Answer: Second difference = 4; $a_5 = 36$

Find the second difference: For a quadratic $an^2 + bn + c$, the constant second difference equals $2a$. Here $a = 2$, so the second difference is $2 \times 2 = 4$. Compute $a_5$: $a_5 = 2(25) - 3(5) + 1 = 50 - 15 + 1 = 36$. Why distractors fail: Option A has the wrong second difference (2 instead of 4). Option C computes $a_5$ incorrectly, perhaps as $2(25) - 3(5) + 1 = 50 - 10 + 1 = 41$, miscalculating $3 \times 5$. Option D has both values wrong.

Answer: Yes, because expanding $(n+1)^3 - (n+1)$ gives $n^3 + 3n^2 + 3n + 1 - n - 1 = n^3 + 3n^2 + 2n$

Expand $(n+1)^3$: $(n+1)^3 = n^3 + 3n^2 + 3n + 1$. Subtract $(n+1)$: $(n+1)^3 - (n+1) = n^3 + 3n^2 + 3n + 1 - n - 1 = n^3 + 3n^2 + 2n$. Compare with the original: This matches $n(n+1)(n+2) = n^3 + 3n^2 + 2n$ exactly, so the claim is valid for all values of $n$, not just positive integers. Why distractors fail: Option A adds an extra $+1$ that should cancel. Option C is wrong because $(n+1)^3 - (n+1)$ is still degree 3. Option D incorrectly limits the algebraic identity to positive integers, when it holds for all real $n$.

Answer: 3

Expand step by step: First, $n(n+1) = n^2 + n$. Then multiply by $(n+2)$: $(n^2 + n)(n + 2) = n^3 + 2n^2 + n^2 + 2n = n^3 + 3n^2 + 2n$. Identify the coefficient: In $n^3 + 3n^2 + 2n$, the coefficient of $n^2$ is $3$. Why distractors fail: Option A (1) is the coefficient of $n^3$. Option B (2) is the coefficient of $n$. Option D (6) may come from multiplying $1 \times 2 \times 3$ or other miscalculation.

Answer: The claim is false; for example, when $x = 3$ the product is odd

Analyze the parity: Parity (even/odd) is defined for integers. When $x$ is an integer, we consider two cases: if $x$ is even, then $20 - x$ is also even, so the product is even. If $x$ is odd, then $20 - x$ is odd (even − odd = odd), so the product is odd × odd = odd. Find a counterexample: When $x = 3$: $x(20 - x) = 3 \times 17 = 51$, which is odd. This disproves the claim. Why the correct answer works: Option B correctly identifies that the claim is false and provides a valid counterexample. Any odd integer value of $x$ will produce an odd result. Why distractors fail: Option A is wrong because $20x - x^2$ is not always even — it is odd whenever $x$ is odd. Option C is wrong because the claim fails specifically when $x$ is an odd integer, so restricting to integers does not make it true. Option D identifies a relevant fact ($x^2$ can be odd) but does not fully explain why the product is odd; the real reason is that both factors are odd when $x$ is odd.

Answer: $n^2 + n$

Expand the product: Distribute $n$ across $(n + 1)$: $n(n + 1) = n \cdot n + n \cdot 1 = n^2 + n$. Why the correct answer works: Option B, $n^2 + n$, is the direct result of distributing $n$ over $(n + 1)$. Why distractors fail: Option A ($n^2 + 1$) incorrectly replaces $n$ with $1$ in the second term. Option C ($2n + 1$) is the sum of two consecutive integers, not their product. Option D ($n^2 + 2n$) would come from $n(n + 2)$, which is the product of integers two apart.

Answer: Because expanding $n(n+1)$ yields $n^2 + n$, a degree 2 polynomial in $n$

Define quadratic: A quadratic expression in $n$ is any polynomial of degree 2 in $n$, meaning the highest power of $n$ is 2. Why the correct answer works: $n(n+1) = n^2 + n$. The highest power of $n$ is 2, so it is a degree 2 polynomial — by definition, a quadratic expression. Why distractors fail: Option A is false; the product of two constants (e.g., $3 \times 4$) is not quadratic. Option C describes a property of the product's value but does not explain why the expression is quadratic. Option D is wrong because $n+1$ is degree 1, i.e., linear.

Answer: $n^2$

Observe the pattern: The sums are 1, 4, 9, 16, 25, … which are $1^2, 2^2, 3^2, 4^2, 5^2$. So the sum of the first $n$ odd numbers is $n^2$. Verify with second differences: The sequence $1, 4, 9, 16, 25$ has first differences $3, 5, 7, 9$ and second differences $2, 2, 2$ — constant, confirming a quadratic rule. Why distractors fail: Option A ($n(n+1)$) gives $2, 6, 12, 20$, which are the products of consecutive integers. Option B ($2n^2 - n$) gives $1, 6, 15, 28$. Option D ($n^2 + 1$) gives $2, 5, 10, 17$, all off by 1.

Answer: The rule is quadratic

Recall the second-difference test: When the first differences of a sequence are not constant but the second differences are constant, the sequence follows a quadratic rule of the form $an^2 + bn + c$. Why the correct answer works: Constant second differences are the defining property of a quadratic sequence, so the rule is quadratic. Why distractors fail: Option A (linear) would require constant first differences. Option C (exponential) would show constant ratios between terms. Option D (cubic) would require constant third differences, not second.

Answer: $x(S - x)$

Set up the relationship: If two numbers add to $S$ and one is $x$, the other must be $S - x$. Form the product: The product of the two numbers is $x(S - x) = Sx - x^2$, which is a quadratic expression in $x$. Why distractors fail: Option A ($xS$) treats $S$ as the second number, ignoring that the other number is $S - x$. Option C ($S(S - x)$) replaces $x$ with $S$ incorrectly. Option D ($x^2 - S$) has no valid derivation from the given conditions.

Answer: $n^2 + 2n$

Use the constant second difference: If the second difference is 2, then $2a = 2$, so $a = 1$. The general form is $n^2 + bn + c$. Find b and c by substitution: For $n = 1$: $1 + b + c = 3$, so $b + c = 2$. For $n = 2$: $4 + 2b + c = 8$, so $2b + c = 4$. Subtracting: $b = 2$, then $c = 0$. The rule is $n^2 + 2n$. Verify: $n = 3$: $9 + 6 = 15$ ✓. $n = 4$: $16 + 8 = 24$ ✓. Why distractors fail: Option A gives $n = 1 \to 3$ ✓ but $n = 2 \to 7 \neq 8$. Option C gives $n = 1 \to 3$ ✓ but $n = 2 \to 9 \neq 8$. Option D gives $n = 1 \to 3$ ✓ but $n = 2 \to 9 \neq 8$.

Answer: The student confused 'five more than a number squared' with 'five more than a number, squared'; the correct expression is $x^2 + 5$

Parse the phrase carefully: 'A number squared' means $x^2$. 'Five more than a number squared' means $x^2 + 5$. Identify the student's error: The student grouped 'five more than a number' first to get $(x + 5)$, then squared the whole thing, yielding $(x + 5)^2$. This misinterprets the order of operations in the verbal phrase. Why distractors fail: Option A ($x^2 - 5$) would translate 'five less than a number squared.' Option C ($5x^2$) translates 'five times a number squared.' Option D is incorrect because $(x + 5)^2 \neq x^2 + 5$.

Answer: The first student is correct; rearranging $x(12 - x) = 35$ gives $x^2 - 12x + 35 = 0$

Verify the first student's work: $x(12 - x) = 35 \Rightarrow 12x - x^2 = 35 \Rightarrow -x^2 + 12x - 35 = 0 \Rightarrow x^2 - 12x + 35 = 0$ (multiply by $-1$). This is correct. Check the factoring: $x^2 - 12x + 35 = (x - 5)(x - 7) = 0$, giving $x = 5$ or $x = 7$. Indeed $5 + 7 = 12$ and $5 \times 7 = 35$. ✓ Evaluate the second student's claim: $x^2 + 12x - 35 = 0$ does not arise from the correct rearrangement. The second student made sign errors on both the $12x$ and $35$ terms. Why other distractors fail: Option B's equation $x^2 - 12x - 35 = 0$ has an incorrect sign on 35. Option D is wrong because the problem clearly has the solution $x = 5$ or $x = 7$.

Answer: $x(14 - x)$; maximum value is 49

Form the expression: If two numbers add to 14 and one is $x$, the other is $14 - x$. Their product is $x(14 - x) = 14x - x^2$. Find the maximum: The expression $-x^2 + 14x$ is a downward-opening parabola. Its vertex is at $x = \frac{14}{2} = 7$, giving a product of $7(14 - 7) = 7 \times 7 = 49$. Why distractors fail: Option A has the correct expression but an incorrect maximum of 48 (likely from trying $6 \times 8$). Option C uses $x(x - 14)$, which gives a negative product for values between 0 and 14. Option D writes $14x - x$ (which simplifies to $13x$, a linear expression) rather than $14x - x^2$.

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

Translate piece by piece: 'The square of a number' → $x^2$. 'Increased by the product of the number and 7' → $+ 7x$. 'Decreased by 10' → $- 10$. Combined: $x^2 + 7x - 10$. Why the correct answer works: Option A, $x^2 + 7x - 10$, matches each part of the verbal phrase exactly. Why distractors fail: Option B reverses the signs on $7x$ and $10$. Option C squares $(x + 7)$ instead of just $x$, misinterpreting the grouping. Option D interprets 'product of the number and 7' as $7x^2$, incorrectly applying the 7 to $x^2$.

Answer: 55

Substitute into the formula: $T_{10} = \frac{10(10 + 1)}{2} = \frac{10 \times 11}{2} = \frac{110}{2} = 55$. Why the correct answer works: Direct substitution of $n = 10$ into the quadratic formula $\frac{n(n+1)}{2}$ gives 55. Why distractors fail: Option A (45) is $T_9 = \frac{9 \times 10}{2}$. Option B (50) may come from computing $\frac{10 \times 10}{2}$, forgetting the $+1$. Option D (60) may result from computing $\frac{10 \times 12}{2}$, adding 2 instead of 1.

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

Interpret the phrase: 'A number' is $x$. '3 more than itself' is $x + 3$. 'A number multiplied by 3 more than itself' is $x(x + 3)$. 'Then 10 is subtracted' gives $x(x + 3) - 10$. Expand and simplify: $x(x + 3) - 10 = x^2 + 3x - 10$. Why distractors fail: Option B ($3x^2 - 10$) misreads '3 more than the number' as '3 times the number.' Option C ($x^2 - 3x - 10$) uses subtraction instead of addition for the $3x$ term. Option D ($(x+3)^2 - 10$) squares $(x+3)$ instead of multiplying $x$ by $(x+3)$.

Answer: Find two numbers that add to 6 and multiply to 8

Connect sum-product problems to quadratic equations: If two numbers add to $S$ and multiply to $P$, they are roots of $x^2 - Sx + P = 0$. For $x^2 - 6x + 8 = 0$, we have $S = 6$ and $P = 8$. Verify Option A: Two numbers adding to 6 and multiplying to 8 are roots of $x^2 - 6x + 8 = 0$, which factors as $(x - 2)(x - 4) = 0$. The numbers are 2 and 4. ✓ Why distractors fail: Option B leads to $n(n+1) = 8$, i.e., $n^2 + n - 8 = 0$, a different equation. Option C translates to $x^2 = 14$, which is not $x^2 - 6x + 8 = 0$. Option D translates to $x(x - 6) + 8$, which equals the expression $x^2 - 6x + 8$ but does not set up an equation to solve — it describes a computation with no condition to satisfy.

Answer: The square of a number decreased by three times the number

Translate each phrase to algebra: Option B says 'the square of a number' → $x^2$, then 'decreased by three times the number' → $- 3x$. Combined: $x^2 - 3x$. Why the correct answer works: Option B directly maps to $x^2 - 3x$ through standard verbal-to-algebraic translation. Why distractors fail: Option A translates to $x^2 - 3$, not $x^2 - 3x$. Option C translates to $(x - 3)^2 = x^2 - 6x + 9$. Option D translates to $2 - 3x$, which is linear, not quadratic.

Answer: $4n^2 + 4n$

Expand using the distributive property: $2n(2n + 2) = 2n \cdot 2n + 2n \cdot 2 = 4n^2 + 4n$. Why the correct answer works: Option B, $4n^2 + 4n$, is the result of correctly distributing $2n$ over both terms. Why distractors fail: Option A ($4n^2 + 2n$) incorrectly computes $2n \times 2 = 2n$ instead of $4n$. Option C ($2n^2 + 4n$) incorrectly computes $2n \times 2n = 2n^2$. Option D ($4n^2 + 2$) drops the $n$ from the second term.