$. Split and group: $12x^2 + 10x - 18x - 15 = 2x(6x + 5) - 3(6x + 5) = (6x + 5)(2x - 3)$. Why distractors fail: Option A: $(4x-5)(3x+3) = 12x^2 - 3x - 15$ and is not fully factored since $3x+3 = 3(x+1)$. Option C: $(4x+3)(3x-5) = 12x^2 - 11x - 15$. Option D: $(12x-5)(x+3) = 12x^2 + 31x - 15$.

Answer: $(5x - 2)(4x - 5)$

Apply the AC method: $ac = 20 \times 10 = 200$. Since $b = -33$ and $c > 0$, both numbers are negative. Find two negative numbers with product $200$ and sum $-33$: $-8$ and $-25$. Split and group: $20x^2 - 8x - 25x + 10 = 4x(5x - 2) - 5(5x - 2) = (5x - 2)(4x - 5)$. Verify: $(5x - 2)(4x - 5) = 20x^2 - 25x - 8x + 10 = 20x^2 - 33x + 10$ ✓. Why distractors fail: Option B: $(10x-5)(2x-2) = 20x^2 - 30x + 10$ and is not fully factored since $10x-5 = 5(2x-1)$ and $2x-2 = 2(x-1)$. Option C: $(4x-1)(5x-10) = 20x^2 - 45x + 10$ and is not fully factored since $5x-10 = 5(x-2)$. Option D: $(20x-1)(x-10) = 20x^2 - 201x + 10$, entirely incorrect.

Answer: $(5x - 2)(6x - 7)$

Apply the AC method: $ac = 30 \times 14 = 420$. Since $b = -47$ and $c > 0$, both numbers are negative. Find two negative numbers with product $420$ and sum $-47$: $-12$ and $-35$. Split and group: $30x^2 - 12x - 35x + 14 = 6x(5x - 2) - 7(5x - 2) = (5x - 2)(6x - 7)$. Verify: $(5x - 2)(6x - 7) = 30x^2 - 35x - 12x + 14 = 30x^2 - 47x + 14$ ✓. Why distractors fail: Option B: $(10x-7)(3x-2) = 30x^2 - 41x + 14$. Option C: $(15x-7)(2x-2) = 30x^2 - 44x + 14$ and is not fully factored. Option D: $(5x-7)(6x-2) = 30x^2 - 52x + 14$ and is not fully factored.

Answer: $3$ and $4$

Check each pair: We need two numbers where the product is $12$ and the sum is $7$. Check: $3 \times 4 = 12$ and $3 + 4 = 7$. ✓ Why distractors fail: Option A: $2 \times 6 = 12$ but $2 + 6 = 8 \neq 7$. Option C: $1 \times 12 = 12$ but $1 + 12 = 13 \neq 7$. Option D: $4 \times 8 = 32 \neq 12$.

Answer: $(4x + 5)(2x + 3)$

Extract the common binomial: From $2x(4x + 5) + 3(4x + 5)$, the common binomial factor is $(4x + 5)$. Factoring it out gives $(4x + 5)(2x + 3)$. Verify: $(4x + 5)(2x + 3) = 8x^2 + 12x + 10x + 15 = 8x^2 + 22x + 15$ ✓. Why distractors fail: Option A: $(2x+5)(4x+3) = 8x^2 + 26x + 15$. Option C: $(8x+5)(x+3) = 8x^2 + 29x + 15$. Option D: $(4x+3)(2x+5) = 8x^2 + 26x + 15$.

Answer: The factorization is correct, and it can also be written as $(2x - 3)^2$

Verify the factorization: $(2x - 3)(2x - 3) = 4x^2 - 6x - 6x + 9 = 4x^2 - 12x + 9$ ✓. This matches the original expression. Recognize the special form: $4x^2 - 12x + 9 = (2x)^2 - 2(2x)(3) + 3^2 = (2x - 3)^2$. It is a perfect square trinomial. Why distractors fail: Option A: $(4x-3)(x-3) = 4x^2 - 15x + 9$, incorrect. Option B is false; the AC method works fine on perfect square trinomials. Option D confuses difference of squares (which requires subtraction of two squares) with a perfect square trinomial.

Answer: $4$ and $35$

Check each pair: We need product $= 140$ and sum $= 39$. Option A: $7 \times 20 = 140$ ✓, $7 + 20 = 27$ ✗. Option B: $4 \times 35 = 140$ ✓, $4 + 35 = 39$ ✓. Option C: $5 \times 28 = 140$ ✓, $5 + 28 = 33$ ✗. Option D: $10 \times 14 = 140$ ✓, $10 + 14 = 24$ ✗. Complete the factorization: $14x^2 + 4x + 35x + 10 = 2x(7x + 2) + 5(7x + 2) = (7x + 2)(2x + 5)$. Why distractors fail: Options A, C, and D all have the correct product of $140$ but their sums are $27$, $33$, and $24$ respectively — none equal $39$.

Answer: Factor out the greatest common factor of $6$ to get $6(2x^2 + 3x + 1)$

Check for a GCF first: The terms $12x^2$, $18x$, and $6$ all share a common factor of $6$. Factoring it out gives $6(2x^2 + 3x + 1)$, which simplifies the subsequent AC method work. Why the correct answer works: Removing the GCF first reduces the size of the coefficients. The AC product for $2x^2 + 3x + 1$ is only $2$, much simpler than working with $ac = 72$. Why distractors fail: Option A skips the GCF step and leads to unnecessarily large numbers. Option B jumps ahead in the process. Option D is unnecessary since the expression is already in standard form.

Answer: The four-term expression is arranged so that each pair of terms shares a factor, and both pairs produce the same remaining binomial

Connect splitting to grouping: Splitting the middle term strategically creates a four-term expression where two pairs each share a factor. After extracting the GCF from each pair, a common binomial emerges. Why the correct answer works: For example, $6x^2 + 9x + 2x + 3$ groups as $(6x^2 + 9x) + (2x + 3) = 3x(2x+3) + 1(2x+3) = (2x+3)(3x+1)$. The shared binomial $(2x+3)$ is what makes grouping possible. Why distractors fail: Option A is incorrect; after splitting you have four terms, not two binomials. Option C is false; grouping only works when the split is chosen so that pairs share a common binomial. Option D is false; the distributive property applies regardless of the number of terms.

Answer: $(2x + 3)(x + 4)$

Apply the AC method: $ac = 2 \times 12 = 24$. Find two numbers with product $24$ and sum $11$: $3$ and $8$. Split and group: $2x^2 + 3x + 8x + 12 = x(2x + 3) + 4(2x + 3) = (2x + 3)(x + 4)$. Why distractors fail: Option B: $(2x+4)(x+3) = 2x^2 + 10x + 12$ and is not fully factored (common factor of $2$ in $2x+4$). Option C: $(2x+6)(x+2) = 2x^2 + 10x + 12$ and is also not fully factored. Option D is false since the expression factors as shown.

Answer: $(3x - 2)(2x + 3)$

Extract the common binomial: From $3x(2x + 3) - 2(2x + 3)$, the common factor is $(2x + 3)$, giving $(2x + 3)(3x - 2)$, or equivalently $(3x - 2)(2x + 3)$. Verify: $(3x - 2)(2x + 3) = 6x^2 + 9x - 4x - 6 = 6x^2 + 5x - 6$ ✓. Why distractors fail: Option B: $(3x+2)(2x-3) = 6x^2 - 5x - 6$, sign error. Option C uses fractions and is not properly factored over the integers. Option D: $(2x+3)(3x+2) = 6x^2 + 13x + 6$, wrong signs.

Answer: $2x^2 + 3x + 5$

Test each with the AC method: For each expression, compute $ac$ and check if there exists an integer pair with the required product and sum. Check Option C: $ac = 2 \times 5 = 10$. Factor pairs of $10$: $(1, 10)$ and $(2, 5)$. Their sums are $11$ and $7$. Neither equals $b = 3$. So $2x^2 + 3x + 5$ cannot be factored over the integers. Verify the others factor: Option A: $ac = 6$, pair $(2, 3)$: sum $= 5$ ✓. Option B: $ac = 12$, pair $(3, 4)$: sum $= 7$ ✓. Option D: $ac = 10$, pair $(1, 10)$: sum $= 11$ ✓.

Answer: When $a = 1$, the AC method reduces to finding two numbers that multiply to $c$ and add to $b$, which is exactly the simpler method

Connect the two methods: When $a = 1$, the AC product is $1 \times c = c$. So you need two numbers that multiply to $c$ and add to $b$, which is exactly the familiar method for $x^2 + bx + c$. Why the correct answer works: The AC method is a generalization. Setting $a = 1$ collapses the method to the simpler factor-pair approach without requiring explicit grouping. Why distractors fail: Option A ignores the clear generalization relationship. Option C is false because the AC method technically works for all values of $a$. Option D is incorrect because when $a = 1$ the factors can be written directly without grouping.

Answer: $(5x - 2)(x - 3)$

Apply the AC method: $ac = 5 \times 6 = 30$. Since $b = -17$ and $c$ is positive, both numbers must be negative: we need two negative numbers whose product is $30$ and sum is $-17$. Those are $-2$ and $-15$. Split and group: $5x^2 - 2x - 15x + 6 = x(5x - 2) - 3(5x - 2) = (5x - 2)(x - 3)$. Why distractors fail: Option A: $(5x-3)(x-2) = 5x^2 - 13x + 6$. Option B: $(5x-6)(x-1) = 5x^2 - 11x + 6$. Option C: $(5x-1)(x-6) = 5x^2 - 31x + 6$. All produce incorrect middle terms.

Answer: $-(3x - 2)(x - 4)$

Factor out the negative leading coefficient: $-3x^2 + 14x - 8 = -(3x^2 - 14x + 8)$. Now factor $3x^2 - 14x + 8$. Apply the AC method: $ac = 3 \times 8 = 24$. Two numbers with product $24$ and sum $-14$: $-2$ and $-12$. Split: $3x^2 - 2x - 12x + 8 = x(3x - 2) - 4(3x - 2) = (3x - 2)(x - 4)$. Combine: $-3x^2 + 14x - 8 = -(3x - 2)(x - 4)$. Why distractors fail: Option B: $(3x-2)(x+4) = 3x^2 + 10x - 8$, wrong signs. Option C: $(-3x+4)(x-2) = -3x^2 + 10x - 8$, incorrect middle term. Option D uses fractions, which means the expression is not factored over the integers.

Answer: $3x^2 - 13x - 10$

Expand the product: $(3x + 2)(x - 5) = 3x^2 - 15x + 2x - 10 = 3x^2 - 13x - 10$. Why the correct answer works: Combining like terms: $-15x + 2x = -13x$. The constant term is $2 \times (-5) = -10$. Why distractors fail: Option A has a positive $13x$ (sign error). Option C: $-17x$ would require $-15x + (-2x)$, meaning the second factor constant would need a different sign. Option D: $+17x$ is entirely wrong.

Answer: $(2x - 3)(4x - 1)$

Apply the AC method: $ac = 8 \times 3 = 24$. Since $b = -14$ and $c > 0$, both numbers are negative. Find: $-2$ and $-12$ give product $24$ and sum $-14$. Split and group: $8x^2 - 2x - 12x + 3 = 2x(4x - 1) - 3(4x - 1) = (4x - 1)(2x - 3)$. Why distractors fail: Option A: $(4x-3)(2x-1) = 8x^2 - 10x + 3$. Option B: $(8x-3)(x-1) = 8x^2 - 11x + 3$. Option D: $(4x+1)(2x-3) = 8x^2 - 10x - 3$.

Answer: Rewriting $bx$ as the sum of two terms whose coefficients multiply to $ac$ and add to $b$

Define splitting the middle term: Splitting the middle term rewrites $bx$ as $mx + nx$ where $m \times n = ac$ and $m + n = b$. This creates a four-term expression that can be factored by grouping. Why the correct answer works: This is the precise definition: the two new coefficients must satisfy both the product and sum conditions derived from the AC method. Why distractors fail: Option A describes a step from completing the square, not the AC method. Option C is an arbitrary grouping with no strategic purpose. Option D splits $b$ equally, which rarely satisfies the product condition $m \times n = ac$.

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

Apply the AC method: Here $a = 2$, $b = 7$, $c = 3$, so $ac = 6$. Find two numbers that multiply to $6$ and add to $7$: those are $1$ and $6$. Split and group: Rewrite: $2x^2 + x + 6x + 3 = x(2x + 1) + 3(2x + 1) = (2x + 1)(x + 3)$. Why distractors fail: Option B: $(2x+3)(x+1) = 2x^2 + 5x + 3$, incorrect middle term. Option C ignores the leading coefficient $2$. Option D: $(2x+7)(x+3) = 2x^2 + 13x + 21$, entirely wrong.

Answer: There is no GCF greater than $1$, so the student's approach is the only option

Check for a GCF: GCF($24$, $50$, $25$). Factors of $24$: $1, 2, 3, 4, 6, 8, 12, 24$. Factors of $25$: $1, 5, 25$. The only common factor is $1$. Why the correct answer works: Since the GCF is $1$, there is nothing to factor out. The student must work with $ac = 600$ directly. The factor pair is $20$ and $30$ since $20 \times 30 = 600$ and $20 + 30 = 50$. Why distractors fail: Option A is misleading because while the answer is the same, the claim about steps is irrelevant here. Option C is false: $24/2 = 12$, $50/2 = 25$, $25/2$ is not an integer. Option D is false: $24/5$ is not an integer.

Answer: $18$

Identify $a$ and $c$: In $6x^2 + 11x + 3$, the leading coefficient is $a = 6$ and the constant term is $c = 3$. Compute the AC product: $ac = 6 \times 3 = 18$. This is the target product for finding the factor pair. Why distractors fail: Option A ($33$) comes from multiplying $b \times c = 11 \times 3$. Option B ($66$) comes from $a \times b = 6 \times 11$. Option D ($9$) comes from $c^2 = 3^2$. Only $a \times c = 18$ is the correct AC product.

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

Apply the AC method: $ac = 4 \times (-3) = -12$. Find two numbers that multiply to $-12$ and add to $4$: $6$ and $-2$. Split and group: $4x^2 + 6x - 2x - 3 = 2x(2x + 3) - 1(2x + 3) = (2x - 1)(2x + 3)$. Why distractors fail: Option A: $(4x+3)(x-1) = 4x^2 - x - 3$, wrong middle coefficient. Option C: $(2x+1)(2x-3) = 4x^2 - 4x - 3$, sign error on the middle term. Option D: $(4x-1)(x+3) = 4x^2 + 11x - 3$, incorrect.

Answer: Each group must share a common binomial factor after factoring out the GCF of each pair

Understand factoring by grouping: After splitting the middle term, the four-term expression is grouped into two pairs. Each pair is factored separately, and if done correctly, both pairs share a common binomial factor. Why the correct answer works: The entire method depends on extracting a common binomial from both groups. For example, $2x(3x+1) + 3(3x+1) = (3x+1)(2x+3)$. Without a shared binomial, the method cannot proceed. Why distractors fail: Option A is false because the $x^2$ term only appears in one group. Option C is false because the GCFs of the two pairs are typically different (that is in fact the point). Option D is irrelevant to the method.

Answer: Yes, but using the difference of squares formula is more direct since $9x^2 - 25 = (3x)^2 - 5^2$

Evaluate the AC method approach: $ac = 9 \times (-25) = -225$. We need two numbers with product $-225$ and sum $0$: $15$ and $-15$. Split: $9x^2 + 15x - 15x - 25 = 3x(3x + 5) - 5(3x + 5) = (3x + 5)(3x - 5)$. This works. Compare with difference of squares: $9x^2 - 25 = (3x)^2 - 5^2 = (3x + 5)(3x - 5)$. This is faster and more direct since it recognizes the special pattern immediately. Why distractors fail: Option A is false; the AC method works with $b = 0$. Option C is false; $9x^2 - 25$ is degree $2$ and is quadratic. Option D incorrectly claims the AC method is always preferred.

Answer: $(3x + 2)(x + 4)$

Apply the AC method: $ac = 3 \times 8 = 24$. Find two numbers that multiply to $24$ and add to $14$: $2$ and $12$. Split and group: $3x^2 + 2x + 12x + 8 = x(3x + 2) + 4(3x + 2) = (3x + 2)(x + 4)$. Why distractors fail: Option A: $(3x+4)(x+2) = 3x^2 + 10x + 8$, wrong middle term. Option B: $(3x+8)(x+1) = 3x^2 + 11x + 8$, wrong middle term. Option D: $(x+8)(3x+1) = 3x^2 + 25x + 8$, wrong middle term.

Answer: No, because $(10x + 3)(x + 2) = 10x^2 + 23x + 6$

Verify by expanding: $(10x+3)(x+2) = 10x^2 + 20x + 3x + 6 = 10x^2 + 23x + 6$. The middle coefficient is $23$, not $19$. Find the correct factorization: $ac = 60$. Factor pair with sum $19$: $4$ and $15$. So $10x^2 + 4x + 15x + 6 = 2x(5x+2) + 3(5x+2) = (5x+2)(2x+3)$. Why distractors fail: Option A is false because the expansion does not match. Option C gives an incorrect expansion. Option D is irrelevant; the GCF of $10$, $19$, and $6$ is $1$.

Answer: No; the pair $9$ and $-8$ has product $-72$ and sum $1$

Verify the claimed pair: $9 \times (-8) = -72$ ✓ and $9 + (-8) = 1$ ✓. So the student made an error in their factor-pair search. Complete the factoring: $6x^2 + 9x - 8x - 12 = 3x(2x + 3) - 4(2x + 3) = (2x + 3)(3x - 4)$. Why distractors fail: Option A is false since a valid pair exists. Option C: $12 + (-6) = 6 \neq 1$. Option D: The number of factor pairs is irrelevant to whether a valid pair exists.

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

Apply the AC method: $ac = 15 \times (-6) = -90$. Find two numbers with product $-90$ and sum $-1$: $9$ and $-10$. Split and group: $15x^2 + 9x - 10x - 6 = 3x(5x + 3) - 2(5x + 3) = (5x + 3)(3x - 2)$. Verify: $(5x + 3)(3x - 2) = 15x^2 - 10x + 9x - 6 = 15x^2 - x - 6$ ✓. Why distractors fail: Option B: $(5x-3)(3x+2) = 15x^2 + 10x - 9x - 6 = 15x^2 + x - 6$, sign error on the middle term. Option C: $(15x+6)(x-1) = 15x^2 - 9x - 6$ and is not fully factored since $15x+6 = 3(5x+2)$. Option D: $(5x+2)(3x-3) = 15x^2 - 9x - 6$ and $3x - 3 = 3(x-1)$, so it is not fully factored.

Answer: It reduces the size of the coefficients, making the factor-pair search simpler

Understand the benefit of factoring out the GCF: When you factor out the GCF, the remaining trinomial has smaller coefficients. The AC product is therefore smaller, and there are fewer factor pairs to test. Why the correct answer works: For example, $8x^2 + 12x + 4$ has $ac = 32$ with many factor pairs. After factoring out $4$, you get $4(2x^2 + 3x + 1)$ where $ac = 2$, making the search trivial. Why distractors fail: Option A is false; the AC method works for any $a \neq 1$. Option C is false; the GCF sits outside the factored binomials. Option D is false; factoring out a GCF does not produce a difference of squares.

Answer: Multiply the leading coefficient $a$ by the constant term $c$

Recall the AC method procedure: The AC method begins by computing the product $a \times c$. This product is then used to find two numbers that multiply to $ac$ and add to $b$. Why the correct answer works: Multiplying $a$ by $c$ is the defining first computational step of the AC method, giving the target product for the factor-pair search. Why distractors fail: Option A (splitting the middle term) is a later step that depends on the result of the AC product. Option C (factoring out a GCF) is a preliminary check done before applying the AC method, not part of the AC method itself. Option D (grouping) is the final stage of the process.

Answer: $6x^2 + 13x - 5$

Test each option using the AC method: For all options, $ac = 6 \times (-5) = -30$. We need a pair of integers with product $-30$ that sums to $b$. Check Option A: $b = 13$. Factor pairs of $-30$: $(15, -2)$ gives $15 + (-2) = 13$ ✓. So $6x^2 + 15x - 2x - 5 = 3x(2x + 5) - 1(2x + 5) = (2x + 5)(3x - 1)$. Check remaining options: Option B: $b = 4$. No pair of $-30$ sums to $4$ (pairs: $\pm1, \pm30$; $\pm2, \pm15$; $\pm3, \pm10$; $\pm5, \pm6$ — sums are $\pm29, \pm13, \pm7, \pm1$). Option C: $b = 8$. No pair sums to $8$. Option D: $b = 11$. No pair sums to $11$.

Answer: Both students, since multiplication is commutative

Verify the factorization: $(2x + 5)(5x + 2) = 10x^2 + 4x + 25x + 10 = 10x^2 + 29x + 10$ ✓. Apply commutativity: Since $A \cdot B = B \cdot A$ for all expressions, $(2x + 5)(5x + 2) = (5x + 2)(2x + 5)$. Both are correct. Why distractors fail: Options A and B each reject one valid answer. Option D is false because the verification shows the binomials are correct.

Answer: The claim is incorrect; $ac = 18$ is not a perfect square, so the example does not test the claim

Compute ac: $ac = 2 \times 9 = 18$. Since $\sqrt{18}$ is not an integer, $18$ is not a perfect square. Evaluate the example: This example does not test the student's claim because $ac$ is not a perfect square. To properly test the claim, we would need an example where $ac$ IS a perfect square but the expression does not factor (e.g., $4x^2 + 3x + 4$ where $ac = 16$ but no pair of factors of $16$ sums to $3$). Why distractors fail: Option A: $18$ is not a perfect square. Option B makes a universal claim that is actually false. Option D incorrectly states $18$ is a perfect square.

Answer: $4x^2 + 14x + 6$

Check each expression for a GCF greater than 1: Option A: GCF(5, 13, 6) = 1. Option B: GCF(3, 10, 7) = 1. Option C: GCF(4, 14, 6) = 2. Option D: GCF(7, 9, 2) = 1. Why the correct answer works: $4x^2 + 14x + 6 = 2(2x^2 + 7x + 3)$. The GCF of $2$ should be factored out first to simplify the AC method. Why distractors fail: Options A, B, and D all have coefficients with GCF = 1, so no preliminary factoring is needed.

Answer: Both approaches are valid and yield the same factored form

Compare the results: Student 1 obtains $(2x + 3)(x + 1)$ and Student 2 obtains $(x + 1)(2x + 3)$. By the commutative property of multiplication, these are the same expression. Why both splits work: Both splits use numbers that multiply to $ac = 6$ and add to $5$ (namely $2$ and $3$). The order in which the terms are written changes the grouping but not the final result. Why distractors fail: Options A and B each reject a valid approach. Option D incorrectly claims that factor order affects the product.

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

Apply the AC method: $ac = 36$. Find two numbers that multiply to $36$ and add to $13$: $4$ and $9$. Split and group: $6x^2 + 4x + 9x + 6 = 2x(3x + 2) + 3(3x + 2) = (3x + 2)(2x + 3)$. Why distractors fail: Option B: $(6x+1)(x+6) = 6x^2 + 37x + 6$. Option C: $(3x+3)(2x+2) = 6x^2 + 12x + 6$ and is not fully factored (contains common factor $6$). Option D: $(6x+3)(x+2) = 6x^2 + 15x + 6$.

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

Determine the second factor: If one factor is $(4x - 1)$, the second factor must be of the form $(x + k)$ for some integer $k$ (since $4 \times 1 = 4 = a$). The product is $(4x - 1)(x + k) = 4x^2 + (4k - 1)x - k$. Check Option A: $4x^2 + 7x - 2$: We need $-k = -2$, so $k = 2$, and $4(2) - 1 = 7$ ✓. So $(4x - 1)(x + 2) = 4x^2 + 7x - 2$ ✓. Why distractors fail: Option B: $c = 2$ means $k = -2$, giving $4(-2) - 1 = -9 \neq 3$. Option C: $c = -1$ means $k = 1$, giving $4(1) - 1 = 3 \neq -7$. Option D: $c = 2$ means $k = -2$, giving $-9 \neq 1$.

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

Factor out the GCF: The GCF of $18$, $21$, and $9$ is $3$. So $18x^2 + 21x - 9 = 3(6x^2 + 7x - 3)$. Apply the AC method to $6x^2 + 7x - 3$: $ac = 6 \times (-3) = -18$. Two numbers with product $-18$ and sum $7$: $9$ and $-2$. Split: $6x^2 + 9x - 2x - 3 = 3x(2x + 3) - 1(2x + 3) = (2x + 3)(3x - 1)$. Final answer: $3(2x + 3)(3x - 1)$. Why distractors fail: Option B is not completely factored since $(6x + 9) = 3(2x + 3)$. Option C: $(18x-3)(x+3) = 18x^2 + 51x - 9$. Option D: $3(6x-1)(x+3) = 3(6x^2 + 17x - 3) = 18x^2 + 51x - 9$.

Answer: The first split uses $8$ and $1$ (product $8$, sum $9$), which satisfies the AC conditions; the second uses $4$ and $5$ (product $20 \neq 8$), which does not

Compute the AC product: $ac = 2 \times 4 = 8$. We need two numbers with product $8$ and sum $9$. Test the splits: First split: $8 \times 1 = 8$ ✓, $8 + 1 = 9$ ✓. Grouping: $2x(x + 4) + 1(x + 4) = (x + 4)(2x + 1)$. Second split: $4 \times 5 = 20 \neq 8$ ✗. This pair does not satisfy the AC condition, so grouping will produce mismatched binomials. Why distractors fail: Option A is wrong because the second split genuinely fails. Option C gives an incorrect reason. Option D is false since the first split works perfectly.

Answer: The student chose the wrong factor pair; $5$ and $5$ do not multiply to $ac = 24$

Identify the error: $ac = 3 \times 8 = 24$. The student used $5$ and $5$, but $5 \times 5 = 25 \neq 24$. The correct pair is $4$ and $6$ since $4 \times 6 = 24$ and $4 + 6 = 10$. Why this causes failure: Using incorrect split numbers means the groups will not share a common binomial factor, which is exactly what happened: $(3x + 5)$ and $(5x + 8)$ are different binomials. Why distractors fail: Option B: Grouping order does not matter if the split is correct. Option C: There is no restriction based on the parity of $b$. Option D: GCF(3, 10, 8) = 1, so no GCF needs to be factored out.

Answer: $x$ and $-12x$

Compute the AC product: $ac = 3 \times (-4) = -12$. We need two numbers that multiply to $-12$ and add to $-11$. Test the pairs: $1 \times (-12) = -12$ ✓ and $1 + (-12) = -11$ ✓. So the middle term splits as $x - 12x$. Why distractors fail: Option A: $(-3)(-8) = 24 \neq -12$. Option C: $(-2)(-9) = 18 \neq -12$. Option D: $(4)(-15) = -60 \neq -12$.