Answer: $1, 0, 0, 0, -16$

Identify all missing terms: $x^4 - 16$ is missing the $x^3$, $x^2$, and $x$ terms. Each must be represented with a coefficient of $0$. Why the correct answer works: The full expansion is $1x^4 + 0x^3 + 0x^2 + 0x - 16$, giving coefficients $1, 0, 0, 0, -16$. Why distractors fail: Option A omits three zero placeholders. Option C only includes one zero, missing two terms. Option D incorrectly inserts $-2$ (the divisor value) into the coefficient list.

Answer: Bring down, multiply, add

Recall the synthetic division cycle: Synthetic division uses a streamlined cycle: bring the coefficient down, multiply it by $c$, then add the result to the next coefficient. Why the correct answer works: The Bring-Multiply-Add cycle is the defining procedure of synthetic division, distinguishing it from polynomial long division's Divide-Multiply-Subtract-Bring Down cycle. Why distractors fail: Option A describes part of the polynomial long division cycle (DMSB). Option C has the operations out of order and uses subtraction instead of addition. Option D includes division, which is not part of synthetic division's cycle.

Answer: Perform synthetic division with $c = 3$ first, then apply synthetic division with $c = 1$ to the quotient

Understand successive division: Dividing by $(x - 3)$ and then dividing the quotient by $(x - 1)$ is equivalent to dividing the original polynomial by $(x - 3)(x - 1)$, but done in two sequential synthetic division steps. Why the correct answer works: Performing synthetic division twice — first with $c = 3$ and then with $c = 1$ on the resulting quotient — correctly factors out both linear binomials in succession. Why distractors fail: Option A incorrectly adds the $c$ values; synthetic division doesn't combine this way. Option C incorrectly uses the product. Option D, while mathematically valid, uses polynomial long division, not synthetic division, and wasn't what the student proposed.

Answer: That the synthetic division was performed correctly

Understand the verification identity: Polynomial division satisfies the identity: Dividend $=$ Divisor $\times$ Quotient $+$ Remainder. If the expansion matches the original dividend, the division was done correctly. Why the correct answer works: The student verified that $(x - 2)(2x^3 + x^2 + 2x + 5) + 5 = 2x^4 - 3x^3 + x - 5$, confirming the accuracy of their synthetic division. Why distractors fail: Option A is wrong because the remainder is $5$, not $0$, so $x = 2$ is not a root. Option C overgeneralizes from one example. Option D is irrelevant — remainders can be any value.

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

Interpret the bottom row: The bottom row $3, 1, 2, -2$ contains the quotient coefficients followed by the remainder. Since the dividend is degree 3, the quotient is degree 2. Why the correct answer works: The first three numbers ($3, 1, 2$) are the coefficients of $3x^2 + x + 2$. The last number ($-2$) is the remainder, not part of the quotient. Why distractors fail: Option A incorrectly makes the quotient degree 3 (same as the dividend). Option C improperly mixes the remainder into the quotient expression. Option D drops the leading coefficient $3$.

Answer: The claim is correct only if $c$ is also an integer

Consider the role of c: In the Bring-Multiply-Add cycle, each step multiplies a result by $c$ and then adds to an integer coefficient. If $c$ is an integer, the results remain integers. If $c$ is a fraction (e.g., dividing by $x - \frac{1}{2}$), non-integer values can appear. Why the correct answer works: When $c$ is an integer and all original coefficients are integers, every multiply and add operation stays within the integers. But the claim is too broad because $c$ can be non-integer. Why distractors fail: Option A ignores the possibility of non-integer $c$ values. Option C confuses the issue; the remainder is computed the same way and would also be integer if $c$ is. Option D is simply false — fractions are not always introduced.

Answer: $x^3 - 2x^2 + 9x - 38$ with remainder $180$

Set up: $(x+5) = (x - (-5))$, so $c = -5$. Coefficients: $1, 3, -1, 7, -10$. Execute the cycle: Bring down $1$. $1 \times (-5) = -5$; $3 + (-5) = -2$. $-2 \times (-5) = 10$; $-1 + 10 = 9$. $9 \times (-5) = -45$; $7 + (-45) = -38$. $-38 \times (-5) = 190$; $-10 + 190 = 180$. Read the result: Bottom row: $1, -2, 9, -38, 180$. Quotient: $x^3 - 2x^2 + 9x - 38$, remainder: $180$. Why distractors fail: Option A uses $c = 5$ instead of $c = -5$, a common sign error with $(x + 5)$. Option C makes a sign error partway through the cycle, computing $-1 - 10 = -11$ instead of $-1 + 10 = 9$. Option D mixes the first half of the $c = 5$ error with a corrected ending, producing an inconsistent result.

Answer: $x + 7$

Check each option against the requirement: Synthetic division requires a monic linear binomial of the form $(x - c)$. Why the correct answer works: $(x + 7) = (x - (-7))$, which is in the form $(x - c)$ with $c = -7$. It is linear and monic. Why distractors fail: Option A ($x^2 - 4$) is degree 2, not linear. Option C ($3x - 6$) has a leading coefficient of $3$, not $1$. Option D ($x^2 + x - 1$) is degree 2.

Answer: A linear binomial $(x - c)$ with leading coefficient 1

Restriction on the divisor: Standard synthetic division requires the divisor to be a monic linear binomial, meaning it must have the form $(x - c)$ where the coefficient of $x$ is $1$. Why the correct answer works: The shortcut relies on the leading coefficient of the divisor being $1$. The form $(x - c)$ is precisely this condition. Why distractors fail: Option A is too broad; quadratic divisors cannot be used in standard synthetic division. Option C fails because $(3x - 2)$ has a leading coefficient of $3$, not $1$. Option D is a monomial, not a binomial, and is not of the form $(x - c)$.

Answer: $2, 0, -4, 7$

Include all terms: In synthetic division, every degree of the polynomial must be represented. If a term is missing, a coefficient of $0$ is used as a placeholder. Why the correct answer works: The polynomial $2x^3 + 0x^2 - 4x + 7$ has coefficients $2, 0, -4, 7$ listed from highest to lowest degree, including the zero for the missing $x^2$ term. Why distractors fail: Option A omits the $0$ placeholder for $x^2$, which would produce an incorrect result. Option C includes $3$ (the divisor value) in the coefficient row, which is placed separately. Option D changes $-4$ to $4$, losing the sign of the coefficient.

Answer: $0$

Set up the synthetic division: Write $c = 2$ and list the coefficients: $1, -6, 11, -6$. Perform the Bring-Multiply-Add cycle: Bring down $1$. Multiply $1 \times 2 = 2$; add to $-6$: $-4$. Multiply $-4 \times 2 = -8$; add to $11$: $3$. Multiply $3 \times 2 = 6$; add to $-6$: $0$. Read the result: The bottom row is $1, -4, 3, 0$. The last entry $0$ is the remainder. The quotient is $x^2 - 4x + 3$. Why distractors fail: Option A ($-6$) is the constant term of the original polynomial, not the remainder. Option C ($2$) is the value of $c$. Option D ($4$) may result from a sign error in the second step.

Answer: The student used the wrong sign for $c$ in every multiply step

Identify the error pattern: $(x + 3) = (x - (-3))$, so $c = -3$. Using $c = 3$ instead means every multiplication produces a value with the wrong sign. Trace the propagation: With $c = 3$: $1 \times 3 = 3$, $2 + 3 = 5$; $5 \times 3 = 15$, $-5 + 15 = 10$; $10 \times 3 = 30$, $3 + 30 = 33$. With $c = -3$: $1 \times (-3) = -3$, $2 + (-3) = -1$; etc. The sign of $c$ is the single root cause. Why distractors fail: Option A is irrelevant — no terms are missing. Option C is wrong because the student did add (the correct operation), just with the wrong value. Option D is wrong because the coefficients $1, 2, -5, 3$ are correct; only $c$ is wrong.

Answer: $2x^2 + x - 2$ with remainder $7$

Rewrite the divisor: $(x + 1) = (x - (-1))$, so $c = -1$. Coefficients of the dividend: $2, 3, -1, 5$. Perform the cycle: Bring down $2$. Multiply $2 \times (-1) = -2$; add to $3$: $1$. Multiply $1 \times (-1) = -1$; add to $-1$: $-2$. Multiply $-2 \times (-1) = 2$; add to $5$: $7$. Read the result: Bottom row: $2, 1, -2, 7$. Quotient: $2x^2 + x - 2$, remainder: $7$. Why distractors fail: Option A results from using $c = 1$ instead of $c = -1$. Options C and D contain sign errors in the multiply-add steps.

Answer: The shortcut nature of the method

Interpret the student's claim: The student is highlighting that synthetic division streamlines the process by stripping away variable notation and operating only on numeric coefficients. Why the correct answer works: This directly describes why synthetic division is considered a shortcut: it reduces notational overhead and simplifies the mechanical steps of division. Why distractors fail: Option A describes a limitation, not the speed advantage. Option C addresses reading the result, not the efficiency of the process. Option D discusses the setup step, not the reason for the method's speed.

Answer: $5$

Apply the remainder principle: When dividing by $(x - 1)$, the remainder equals $p(1)$. So $p(1) = 1 + a - 4 + 6 = a + 3$. Solve for a: Setting $p(1) = 8$: $a + 3 = 8$, so $a = 5$. Verify with synthetic division: Using $c = 1$ and coefficients $1, 5, -4, 6$: bring down $1$; $1 \times 1 = 1$, $5 + 1 = 6$; $6 \times 1 = 6$, $-4 + 6 = 2$; $2 \times 1 = 2$, $6 + 2 = 8$. Remainder is $8$. ✓ Why distractors fail: Option A ($a = 3$) gives remainder $6$. Option C ($a = -5$) gives remainder $-2$. Option D ($a = 1$) gives remainder $4$.

Answer: The very first multiplication in the cycle

Trace the error: The first step brings down the leading coefficient, which is unaffected by $c$. The next step multiplies this coefficient by $c$, which is where the wrong value of $c$ first propagates an error. Why the correct answer works: After bringing down the first coefficient, the student multiplies it by $-4$ instead of $4$. This is the first operation that uses $c$, so it is the first point of error. Why distractors fail: Option A is wrong because the dividend coefficients are independent of $c$. Option C is wrong because the degree of the quotient depends only on the degree of the dividend and divisor, not $c$'s value. Option D is wrong because the number of entries depends on the number of coefficients, not $c$.

Answer: Yes, because the remainder (last entry) is 0

Meaning of remainder 0: In polynomial division, a remainder of $0$ means the divisor divides the dividend exactly — i.e., the divisor is a factor of the dividend. Why the correct answer works: The last entry in the bottom row is the remainder. Since it is $0$, the division has no remainder, confirming exact divisibility. Why distractors fail: Option A is irrelevant; negative coefficients do not affect divisibility. Option C misinterprets $0$ as meaning undefined, which is incorrect. Option D cites integer entries, which have nothing to do with exact divisibility.

Answer: No, because the divisor's leading coefficient is not 1

Check the divisor form: Standard synthetic division requires a divisor of the form $(x - c)$, where the coefficient of $x$ is $1$. Why the correct answer works: The divisor $(2x - 1)$ has a leading coefficient of $2$, not $1$. Therefore, standard synthetic division cannot be applied directly. Why distractors fail: Options A and C incorrectly claim it can be used by choosing some value of $c$, ignoring the leading coefficient issue. Option D is wrong because the number of terms in the dividend does not restrict synthetic division.

Answer: $5$

Identify the divisor form: Synthetic division works with divisors of the form $(x - c)$. You extract the value $c$ and place it in the divisor box. Why the correct answer works: Since the divisor is $(x - 5)$, we have $c = 5$. The value $5$ is placed in the box. Why distractors fail: Option A ($-5$) confuses the sign; students sometimes negate $c$ again. Option C places the entire binomial, which is not how synthetic division is set up. Option D uses the reciprocal, which is irrelevant to the procedure.

Answer: The remainder

Reading the bottom row: After completing synthetic division, the bottom row contains the quotient coefficients followed by the remainder as the final entry. Why the correct answer works: By convention, the last number in the bottom row is always the remainder of the division. Why distractors fail: Option A is incorrect because the leading coefficient of the quotient is the first number, not the last. Option C is wrong because $c$ is placed in the divisor box, not in the result row. Option D confuses a number with a degree; the degree of the quotient is one less than the dividend's degree.

Answer: It is exactly one degree lower than the dividend

Understand degree reduction: When you divide a polynomial of degree $n$ by a linear polynomial (degree 1), the quotient has degree $n - 1$. Why the correct answer works: Since synthetic division divides by a degree-1 polynomial $(x - c)$, the quotient is always one degree less than the dividend. For example, dividing a cubic yields a quadratic. Why distractors fail: Option A is wrong because dividing by a non-constant polynomial always reduces the degree. Option C would be correct only if dividing by a degree-2 polynomial. Option D is incorrect because the degree relationship depends on the degree of the divisor, not the value of $c$.

Answer: $22$

Trace the cycle so far: Coefficients: $1, 5, -2, 8$ with $c = 3$. Bring down $1$. Multiply $1 \times 3 = 3$; add to $5$: $8$. So far the bottom row is $1, 8$. Compute the next step: Multiply $8 \times 3 = 24$. Add to $-2$: $-2 + 24 = 22$. Why the correct answer works: Following the Bring-Multiply-Add cycle correctly yields $22$ as the next entry. Why distractors fail: Option A ($-2$) is the original coefficient without adding the product. Option C ($26$) results from adding to $2$ instead of $-2$. Option D ($6$) may come from multiplying $2 \times 3$ instead of $8 \times 3$.