Answer: $p(c)$

State the Remainder Theorem: The Remainder Theorem says that when a polynomial $p(x)$ is divided by the linear divisor $(x - c)$, the remainder is exactly $p(c)$. Why the correct answer works: By direct statement of the theorem, evaluating the polynomial at $x = c$ gives the remainder. So $p(c)$ is correct. Why distractors fail: $p(0)$ would only be the remainder if $c = 0$, i.e., dividing by $x$ itself. $p(-c)$ confuses the sign — the divisor is $(x - c)$, so we substitute $c$, not $-c$. The constant $c$ alone has no general relationship to the remainder.

Answer: $-5$

Identify $c$: The divisor is $(x + 1) = (x - (-1))$, so $c = -1$. Evaluate $p(-1)$: $p(-1) = (-1)^4 - 2(-1)^3 + (-1) - 7 = 1 + 2 - 1 - 7 = -5$. Conclusion: The remainder is $-5$. Why distractors fail: $-7$ is the constant term $p(0)$, not $p(-1)$. $-11$ results from incorrectly computing $(-1)^4 = -1$ instead of $1$. $3$ comes from sign errors in multiple terms.

Answer: $p(x)$ can be written as $(x - c) \cdot q(x)$ where $q(x)$ has degree $n - 1$.

Apply the Factor Theorem: The Factor Theorem guarantees that $p(c) = 0$ implies $(x - c)$ divides $p(x)$, so $p(x) = (x - c) \cdot q(x)$ where $\deg(q) = n - 1$. Why the correct answer works: This is precisely what the Factor Theorem asserts: one factor of $(x-c)$ is guaranteed, yielding a quotient of degree one less. Why distractors fail: $(x - c)^2$ divides $p(x)$ only if $c$ is a repeated root, which is not guaranteed. $c$ can be any real number, not necessarily a positive integer. There is no requirement on divisibility of coefficients by $c$.

Answer: $0$

Apply the Remainder Theorem: The remainder when dividing by $(x - 2)$ is $p(2)$. Evaluate $p(2)$: $p(2) = 3(8) - 7(4) + 4 = 24 - 28 + 4 = 0$. Conclusion: Since $p(2) = 0$, the remainder is 0. This also means $(x - 2)$ is a factor of $p(x)$ by the Factor Theorem. Why distractors fail: $4$ is the constant term $p(0)$. $-8$ and $12$ arise from arithmetic mistakes in evaluating $p(2)$.

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

Test $c = 1$: $p(1) = 1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor. Perform synthetic division by $(x - 1)$: Coefficients $[1, -6, 11, -6]$ with $c = 1$: bottom row gives $[1, -5, 6, 0]$. Quotient: $x^2 - 5x + 6$. Factor the quadratic: $x^2 - 5x + 6 = (x - 2)(x - 3)$, since $(-2)(-3) = 6$ and $(-2) + (-3) = -5$. Why distractors fail: Options with $+3$, $+2$, or $+1$ produce incorrect products. For example, $(x-1)(x-2)(x+3)$ expands to $x^3 - 13x + 6$, not matching $p(x)$.

Answer: The remainder is 0, so $(x + 1)$ is a factor and the quotient is $x^2 + x - 2$.

Read the synthetic division output: The bottom row $[1, 1, -2, 0]$ has the last entry as the remainder (0) and the first three entries as the quotient coefficients: $x^2 + x - 2$. Conclude using the Factor Theorem: Remainder = 0 means $p(-1) = 0$, so $(x - (-1)) = (x + 1)$ is a factor, and $p(x) = (x + 1)(x^2 + x - 2)$. Why distractors fail: The remainder is the last number (0), not 1 or $-2$. The quotient coefficient for $x^0$ is $-2$, not $+2$.

Answer: The Factor Theorem is the special case of the Remainder Theorem where the remainder is zero.

Connect the two theorems: The Remainder Theorem states that dividing $p(x)$ by $(x - c)$ leaves remainder $p(c)$. Setting that remainder equal to zero gives the Factor Theorem: $p(c) = 0 \Leftrightarrow (x - c)$ is a factor. Why the correct answer works: The Factor Theorem is literally the Remainder Theorem with the extra condition that the remainder is 0, making it a special case. Why distractors fail: Both theorems apply to any polynomial, not just quadratics. They are closely related, not unrelated. The logical derivation goes from the Remainder Theorem to the Factor Theorem, not the reverse.

Answer: $p(c) = 0$

State the Factor Theorem: The Factor Theorem is a special case of the Remainder Theorem: $(x - c)$ is a factor of $p(x)$ if and only if $p(c) = 0$. Why the correct answer works: A factor divides evenly, meaning the remainder is zero. Since the remainder equals $p(c)$, the condition is $p(c) = 0$. Why distractors fail: $p(c) = 1$ would mean a nonzero remainder of 1. $p(c) = c$ is an arbitrary condition unrelated to factoring. $p(0) = c$ evaluates the polynomial at the wrong input.

Answer: Both methods are valid and will reach the same conclusion, but Student B's method additionally provides the quotient polynomial for further factoring.

Compare the two approaches: Student A uses the Remainder Theorem: if $p(4) = 0$, then $(x-4)$ is a factor. Student B uses synthetic division, which also produces the remainder and additionally gives the quotient polynomial. Advantage of synthetic division: Both approaches correctly test factorhood. However, synthetic division also outputs the quotient, enabling further factoring — a practical advantage. Why distractors fail: Both methods are valid, so options claiming only one works are wrong. The methods always agree regardless of degree.

Answer: $1$

Apply the Remainder Theorem: When dividing by $(x - 1)$, the remainder is $p(1)$. Evaluate $p(1)$: $p(1) = 1^3 + 2(1)^2 - 5(1) + 3 = 1 + 2 - 5 + 3 = 1$. Why distractors fail: $0$ would mean $(x-1)$ is a factor, but the remainder is 1, not 0. $3$ is the constant term $p(0)$, not $p(1)$. $-1$ results from a sign error in the evaluation.

Answer: Yes in general, but for a simple value like $c = -1$ the computation is comparable; synthetic division and direct substitution via the Remainder Theorem give the same result either way.

Clarify the connection: By the Remainder Theorem, synthetic division with $c = -1$ yields $p(-1)$ as the remainder, so it is a valid method for evaluating a polynomial. Compare efficiency: Synthetic division uses Horner's method (multiply and add), which is generally efficient. However, for $c = -1$, multiplication is trivial (just sign changes), so direct substitution is equally easy. Why distractors fail: Synthetic division does not always require fewer operations — it depends on $c$. Synthetic division absolutely can evaluate a polynomial (that's the Remainder Theorem). Missing terms are handled by inserting zero coefficients.

Answer: 2

Identify the factorization: $p(x) = (x - 1)(x + 2)(x^2 + x + 1)$, where $x^2 + x + 1$ has discriminant $1 - 4 = -3 < 0$, so no real roots. Count real roots: The factors $(x - 1)$ and $(x + 2)$ each contribute one real root ($x = 1$ and $x = -2$). The irreducible quadratic contributes none. Total: 2 real roots. Why distractors fail: 4 assumes all roots are real, ignoring the irreducible quadratic. 3 may stem from miscounting. 0 ignores the two known real roots.

Answer: $p(x) = q(x) \cdot (x - c) + r$

State the division algorithm for polynomials: Just as with integers, polynomial division satisfies: dividend = divisor × quotient + remainder, i.e., $p(x) = (x - c) \cdot q(x) + r$. Connection to the Remainder Theorem: Substituting $x = c$: $p(c) = (c - c) \cdot q(c) + r = 0 + r = r$. This is exactly the Remainder Theorem. Why distractors fail: Option A mixes up addition and multiplication roles. Option C reverses quotient and remainder placement. Option D has $p(x)$ as a factor of $q(x)$, which inverts the relationship.

Answer: $x^3 - 2$

Set up synthetic division: Dividing by $(x + 3) = (x - (-3))$, use $c = -3$ with coefficients $[1, 3, 0, -2, -6]$. Note the $0$ placeholder for the missing $x^2$ term. Perform synthetic division: Bring down 1. $1 \times (-3) = -3$; $3 + (-3) = 0$. $0 \times (-3) = 0$; $0 + 0 = 0$. $0 \times (-3) = 0$; $-2 + 0 = -2$. $(-2) \times (-3) = 6$; $-6 + 6 = 0$. Bottom row: $[1, 0, 0, -2, 0]$. Read the result: The remainder is 0, confirming $(x + 3)$ is a factor. The quotient coefficients are $[1, 0, 0, -2]$, giving $x^3 + 0x^2 + 0x - 2 = x^3 - 2$. Verify: $(x + 3)(x^3 - 2) = x^4 - 2x + 3x^3 - 6 = x^4 + 3x^3 - 2x - 6$ ✓. Why distractors fail: $(x+3)(x^3 - x + 2) = x^4 + 3x^3 - x^2 - x + 6 \neq p(x)$. $(x+3)(x^3 + 6x^2 + 17x + 42) $ has degree 4 with much larger coefficients. $(x+3)(x^3 - x - 6) = x^4 + 3x^3 - x^2 - 9x - 18 \neq p(x)$.

Answer: $k = 2$

Apply the Factor Theorem condition: $(x - 3)$ is a factor if and only if $p(3) = 0$. Set up and solve for $k$: $p(3) = 27 - 45 + 3k + 12 = 3k - 6 = 0$, so $k = 2$. Why distractors fail: $k = -2$ gives $p(3) = -12 \neq 0$. $k = 8$ gives $p(3) = 18 \neq 0$. $k = -4$ gives $p(3) = -18 \neq 0$.

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

Use the Factor Theorem on each root: Since $p(1) = 0$, $p(2) = 0$, and $p(3) = 0$, the Factor Theorem gives us factors $(x-1)$, $(x-2)$, and $(x-3)$. Check degrees: $p(x)$ is degree 3 and the product $(x-1)(x-2)(x-3)$ is also degree 3 with leading coefficient 1, matching $p(x)$. So the factorization is complete. Why distractors fail: $(x-1)(x^2-5x+6)$ is partially factored but not complete since $x^2-5x+6 = (x-2)(x-3)$. $(x+1)(x+2)(x+3)$ uses the wrong signs. $(x-1)(x-2)(x+3)$ incorrectly uses $+3$ instead of $-3$.

Answer: The Factor Theorem only tells us about linear factors of the form $(x - c)$; the polynomial may still have factors over extensions of the real numbers or may simply be irreducible over the reals.

Identify the logical gap: Testing only two values does not rule out all possible roots. More importantly, the Factor Theorem identifies linear factors $(x - c)$ where $c$ is a number in the domain being considered. Clarify the scope: $x^2 + 1$ is indeed irreducible over the real numbers, but the student's reasoning — testing just two values — is insufficient. The correct conclusion requires showing that $x^2 + 1 > 0$ for all real $x$. Why distractors fail: Testing $p(0) = 1 \neq 0$ wouldn't help either. The arithmetic is correct. The Factor Theorem applies to polynomials of any degree.

Answer: Evaluate $p(3) = 2(3)^3 - 3(3)^2 + 4(3) - 5$

Identify the correct substitution: The divisor is $(x - 3)$, so $c = 3$. By the Remainder Theorem, the remainder is $p(3)$. Compute $p(3)$: $p(3) = 2(27) - 3(9) + 4(3) - 5 = 54 - 27 + 12 - 5 = 34$. Why distractors fail: $p(-3)$ would be the remainder when dividing by $(x + 3)$, not $(x - 3)$. $p(0)$ gives the remainder for division by $x$. $p(1)$ gives the remainder for division by $(x - 1)$.

Answer: The polynomial $p(x)$ can be written as $(x - c) \cdot q(x)$ for some polynomial $q(x)$ with no remainder.

Define what 'factor' means: If $(x - c)$ is a factor of $p(x)$, then $p(x)$ divides evenly by $(x - c)$, meaning $p(x) = (x - c) \cdot q(x)$ with remainder 0. Why the correct answer works: This is the precise definition: exact division with no leftover, so $p(x)$ is the product of $(x - c)$ and a quotient polynomial $q(x)$. Why distractors fail: A remainder of $c$ means $(x-c)$ is not a factor. The value $c$ has nothing to do with the leading coefficient. The degree of $p(x)$ is unrelated to the value $c$.

Answer: The last number in the synthetic division row is 0.

Recall synthetic division output: In synthetic division, the last value in the bottom row is the remainder when dividing by $(x - c)$. Apply the Factor Theorem: If that remainder is 0, then $p(2) = 0$, confirming $x = 2$ is a root by the Factor Theorem. Why distractors fail: The quotient always has degree one less regardless of whether $c$ is a root. The signs of the bottom row coefficients are irrelevant to whether $c$ is a root. The first coefficient in the bottom row simply carries down from the leading coefficient.

Answer: Yes, because if $q(x) = p(x) - 5$, then $q(3) = p(3) - 5 = 0$, so $(x - 3)$ is a factor of $q(x)$.

Analyze the claim: Define $q(x) = p(x) - 5$. Then $q(3) = p(3) - 5 = 5 - 5 = 0$. Apply the Factor Theorem to $q(x)$: Since $q(3) = 0$, the Factor Theorem guarantees that $(x - 3)$ is a factor of $q(x) = p(x) - 5$. Why distractors fail: The Remainder Theorem applies to any polynomial, including $p(x) - 5$. Saying no factoring can be done ignores the valid algebraic manipulation. The theorem works regardless of whether coefficients are integers.

Answer: $x = 2$ is a root and $p(x) = (x - 2)(2x^2 + 5x - 3)$.

Interpret the zero remainder: A remainder of 0 from synthetic division by $(x - 2)$ means $p(2) = 0$, so $x = 2$ is a root and $(x - 2)$ is a factor. Verify the quotient: Synthetic division with $c = 2$ on coefficients $[2, 1, -13, 6]$: bring down 2; $2 \times 2 = 4$, $1 + 4 = 5$; $5 \times 2 = 10$, $-13 + 10 = -3$; $-3 \times 2 = -6$, $6 + (-6) = 0$. Quotient: $2x^2 + 5x - 3$. Why distractors fail: $x = -2$ was not tested; synthetic division tested $c = 2$. Saying $p(x)$ has no real roots contradicts the fact that $x = 2$ is a root. $(x + 2)$ being a factor would require $p(-2) = 0$, which was not established.

Answer: Yes, because $p(2) = 0$.

Apply the Factor Theorem: $(x - 2)$ is a factor if and only if $p(2) = 0$. Evaluate $p(2)$: $p(2) = 8 - 16 + 2 + 6 = 0$. Since $p(2) = 0$, $(x - 2)$ is indeed a factor. Why distractors fail: $p(2) = 4$ and $p(2) = -2$ are incorrect evaluations. $p(-2) = 0$ tests the wrong value — that would check whether $(x + 2)$ is a factor.

Answer: $2x^3 - 4x^2 - 10x + 12$

Build from the Factor Theorem: Roots at $x = 1, -2, 3$ mean the factors are $(x - 1)(x + 2)(x - 3)$. To get leading coefficient 2, multiply by 2: $p(x) = 2(x - 1)(x + 2)(x - 3)$. Expand step by step: $(x - 1)(x + 2) = x^2 + x - 2$. Then $(x^2 + x - 2)(x - 3) = x^3 - 3x^2 + x^2 - 3x - 2x + 6 = x^3 - 2x^2 - 5x + 6$. Finally, $2(x^3 - 2x^2 - 5x + 6) = 2x^3 - 4x^2 - 10x + 12$. Verify roots: $p(1) = 2 - 4 - 10 + 12 = 0$ ✓. $p(-2) = -16 - 16 + 20 + 12 = 0$ ✓. $p(3) = 54 - 36 - 30 + 12 = 0$ ✓. Why distractors fail: Option B: $p(1) = 2 + 4 - 10 - 12 = -16 \neq 0$. Option C: has leading coefficient 1, not 2. Option D: $p(1) = 2 - 4 + 10 + 12 = 20 \neq 0$.

Answer: $a = -2, \; b = -5$

Set up equations using the Factor Theorem: $(x - 1)$ is a factor $\Rightarrow p(1) = 0$: $1 + a + b + 6 = 0 \Rightarrow a + b = -7$. $(x + 2)$ is a factor $\Rightarrow p(-2) = 0$: $-8 + 4a - 2b + 6 = 0 \Rightarrow 4a - 2b = 2 \Rightarrow 2a - b = 1$. Solve the system: From $a + b = -7$ and $2a - b = 1$: adding gives $3a = -6$, so $a = -2$. Then $b = -7 - (-2) = -5$. Verify: $p(x) = x^3 - 2x^2 - 5x + 6$. Check: $p(1) = 1 - 2 - 5 + 6 = 0$ ✓. $p(-2) = -8 - 8 + 10 + 6 = 0$ ✓. The third root is $x = 3$ since $(x-1)(x+2)(x-3) = x^3 - 2x^2 - 5x + 6$. Why distractors fail: $a = 2, b = -5$: $p(1) = 1 + 2 - 5 + 6 = 4 \neq 0$. $a = -2, b = -1$: $p(1) = 1 - 2 - 1 + 6 = 4 \neq 0$. $a = 2, b = 1$: $p(1) = 1 + 2 + 1 + 6 = 10 \neq 0$.