Answer: m × n = c and m + n = b

Recall the Product-Sum Method: To factor $x^2 + bx + c$, we look for two numbers $m$ and $n$ such that $(x + m)(x + n) = x^2 + (m+n)x + mn$. Match coefficients: Comparing $x^2 + (m+n)x + mn$ with $x^2 + bx + c$, we need $m \times n = c$ (the constant term) and $m + n = b$ (the coefficient of $x$). Why distractors fail: Option A reverses the roles of $b$ and $c$. Options C and D introduce subtraction, which is not part of the standard product-sum conditions.

Answer: $(x + 5)(x + 5)$

Apply the Product-Sum Method: We need two numbers that multiply to $25$ and add to $10$. The pair is $5$ and $5$. Write the factored form: $x^2 + 10x + 25 = (x + 5)(x + 5) = (x + 5)^2$. Why distractors fail: Option B: $10 \times 15 = 150 \neq 25$. Option C: $2 \times 13 = 26 \neq 25$. Option D: $1 + 25 = 26 \neq 10$.

Answer: Only the first and second students

Verify the first factorization: $(x - 6)(x + 8) = x^2 + 8x - 6x - 48 = x^2 + 2x - 48$. Correct. Compare the second factorization: $(x + 8)(x - 6)$ is the same as $(x - 6)(x + 8)$ by the commutative property. Also correct. Check the third factorization: $(x + 6)(x - 8) = x^2 - 8x + 6x - 48 = x^2 - 2x - 48$. The middle term is $-2x$, not $+2x$. Incorrect. Why distractors fail: Option A ignores the commutative property. Option C: the third student swapped the signs incorrectly. Option D: the third student's expansion gives the wrong middle term.

Answer: Negative pairs allow the factorization to produce negative or mixed-sign middle terms

Role of negative factor pairs: The sum $m + n = b$ can be negative or have mixed contributions only if at least one of $m, n$ is negative. Without considering negative pairs, we could not factor expressions with negative $b$ or negative $c$. Why distractors fail: Option A: negative pairs are needed whenever $b < 0$ or $c < 0$, not just when $b = 0$. Option C: primality of $c$ is irrelevant to the need for negative pairs. Option D: the signs must be determined as part of identifying the pair, not after.

Answer: $(x + m)(x + n)$

Recall the factored form: The Product-Sum Method produces the factorization $x^2 + bx + c = (x + m)(x + n)$ where $mn = c$ and $m + n = b$. Why the correct answer works: The numbers $m$ and $n$ appear as the constant terms in each binomial factor. Why distractors fail: Option A places $m$ and $n$ as coefficients of $x$, which applies when $a \neq 1$. Option C also misplaces them as leading coefficients. Option D incorrectly combines $m$ and $n$ into one factor.

Answer: The two numbers have opposite signs

Apply sign rules for multiplication: For the product $m \times n = c < 0$, one number must be positive and the other negative, since only opposite signs yield a negative product. Why distractors fail: Options A and B describe same-sign pairs, which always give a positive product. Option D: if either number were zero, $c$ would be zero, not negative.

Answer: Yes, because a perfect-square discriminant guarantees rational roots, and since $a = 1$ with integer $b$ and $c$, those roots are integers

Connect discriminant to factorability: For $x^2 + bx + c$, the roots are $\frac{-b \pm \sqrt{b^2 - 4c}}{2}$. If $b^2 - 4c$ is a perfect square, say $d^2$, the roots are $\frac{-b \pm d}{2}$. Why integer roots follow: Since $b$ and $d$ are integers and $a = 1$, if $b$ and $d$ have the same parity (both even or both odd), then $\frac{-b \pm d}{2}$ are integers. With integer $b$ and $c$, $b^2 - 4c \equiv b^2 \pmod{2}$, ensuring same parity. Thus the roots are integers, and the expression factors as $(x - r_1)(x - r_2)$. Why distractors fail: Option A: the discriminant is directly related to factorability. Option C: when the discriminant is a perfect square, the roots are specifically rational (and integer for $a=1$). Option D: the result holds regardless of the sign of $c$.

Answer: $(x - 8)(x - 7)$

Determine signs: $c = 56 > 0$ and $b = -15 < 0$, so both factors are negative. Find the pair: $(-8)(-7) = 56$ and $(-8) + (-7) = -15$. So $(x - 8)(x - 7)$. Why distractors fail: Option A: $(-4) + (-14) = -18 \neq -15$. Option C: positive factors give $+15x$. Option D: $(-2) + (-28) = -30 \neq -15$.

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

Determine signs: $c = 42 > 0$, $b = -13 < 0$: both numbers are negative. Find the pair: $(-6)(-7) = 42$ and $(-6) + (-7) = -13$. Hence $(x - 6)(x - 7)$. Why distractors fail: Option A: $(-3)(-14) = 42$ but $-3 + (-14) = -17 \neq -13$. Option C: positive signs give $+13x$. Option D: $(-2)(-21) = 42$ but $-2 + (-21) = -23 \neq -13$.

Answer: Expanding $(x + 2)(x + 3)$ using the distributive property

The role of expanding: Expanding (FOILing) the factored form and simplifying should reproduce the original expression if the factorization is correct. Verify the claim: $(x + 2)(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$, confirming the factorization. Why distractors fail: Option A: checking one value doesn't prove identity for all $x$. Option C: sharing a vertex doesn't confirm algebraic equivalence. Option D: factor pairs of $5$ are irrelevant; the constant term is $6$.

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

Identify the target values: We need two numbers that multiply to $12$ and add to $7$. Test factor pairs of 12: The factor pairs of $12$ are $(1, 12)$, $(2, 6)$, and $(3, 4)$. Only $3 + 4 = 7$. Why distractors fail: Option A: $2 + 6 = 8 \neq 7$. Option C: $1 + 12 = 13 \neq 7$. Option D: $5 + 2 = 7$ but $5 \times 2 = 10 \neq 12$.

Answer: $b > 0$ and $c < 0$

Determine the sign of c: $c = pq$. Since $p > 0$ and $q < 0$, their product is negative: $c < 0$. Determine the sign of b: $b = p + q$. Since $|p| > |q|$ and $p > 0$, the positive number has greater absolute value, so $b = p + q > 0$. Why distractors fail: Option A: $c$ must be negative since the factors have opposite signs. Option C: $b$ is positive due to $|p| > |q|$. Option D: both conclusions are wrong — $c < 0$ and $b > 0$.

Answer: Both the factorization and the verification are correct

Check the factorization: $(-3)(-5) = 15$ and $(-3) + (-5) = -8$. The factor pair is correct. Check the expansion: $(x - 3)(x - 5) = x^2 - 5x - 3x + 15 = x^2 - 8x + 15$. Every step is correct. Why distractors fail: Option A: the arithmetic is correct. Option C: $(x + 3)(x + 5) = x^2 + 8x + 15$, which has the wrong sign on the middle term. Option D: expansion is a perfectly valid verification method.

Answer: $3$ and $-4$

Identify targets: We need two numbers whose product is $-12$ and whose sum is $-1$. Test pairs: $3 \times (-4) = -12$ and $3 + (-4) = -1$. This pair satisfies both conditions. Why distractors fail: Option B: $(-3) + 4 = 1 \neq -1$. Option C: $2 \times (-6) = -12$ but $2 + (-6) = -4 \neq -1$. Option D: $(-2) + 6 = 4 \neq -1$.

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

Determine sign pattern: $c = 20 > 0$ and $b = -9 < 0$, so both numbers are negative. Find the pair: $(-4) \times (-5) = 20$ and $(-4) + (-5) = -9$. Therefore $x^2 - 9x + 20 = (x - 4)(x - 5)$. Why distractors fail: Option A: $(-2) + (-10) = -12 \neq -9$. Option C: $(4)(5) = 20$ but $4 + 5 = 9 \neq -9$. Option D: $(-1) + (-20) = -21 \neq -9$.

Answer: Swap which constant carries the positive sign, writing $(x + 6)(x - 4)$

Diagnose the error: The student wrote $(x - 6)(x + 4)$, giving $-6 + 4 = -2$ for the middle term. They need $+2$, so the $6$ should carry the positive sign and the $4$ should carry the negative sign. Correct the factorization: $(x + 6)(x - 4) = x^2 - 4x + 6x - 24 = x^2 + 2x - 24$. Swapping which constant carries which sign fixes the middle term. Why distractors fail: Option A: Using $-12$ and $2$ gives a middle term of $-10x$. Option B: Making both positive gives $6 \times 4 = 24$, not $-24$. Option D: Making both negative gives $(-6)(-4) = 24$, which is also incorrect.

Answer: $(x - 5)(x + 8)$, verified: $x^2 + 8x - 5x - 40 = x^2 + 3x - 40$

Find the factor pair: Need $m \times n = -40$ and $m + n = 3$. Since $c < 0$, signs are opposite with the positive factor larger. $(-5)(8) = -40$ and $-5 + 8 = 3$. Verify by expanding: $(x - 5)(x + 8) = x^2 + 8x - 5x - 40 = x^2 + 3x - 40$. Confirmed. Why distractors fail: Option B gives middle term $-3x$. Option C gives $+6x$. Option D gives $-6x$. None match $+3x$.

Answer: $(1, 18)$, $(2, 9)$, $(3, 6)$ and their negatives $(-1, -18)$, $(-2, -9)$, $(-3, -6)$

List positive factor pairs: The positive integer pairs whose product is $18$ are $(1, 18)$, $(2, 9)$, and $(3, 6)$. Include negative pairs: Since $(-m)(-n) = mn$, the pairs $(-1, -18)$, $(-2, -9)$, and $(-3, -6)$ also multiply to $18$. These give negative values of $b$. Why distractors fail: Option A omits the negative pairs. Option C uses mixed-sign pairs like $(-1, 18)$, which multiply to $-18$, not $18$. Option D includes $(4, 5)$, but $4 \times 5 = 20 \neq 18$.

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

Set up the conditions: We need $m \times n = -15$ and $m + n = 2$. Since $c < 0$, the signs are opposite. Find the correct pair: $(-3) \times 5 = -15$ and $(-3) + 5 = 2$. So $x^2 + 2x - 15 = (x - 3)(x + 5)$. Why distractors fail: Option A: $3 + (-5) = -2 \neq 2$. Option C: $1 + (-15) = -14 \neq 2$. Option D: $(-1) + 15 = 14 \neq 2$.

Answer: $(x - 7)(x - 4)$

Determine signs: $c = 28 > 0$ and $b = -11 < 0$, so both numbers are negative. Find the pair: $(-7) \times (-4) = 28$ and $(-7) + (-4) = -11$. So $x^2 - 11x + 28 = (x - 7)(x - 4)$. Why distractors fail: Option B: $(-14)(-2) = 28$ but $-14 + (-2) = -16 \neq -11$. Option C: positive signs give $+11x$ instead of $-11x$. Option D: $(-7)(4) = -28 \neq 28$.

Answer: 5

List positive factor pairs of 36: $(1, 36), (2, 18), (3, 12), (4, 9), (6, 6)$ — these are all pairs of positive integers with product 36. Compute possible values of k: Each pair gives $k = m + n$: $37, 20, 15, 13, 12$. That is 5 distinct values. Why distractors fail: Option A (3) and Option B (4) undercount. Option D (6) would require an additional pair, but we have exhausted all positive factor pairs of 36.

Answer: Assign opposite signs to each pair and check which gives a sum of $-2$

Recognize the sign of c: Since $c = -35 < 0$, the two numbers must have opposite signs. The student already found the magnitude pairs; the next logical step is to test sign assignments. Apply sign assignment: For $(5, 7)$: try $5$ and $-7$, giving $5 + (-7) = -2$. This matches $b = -2$, so the factorization is $(x + 5)(x - 7)$. Why distractors fail: Option B ignores the negative constant. Option C: multiplying both by $-1$ changes the product to $+35$, which is incorrect. Option D introduces an irrelevant heuristic.

Answer: $b < 0$ and $c > 0$

Apply sign rules: If both $m$ and $n$ are negative: $mn > 0$ (so $c > 0$) and $m + n < 0$ (so $b < 0$). Why the correct answer works: $b < 0$ and $c > 0$ is the unique combination produced by two negative factors. Why distractors fail: Option A ($b > 0, c > 0$): both factors positive. Option B ($b < 0, c < 0$): opposite-sign factors. Option D ($b > 0, c < 0$): opposite-sign factors with the positive factor dominating.

Answer: Swap the signs of the two constant terms so the larger absolute value carries the positive sign

Diagnose the error: The student likely wrote $(x - 5)(x + 4)$, giving $-5 + 4 = -1$ for the middle term. They need $+1$, so the $5$ should be positive and the $4$ negative. Correct the factorization: $(x + 5)(x - 4) = x^2 - 4x + 5x - 20 = x^2 + x - 20$. Swapping which term carries which sign fixes the middle term. Why distractors fail: Option A: swapping constants $5$ and $4$ changes the factor pair entirely. Option B: swapping both signs gives $(x + 5)(x - 4)$ from $(x - 5)(x + 4)$, which is the correct fix — but this option says 'both binomials' which would give $(x + 5)(x - 4)$ only if interpreted correctly. The precise description is in Option C. Option D: making both positive gives $5 \times 4 = 20$, not $-20$.

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

Determine the signs: Since $c = 6$ is positive and $b = -5$ is negative, both numbers in the factor pair must be negative (negative × negative = positive, and negative + negative = negative). Find the pair: $(-2) \times (-3) = 6$ and $(-2) + (-3) = -5$. So $x^2 - 5x + 6 = (x - 2)(x - 3)$. Why distractors fail: Option A: $(-1)(-6) = 6$ but $-1 + (-6) = -7 \neq -5$. Option B uses positive signs, giving $+5x$ instead of $-5x$. Option D: $(-2)(3) = -6 \neq 6$.

Answer: No, because $(x + 4)(x - 4)$ is correct and does use the Product-Sum Method with $b = 0$

Analyze the expression: $x^2 - 16$ can be rewritten as $x^2 + 0x - 16$. In this form, $b = 0$ and $c = -16$. We need two numbers that multiply to $-16$ and add to $0$. Verify the pair: $4 \times (-4) = -16$ and $4 + (-4) = 0$. The Product-Sum Method strictly applies, giving $(x + 4)(x - 4)$. Why distractors fail: Option A: The method works for any sign of $c$. Option C: $4 \times (-4) = -16$ is a true statement, not false. Option D: A quadratic binomial like $x^2 - 16$ is mathematically equivalent to a trinomial with a $0x$ term, so the method is perfectly valid.

Answer: Both numbers are positive

Analyze the sign of c: If $c > 0$, the two numbers must have the same sign (both positive or both negative), since only same-sign numbers yield a positive product. Analyze the sign of b: If $b > 0$, the sum of the two numbers is positive. Since they share the same sign and their sum is positive, both must be positive. Why distractors fail: Option A: two negatives would give a negative sum, contradicting $b > 0$. Option C: opposite signs would give a negative product, contradicting $c > 0$. Option D: the signs are fully determined by $b$ and $c$.

Answer: When $c < 0$, the factor pair has opposite signs; when $c > 0$, the factor pair has the same sign

Sign of the product determines sign pattern: If $c < 0$, the product of the two numbers is negative, which only happens with opposite signs. If $c > 0$, the product is positive, which requires same signs (both positive or both negative, determined by $b$). Why distractors fail: Option B: the Product-Sum Method works for negative $c$ as well. Option C: a factor of $x$ only occurs when $c = 0$. Option D: the sign assignment depends on $b$, not the magnitude of the factor.

Answer: Both are correct because multiplication is commutative

Apply the commutative property: Multiplication of binomials is commutative: $(x - 2)(x - 4) = (x - 4)(x - 2)$. The order does not affect the product. Verify the factorization: $(-2)(-4) = 8$ and $(-2) + (-4) = -6$. Both expressions correctly factor $x^2 - 6x + 8$. Why distractors fail: Options A and B impose an ordering rule that does not exist. Option D: with $c > 0$ and $b < 0$, the signs must indeed be negative.

Answer: $x^2 + 7x + 10$

Check each constraint: We need: $c$ is two-digit ($10 \leq c \leq 99$), $b$ is single-digit positive ($1 \leq b \leq 9$), both constants in the factors are positive, and the expression factors over the integers. Verify Option A: $x^2 + 7x + 10$: $c = 10$ (two-digit), $b = 7$ (single-digit positive). Factors: $(x+2)(x+5)$ since $2 \times 5 = 10$ and $2 + 5 = 7$. Both constants are positive. Why distractors fail: Option B: $c = -10$ is not a two-digit positive number and requires opposite-sign factors. Option C: $b = 11$ is not a single-digit number. Option D: $b = -7$ is negative, so the factor constants would be negative, not positive.

Answer: $(x - 3)(x + 7)$

Identify conditions: Need $m \times n = -21$ and $m + n = 4$. Since $c < 0$, the signs are opposite, and since $b > 0$, the positive number has the larger absolute value. Find the pair: $(-3) \times 7 = -21$ and $(-3) + 7 = 4$. So $x^2 + 4x - 21 = (x - 3)(x + 7)$. Why distractors fail: Option A: $3 + (-7) = -4 \neq 4$. Option C: $21 + (-1) = 20 \neq 4$. Option D: $(-21) + 1 = -20 \neq 4$.

Answer: $15$

Set up the consecutive integer condition: If $m$ and $n$ are consecutive positive integers with $mn = 56$, then $n = m + 1$. So $m(m+1) = 56$. Solve for m: $m^2 + m - 56 = 0$. Testing: $7 \times 8 = 56$, so $m = 7$ and $n = 8$. Find b: $b = m + n = 7 + 8 = 15$. Why distractors fail: Option A (13): no consecutive pair multiplies to 56 and sums to 13. Option C (17): $8 + 9 = 17$ but $8 \times 9 = 72 \neq 56$. Option D (11): $5 + 6 = 11$ but $5 \times 6 = 30 \neq 56$.

Answer: It reverses the factoring process, so if the original expression is recovered, the factorization must be correct

Expanding undoes factoring: Factoring rewrites a polynomial as a product. Expanding distributes that product back. If the result matches the original, the factorization is algebraically verified for all values of $x$. Why other options are weaker: Option A describes a graphical check, not an algebraic proof. Option C checks only specific values—two different expressions can agree at finitely many points without being identical. Option D addresses roots, not the correctness of a factorization.

Answer: $x^2 - 3x - 18$

Identify the constraints: We need $c < 0$ (opposite-sign factors) and $b < 0$ (the negative factor has the larger absolute value). Check the correct answer: $x^2 - 3x - 18$: $c = -18 < 0$ and $b = -3 < 0$. It factors as $(x + 3)(x - 6)$, where $3 \times (-6) = -18$ and $3 + (-6) = -3$. Why distractors fail: Option A: $b = 3 > 0$. Option C: $c = 18 > 0$, so both factors have the same sign. Option D: $b = 3 > 0$ and $c = 18 > 0$.

Answer: 9

List all opposite-sign integer pairs with product $-36$: Since $c = -36 < 0$, the two numbers must have opposite signs. The pairs $(m, n)$ with $mn = -36$ are: $(1, -36),\ (2, -18),\ (3, -12),\ (4, -9),\ (6, -6)$ and their reverses $(-1, 36),\ (-2, 18),\ (-3, 12),\ (-4, 9),\ (-6, 6)$. Compute the sums: The sums $m + n$ for each pair are: $-35,\ -16,\ -9,\ -5,\ 0,\ 35,\ 16,\ 9,\ 5,\ 0$. The value $0$ appears twice — from $(6, -6)$ and $(-6, 6)$ — so it is counted only once. Count distinct values: The distinct sums are $\{-35,\ -16,\ -9,\ -5,\ 0,\ 5,\ 9,\ 16,\ 35\}$, giving $\mathbf{9}$ distinct integer values of $b$. Why distractors fail: Option A (5) counts only the positive sums, ignoring negative ones. Option C (10) counts the 10 ordered pairs without recognising that $(6,-6)$ and $(-6,6)$ produce the same sum $b = 0$. Option D (18) overcounts by treating order within each binomial as significant.

Answer: They placed the signs incorrectly in the binomials

Analyze the error: The correct pair is $(-5)$ and $2$ since $(-5)(2) = -10$ and $(-5) + 2 = -3$. The student swapped the signs, writing $(x + 5)(x - 2)$ instead of $(x - 5)(x + 2)$. Expanding confirms the mistake: $(x + 5)(x - 2) = x^2 - 2x + 5x - 10 = x^2 + 3x - 10$. The middle term has the wrong sign. Why distractors fail: Option A: the factor pair values $5$ and $2$ are correct — only the signs are wrong. Option C: the $x^2$ term is present, so the first terms were multiplied. Option D: factor pairs of $c = -10$ were used, not of $b$.

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

Check each expression: For each, we need integer pairs that multiply to $c$ and add to $b$. Test $x^2 + 3x + 5$: Factor pairs of $5$: $(1, 5)$ with sum $6$, and $(-1, -5)$ with sum $-6$. Neither sum equals $3$, so this expression is not factorable over the integers. Why distractors fail: Option A: $2 + 3 = 5$ and $2 \times 3 = 6$. Option B: $(-2) + (-5) = -7$ and $(-2)(-5) = 10$. Option D: $3 + (-4) = -1$ and $3(-4) = -12$. All are factorable.

Answer: Expand $(x + 5)(x + 7)$ fully and confirm it equals $x^2 + 12x + 35$

Complete verification requires full expansion: While checking that $5 + 7 = 12$ and $5 \times 7 = 35$ confirms the factor pair, the most complete verification is to expand the entire product and confirm all three terms match. Perform the expansion: $(x+5)(x+7) = x^2 + 7x + 5x + 35 = x^2 + 12x + 35$. This confirms the factorization. Why other options are incomplete: Options A and B each check only one condition. Option C checks both conditions numerically but does not fully verify the algebraic identity — expanding does both and also confirms the $x^2$ coefficient is $1$.

Answer: Start with factor pairs of 72 near $\sqrt{72} \approx 8.5$ and assign opposite signs so the sum is $-1$

Efficiency of searching near the square root: Factor pairs closest to $\sqrt{c}$ tend to have the smallest difference, which is useful when $|b|$ is small relative to $c$. Since $b = -1$ is small, the two numbers are close in magnitude. Apply the strategy: Near $\sqrt{72} \approx 8.5$: try $8$ and $9$. Then $8 \times 9 = 72$, and with opposite signs, $8 + (-9) = -1 = b$. So $x^2 - x - 72 = (x + 8)(x - 9)$. Why distractors fail: Option A works but is inefficient — checking all 12 pairs is unnecessary. Options C and D use more advanced techniques that are overkill for a simple $a = 1$ factoring.