Answer: Evaluate the dividend at $x = 3$ and check if the result equals $31$

Apply the Remainder Theorem connection: When dividing by a linear divisor $(x - c)$, the remainder equals $P(c)$. Here $c = 3$, so the remainder should equal $P(3) = 27 + 18 - 15 + 1 = 31$. Why the correct answer works: Option B uses the Remainder Theorem: $P(3) = 3^3 + 2(3^2) - 5(3) + 1 = 27 + 18 - 15 + 1 = 31$, which matches the claimed remainder. This is a quick verification without redoing the full division. Why distractors fail: Option A only checks the degree, not the correctness of coefficients. Options C and D propose invalid verification methods that have no mathematical basis.

Answer: They incorrectly computed the exponent; $\frac{6x^3}{3x} = 2x^2$, not $2x^3$

Identify the error: When dividing monomials, both coefficients and exponents are handled separately: $\frac{6x^3}{3x} = \frac{6}{3} \cdot \frac{x^3}{x} = 2x^{3-1} = 2x^2$. Why the correct answer works: Option B correctly identifies that the student failed to subtract the exponents. The rule $\frac{x^a}{x^b} = x^{a-b}$ gives $x^{3-1} = x^2$, not $x^3$. Why distractors fail: Option A is wrong because you always divide the leading terms, not the constant terms. Option C is wrong because you must start with the highest-degree term. Option D is wrong because only the leading term of the divisor is used in the division step (the full divisor is used later in the multiply step).

Answer: $x^2 + 2x + 1$

Recognize the structure: Notice that $x^2 + 2x + 1 = (x+1)^2$ and $x^4 + 4x^3 + 6x^2 + 4x + 1 = (x+1)^4$. Divide: $\frac{(x+1)^4}{(x+1)^2} = (x+1)^2 = x^2 + 2x + 1$. Verification by long division: Alternatively, dividing leading terms: $x^4 \div x^2 = x^2$. Multiply: $x^2(x^2+2x+1)=x^4+2x^3+x^2$. Subtract: $2x^3+5x^2+4x+1$. Next: $2x^3 \div x^2 = 2x$. Multiply: $2x(x^2+2x+1)=2x^3+4x^2+2x$. Subtract: $x^2+2x+1$. Next: $x^2 \div x^2 = 1$. Multiply: $x^2+2x+1$. Subtract: $0$. Quotient is $x^2+2x+1$. Why distractors fail: Options A, C, and D, when multiplied by $x^2+2x+1$, do not produce the original dividend.

Answer: $2x^2 - x + 1$

Perform the first division step: Divide $2x^3$ by $x$ to get $2x^2$. Multiply: $2x^2(x+2)=2x^3+4x^2$. Subtract from $2x^3+3x^2$: $(2x^3+3x^2)-(2x^3+4x^2)=-x^2$. Bring down $-x$: $-x^2-x$. Second division step: Divide $-x^2$ by $x$ to get $-x$. Multiply: $-x(x+2)=-x^2-2x$. Subtract: $(-x^2-x)-(-x^2-2x)=x$. Bring down $+5$: $x+5$. Third division step: Divide $x$ by $x$ to get $1$. Multiply: $1(x+2)=x+2$. Subtract: $(x+5)-(x+2)=3$. The remainder is $3$ and the quotient is $2x^2 - x + 1$. Why distractors fail: Option A results from sign errors in the subtraction steps. Option C results from incorrectly adding instead of subtracting. Option D has an incorrect constant term from a computation error.

Answer: $2x^3 - x^2 + 1$

Use the division identity: $P(x) = D(x) \cdot Q(x) + R(x) = (x^2+x+1)(2x-3) + (x+4)$. Expand the product: $(x^2+x+1)(2x-3) = 2x^3 - 3x^2 + 2x^2 - 3x + 2x - 3 = 2x^3 - x^2 - x - 3$. Add the remainder: $P(x) = 2x^3 - x^2 - x - 3 + x + 4 = 2x^3 - x^2 + 0x + 1 = 2x^3 - x^2 + 1$. Why the correct answer works: Option B matches the computed result $2x^3 - x^2 + 1$ exactly. Why distractors fail: Option A ($2x^3 - x^2 - x + 4$) fails to combine the $x$ terms ($-x + x = 0$) and the constants ($-3 + 4 = 1$) correctly. Option C ($2x^3 - x^2 - x + 1$) correctly computes the constant but does not cancel the $x$ terms. Option D ($2x^3 + x^2 + x + 4$) has sign errors in multiple terms.

Answer: $x^4 + 2x^3 + 2x^2 - 6x - 15$

Multiply divisor and quotient: $(x^2-3)(x^2+2x+5) = x^4 + 2x^3 + 5x^2 - 3x^2 - 6x - 15 = x^4 + 2x^3 + 2x^2 - 6x - 15$. Why the correct answer works: Option A matches the expanded product exactly. Why distractors fail: Option B has $5x^2$ instead of $2x^2$ (forgot to combine $5x^2 - 3x^2$). Option C has $+6x$ instead of $-6x$ (sign error). Option D has multiple sign errors.

Answer: The result is correct because $(x-1)(2x^2-3x+1) = 2x^3-5x^2+4x-1$

Verify by multiplication: $(x-1)(2x^2-3x+1) = 2x^3 - 3x^2 + x - 2x^2 + 3x - 1 = 2x^3 - 5x^2 + 4x - 1$. Check against the original: The product equals the original dividend exactly, confirming the remainder is $0$ and the division is correct. Why distractors fail: Option B incorrectly claims a non-zero remainder. Option C incorrectly challenges the constant term. Option D is wrong because verification by multiplication is always valid.

Answer: Multiply $(x - 2)(x^2 - 4x + 3)$ and verify it equals $x^3 - 6x^2 + 11x - 6$

Division verification formula: The division algorithm states $P(x) = D(x) \cdot Q(x) + R(x)$. When $R(x) = 0$, this simplifies to $P(x) = D(x) \cdot Q(x)$. Why the correct answer works: Option B applies the verification formula directly: $(x-2)(x^2-4x+3) = x^3 - 4x^2 + 3x - 2x^2 + 8x - 6 = x^3 - 6x^2 + 11x - 6$, which matches the original dividend. Why distractors fail: Option A checks for common factors, which is unrelated to verifying the division result. Option C only checks one value and doesn't confirm the division in general. Option D adds instead of multiplies, which is not the correct relationship.

Answer: The first student is correct; when dividing by a degree-2 polynomial, the remainder can have degree up to 1

Remainder degree rule: The remainder $R(x)$ must have degree strictly less than the divisor $D(x)$. Since $D(x) = x^2 + x - 1$ has degree 2, the remainder can be degree 0 (a constant) or degree 1. Verify the student's answer: Check using $P(x) = D(x) \cdot Q(x) + R(x)$: $(x^2+x-1)(x^2+2x-3) + (6x-8)$. Expanding: $(x^2+x-1)(x^2+2x-3) = x^4+2x^3-3x^2+x^3+2x^2-3x-x^2-2x+3 = x^4+3x^3-2x^2-5x+3$. Adding $6x-8$: $x^4+3x^3-2x^2+x-5$. This matches the original dividend. ✓ Why the correct answer works: Option B correctly states that a linear remainder ($6x - 8$, degree 1) is valid when dividing by a quadratic, since $1 < 2$. Why distractors fail: Options A and D reflect the misconception that remainders are always constants — this is only true when dividing by a linear polynomial. Option C states a false rule about remainder degree.

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

Set up with placeholders: Write $x^3 - 8$ as $x^3 + 0x^2 + 0x - 8$. Perform the division: Step 1: $x^3 \div x = x^2$. Multiply: $x^2(x-2)=x^3-2x^2$. Subtract: $2x^2+0x-8$. Step 2: $2x^2 \div x = 2x$. Multiply: $2x(x-2)=2x^2-4x$. Subtract: $4x-8$. Step 3: $4x \div x = 4$. Multiply: $4(x-2)=4x-8$. Subtract: $0$. Result: The quotient is $x^2 + 2x + 4$ with remainder $0$. This is the standard factorization $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x, b=2$. Why distractors fail: Option B uses $-2x$ instead of $+2x$ (confusing with the sum-of-cubes pattern). Option C is $(x+2)^2$, not the correct quotient. Option D is $(x-2)^2$.

Answer: $2x^2 + x + 1$

Set up with placeholders: The dividend $4x^4 + 2x^3 - 3x + 1$ is missing the $x^2$ term. Rewrite as $4x^4 + 2x^3 + 0x^2 - 3x + 1$. First DMSB cycle: Divide $4x^4$ by $2x^2$ to get $2x^2$. Multiply: $2x^2(2x^2 - 1) = 4x^4 - 2x^2$. Subtract: $(4x^4 + 2x^3 + 0x^2) - (4x^4 - 2x^2) = 2x^3 + 2x^2$. Bring down $-3x$: $2x^3 + 2x^2 - 3x$. Second DMSB cycle: Divide $2x^3$ by $2x^2$ to get $x$. Multiply: $x(2x^2-1)=2x^3-x$. Subtract: $(2x^3+2x^2-3x)-(2x^3-x)=2x^2-2x$. Bring down $+1$: $2x^2-2x+1$. Third DMSB cycle: Divide $2x^2$ by $2x^2$ to get $1$. Multiply: $1(2x^2-1)=2x^2-1$. Subtract: $(2x^2-2x+1)-(2x^2-1)=-2x+2$. The remainder is $-2x+2$ and the quotient is $2x^2+x+1$. Why distractors fail: Option B has an incorrect constant term. Option C has a sign error on the $x$ term. Option D is missing the constant term of the quotient.

Answer: $2x^2 - 3x + 1 - \frac{x - 1}{2x^2 - x + 1}$

Express using the division algorithm: $\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)} = 2x^2 - 3x + 1 + \frac{-x+1}{2x^2 - x + 1}$. Simplify the sign: $\frac{-x+1}{2x^2-x+1} = -\frac{x-1}{2x^2-x+1}$, so the expression becomes $2x^2 - 3x + 1 - \frac{x-1}{2x^2-x+1}$. Verify the division: Confirm $D(x) \cdot Q(x) + R(x) = P(x)$: $(2x^2-x+1)(2x^2-3x+1) = 4x^4-6x^2+2x^2-2x^3+3x^2-x+2x^2-3x+1$. Collecting: $4x^4-8x^3+8x^2-4x+1$. Wait — let us expand carefully: $2x^2(2x^2-3x+1)=4x^4-6x^3+2x^2$; $-x(2x^2-3x+1)=-2x^3+3x^2-x$; $1(2x^2-3x+1)=2x^2-3x+1$. Sum: $4x^4-8x^3+7x^2-4x+1$. Adding $R(x)=-x+1$: $4x^4-8x^3+7x^2-5x+2$. ✓ Matches $P(x)$. Why distractors fail: Option B has the wrong sign on the remainder fraction — since $R(x) = -x+1$ is negative, the fraction must be subtracted, not added. Option C rewrites the expression as a different division entirely. Option D drops the $x$ from the remainder, writing $-1$ instead of $-(x-1)$.

Answer: The remainder is $0$

Apply the Remainder Theorem: When dividing $P(x)$ by $(x - c)$, the remainder equals $P(c)$. Here, $P(x) = x^5 - 32$ and $c = 2$. Evaluate P(2): $P(2) = 2^5 - 32 = 32 - 32 = 0$. Conclusion: The remainder is $0$, meaning $(x-2)$ divides $x^5 - 32$ evenly. Why distractors fail: Options A and B give incorrect numerical values. Option D is wrong because the Remainder Theorem allows us to find the remainder without full division.

Answer: The dividend needs a $0x^2$ placeholder term

Identify missing terms in the dividend: The dividend $2x^4 - 3x^3 + x - 7$ jumps from $x^3$ to $x$, skipping the $x^2$ term. The correct form is $2x^4 - 3x^3 + 0x^2 + x - 7$. Why the correct answer works: Option B correctly identifies that the $x^2$ term is missing from the dividend. Without it, the subtraction steps will misalign columns of different degrees. Why distractors fail: Option A: rewriting the divisor in a different order is not necessary and would cause confusion. Option C: factoring is not a prerequisite. Option D: while the divisor $x^2 - 2$ also lacks an $x$ term, many students write it as $x^2 + 0x - 2$, but the critical missing placeholder is in the dividend.

Answer: $\deg(Q) = 3$, $\deg(R) = 1$

Degree of the quotient: When dividing degree $n$ by degree $m$, the quotient has degree $n - m = 5 - 2 = 3$. Degree of the remainder: The remainder must have degree strictly less than $m = 2$, so $\deg(R)$ can be $0$ or $1$ (or the remainder could be $0$). Why the correct answer works: Option B has $\deg(Q) = 3$ and $\deg(R) = 1$, both satisfying the constraints. Why distractors fail: Option A: $\deg(R) = 2$ is not less than $\deg(D) = 2$. Option C: $\deg(Q) = 2$ is too small and $\deg(R) = 3$ is too large. Option D: $\deg(Q) = 4$ would imply the dividend has degree $4 + 2 = 6$, not $5$.

Answer: Neither student is correct

Verify Student A's answer: $(x^2-1)(x^2+2x) + (3x+3) = x^4+2x^3-x^2-2x+3x+3 = x^4+2x^3-x^2+x+3$. This gives an extra $x$ term, so it does not match the dividend $x^4+2x^3-x^2+3$. Verify Student B's answer: $(x^2-1)(x^2+2x) + 3 = x^4+2x^3-x^2-2x+3$. This produces a $-2x$ term, which is not in the original dividend. Student B is also incorrect. Find the correct result: Performing the division properly: first term $x^2$, giving remainder $2x^3+0x^2+0x+3$ (after inserting the placeholder $0x$). Second term $2x$, multiply $(2x)(x^2-1)=2x^3-2x$. Subtract to get $2x+3$. Since degree of $2x+3$ is less than 2, the quotient is $x^2+2x$ and remainder is $2x+3$. Verify: $(x^2-1)(x^2+2x)+(2x+3) = x^4+2x^3-x^2-2x+2x+3 = x^4+2x^3-x^2+3$. This matches the original dividend. Why distractors fail: Neither Student A nor Student B obtained the correct remainder of $2x + 3$, so Option C is correct. Option D is wrong because the quotient and remainder from polynomial division are unique.

Answer: $x^3 + x^2 + x + 1$

Set up with placeholders: Write the dividend as $x^4 + 0x^3 + 0x^2 + 0x - 1$ to account for all missing terms. Execute the DMSB cycles: Step 1: $x^4 \div x = x^3$. Multiply and subtract to get $x^3 + 0x^2 + 0x - 1$. Step 2: $x^3 \div x = x^2$. Continue to get $x^2 + 0x - 1$. Step 3: $x^2 \div x = x$. Continue to get $x - 1$. Step 4: $x \div x = 1$. Subtract to get remainder $0$. Final result: The quotient is $x^3 + x^2 + x + 1$ with remainder $0$, confirming $x^4 - 1 = (x-1)(x^3+x^2+x+1)$. Why distractors fail: Option B has alternating signs, which would result from dividing by $x+1$, not $x-1$. Options C and D are missing terms or have incorrect signs.

Answer: All subsequent quotient terms and the remainder will have incorrect coefficients

Impact of the sign error: Adding instead of subtracting the product doubles the leading term's contribution rather than cancelling it. This cascading error corrupts the partial remainder, which feeds into every subsequent division step. Why the correct answer works: Option B correctly identifies that since each subsequent step uses the previous partial remainder, an incorrect subtraction in the first cycle propagates errors through all remaining quotient terms and the final remainder. Why distractors fail: Option A is wrong because the degrees of quotient terms are determined by leading-term division, which follows the correct exponent pattern regardless. Option C is wrong because the degree of the remainder is still bounded by the process's stopping criterion. Option D is wrong because the error propagates to all subsequent quotient terms, not just the remainder.

Answer: Both methods produce identical results; synthetic division is a shortcut that works specifically for linear divisors of the form $x - c$

Comparison of methods: Polynomial long division is a general algorithm that works for divisors of any degree. Synthetic division is an abbreviated version that only works for divisors of the form $x - c$ (linear binomials). Why the correct answer works: Option B correctly states that both methods yield identical results and that synthetic division is a streamlined shortcut restricted to linear divisors. Why distractors fail: Option A reverses the capabilities: long division works for any divisor, while synthetic division is limited. Option C is false—both methods are equally accurate when performed correctly. Option D is false—long division handles any divisor.

Answer: The polynomial part of the division result plus the proper fractional remainder over the divisor

Division algorithm result: When dividing a polynomial $P(x)$ by $D(x)$, the result is expressed as $P(x) / D(x) = Q(x) + R(x)/D(x)$, where the degree of $R(x)$ is less than the degree of $D(x)$. Why the correct answer works: Option B correctly interprets the expression: $Q(x)$ is the polynomial (whole) part of the result, and $R(x)/D(x)$ is a proper rational expression (analogous to a proper fraction in numeric division). Why distractors fail: Option A describes the reconstruction formula $P(x) = Q(x) \cdot D(x) + R(x)$, not the division result itself. Option C is wrong because this is not a factorization. Option D is wrong because the expression is exact, not an approximation.

Answer: $x^4 + 5x^3 + 7x^2 + 4x + 3$

Approach: We need a degree-3 quotient $Q(x)$ with all positive integer coefficients such that $(x+3)Q(x)$ produces a degree-4 polynomial with remainder $0$. Check Option A: First verify $P(-3) = 0$: $(-3)^4 + 5(-3)^3 + 7(-3)^2 + 4(-3) + 3 = 81 - 135 + 63 - 12 + 3 = 0$. ✓ Now divide: $(x^4 + 5x^3 + 7x^2 + 4x + 3) \div (x+3)$. The quotient is $x^3 + 2x^2 + x + 1$, which has all positive integer coefficients ($1, 2, 1, 1$). ✓ Verification: $(x+3)(x^3 + 2x^2 + x + 1) = x^4 + 2x^3 + x^2 + x + 3x^3 + 6x^2 + 3x + 3 = x^4 + 5x^3 + 7x^2 + 4x + 3$. ✓ Why distractors fail: Option B: $P(-3) = 81 + 81 + 9 + 9 = 180 \neq 0$, so $x+3$ is not a factor. Option C: $P(-3) = 81 - 162 + 99 - 18 = 0$, but dividing gives quotient $x^3 + 3x^2 + 2x$, which has a $0$ constant term — not all positive. Option D: $P(-3) = 81 - 108 + 36 - 3 - 6 = 0$, but dividing gives quotient $x^3 + x^2 + x - 2$, which has a negative coefficient.

Answer: Yes, the formula $P(x) = D(x) \cdot Q(x) + R(x)$ gives $(x^2 - x + 1)(x^2 - 2x - 1) + 2x = x^4 - 3x^3 + 2x^2 + x - 1$, which matches the dividend

Apply the division algorithm: The division algorithm states $P(x) = D(x) \cdot Q(x) + R(x)$. The student's claim is that $x^4 - 3x^3 + 2x^2 + x - 1 = (x^2 - x + 1)(x^2 - 2x - 1) + 2x$. Verify by expanding: $(x^2 - x + 1)(x^2 - 2x - 1) = x^4 - 2x^3 - x^2 - x^3 + 2x^2 + x + x^2 - 2x - 1 = x^4 - 3x^3 + 2x^2 - x - 1$. Adding the remainder $2x$: $x^4 - 3x^3 + 2x^2 - x - 1 + 2x = x^4 - 3x^3 + 2x^2 + x - 1$. This matches the original dividend. ✓ Why the correct answer works: Option B is correct because the reconstruction formula $P(x) = D(x) \cdot Q(x) + R(x)$ holds for any valid polynomial division, regardless of whether $R(x)$ is zero, a constant, or a polynomial of degree less than the divisor. Why distractors fail: Option A is wrong because the formula uses addition of $R(x)$, and $R(x) = 2x$ is positive here. Option C is wrong because the remainder can be any polynomial of degree less than the divisor's degree (here, degree less than 2). Option D is wrong because the formula holds for any remainder, not just zero.

Answer: Divide, Multiply, Subtract, Bring down

DMSB cycle definition: DMSB stands for Divide–Multiply–Subtract–Bring Down. This cycle is repeated until the remainder's degree is less than the divisor's degree. Why the correct answer works: Option B gives the standard expansion of the DMSB mnemonic used in long division for both numerical and polynomial cases. Why distractors fail: Options A, C, and D use words that do not correspond to the actual steps of the long division algorithm.

Answer: To find a term that, when multiplied by the full divisor, cancels the leading term of the current dividend

Purpose of the leading-term division: At each step, dividing the leading term of the current dividend by the leading term of the divisor produces a quotient term. When this quotient term is multiplied by the entire divisor, the product's leading term matches and cancels the current dividend's leading term upon subtraction. Why the correct answer works: Option A captures the exact purpose: the quotient term is chosen so that multiplying it by the divisor produces a polynomial whose leading term equals the current dividend's leading term, allowing it to be eliminated. Why distractors fail: Option B is wrong because the divisor's degree never changes. Option C is wrong because the quotient's leading coefficient depends on both polynomials, not just the dividend. Option D is nonsensical in this context.

Answer: $3$

Reconstruct the dividend: By the division algorithm, $P(x) = (x^2+1)(x^2+3x-1) + (2x+5)$. Expand the product: $(x^2+1)(x^2+3x-1) = x^4 + 3x^3 - x^2 + x^2 + 3x - 1 = x^4 + 3x^3 + 0x^2 + 3x - 1$. Add the remainder: $P(x) = x^4 + 3x^3 + 0x^2 + 3x - 1 + 2x + 5 = x^4 + 3x^3 + 0x^2 + 5x + 4$. Identify the coefficient: Comparing with $x^4 + ax^3 + bx^2 + cx + d$, we see $a = 3$. Why distractors fail: Options A, C, and D result from errors in the expansion or from confusing coefficients of different terms.

Answer: When the degree of the remainder is less than the degree of the divisor

Stopping condition: The polynomial long division algorithm terminates when the degree of the current remainder is strictly less than the degree of the divisor. At that point, no further division of leading terms is possible. Why the correct answer works: Option C correctly states the general stopping criterion. A zero remainder is a special case (degree $-\infty$), but the process also stops with a non-zero remainder of lower degree. Why distractors fail: Option A is incomplete—the remainder need not be zero; it just needs degree less than the divisor. Option B is not the formal criterion; leftover terms are handled during bring-down steps. Option D is incorrect because the quotient's degree is typically less than the dividend's degree.

Answer: Correct; $(x^2-1)(x^4+x^2+1) = x^6 - 1$

Verify by multiplication: $(x^2-1)(x^4+x^2+1) = x^6 + x^4 + x^2 - x^4 - x^2 - 1 = x^6 - 1$. ✓ Pattern recognition: This uses the factorization $a^3 - b^3 = (a-b)(a^2+ab+b^2)$ with $a = x^2$ and $b = 1$: $x^6 - 1 = (x^2)^3 - 1^3 = (x^2-1)(x^4+x^2+1)$. Why distractors fail: Option A gives $x^4 - x^2 + 1$, and $(x^2-1)(x^4-x^2+1)=x^6-x^4+x^2-x^4+x^2-1=x^6-2x^4+2x^2-1 \neq x^6-1$. Option C is wrong because the division is exact. Option D gives the quotient of $x^5-1$ divided by $x-1$, not this problem.

Answer: No, the quotient must have degree $5 - 2 = 3$

Degree relationship in polynomial division: When dividing a polynomial of degree $n$ by a polynomial of degree $m$ (where $n \geq m$), the quotient has degree $n - m$. Apply to this problem: Here $n = 5$ and $m = 2$, so the quotient must have degree $5 - 2 = 3$, not degree 2. Why distractors fail: Option A is self-contradictory. Option C states a false rule about halving degrees. Option D incorrectly adds the degrees instead of subtracting.

Answer: By dividing the leading term of the current dividend by the leading term of the divisor

Core principle of the divide step: In each iteration of the DMSB cycle, the leading term of the current dividend (or partial remainder) is divided by the leading term of the divisor to produce the next quotient term. Why the correct answer works: Option B states exactly this process: for example, if the current leading term is $6x^3$ and the divisor's leading term is $2x$, then the quotient term is $\frac{6x^3}{2x} = 3x^2$. Why distractors fail: Option A uses constant terms rather than leading terms. Option C describes subtraction, which is a different step in the cycle. Option D describes multiplication, which is also a different step.

Answer: $-5x + 2$

First DMSB cycle: Divide $3x^3$ by $x^2$ to get $3x$. Multiply: $3x(x^2+1)=3x^3+3x$. Subtract: $(3x^3+5x^2-2x+7)-(3x^3+3x) = 5x^2 - 5x + 7$. Second DMSB cycle: Divide $5x^2$ by $x^2$ to get $5$. Multiply: $5(x^2+1) = 5x^2+5$. Subtract: $(5x^2 - 5x + 7) - (5x^2 + 5) = -5x + 2$. Check the stopping condition: The degree of $-5x + 2$ is $1$, which is less than the degree of $x^2 + 1$ (degree $2$), so the division terminates and the remainder is $-5x + 2$. Verification: Verify: $(3x + 5)(x^2 + 1) + (-5x + 2) = 3x^3 + 3x + 5x^2 + 5 - 5x + 2 = 3x^3 + 5x^2 - 2x + 7$. This matches the dividend, confirming the remainder is $-5x + 2$. Why distractors fail: Option B ($-5x + 12$) results from an arithmetic error in the constant during the second subtraction. Option C ($5x + 12$) has the wrong sign on the $x$ term. Option D ($-2x + 7$) uses the last two terms of the original dividend without performing any division.

Answer: As $Q(x) + \frac{R}{2x - 1}$ where $Q(x)$ is a polynomial and $R$ is a constant

Division result form: When dividing a degree-3 polynomial by a degree-1 polynomial, the quotient is degree 2 and the remainder is degree 0 (a constant) or zero. The result is expressed as $Q(x) + \frac{R}{D(x)}$. Why the correct answer works: Option B correctly describes the standard form: a polynomial quotient plus a proper rational expression with the remainder over the divisor. Why distractors fail: Option A describes factoring, not division. Option C would only be true if the remainder is exactly $0$, which isn't guaranteed. Option D puts the quotient in the numerator rather than the standard mixed form.

Answer: Write both polynomials in descending order of degree and insert placeholders for any missing terms

Identify the key principle: Polynomial long division requires that both polynomials be arranged in standard form (descending powers) with $0$-coefficient placeholders for any missing degrees, so the alignment of like terms is correct throughout the process. Why the correct answer works: Option B correctly describes the essential setup: writing polynomials in descending order and inserting terms like $0x^2$ if that degree is absent. This ensures no alignment errors occur during the DMSB cycle. Why distractors fail: Option A is wrong because factoring is not required before performing long division. Option C is wrong because you do not multiply the dividend by the divisor—that is not part of the algorithm. Option D is wrong because cancelling common factors is simplification, not part of the setup step.

Answer: To maintain proper alignment of like terms during subtraction steps

Purpose of placeholders: The dividend $x^3 + 5$ is missing the $x^2$ and $x$ terms. Writing it as $x^3 + 0x^2 + 0x + 5$ ensures that when you multiply and subtract, each column contains terms of the same degree. Why the correct answer works: Option C correctly identifies the reason: without placeholders, the subtraction step would misalign terms of different degrees, leading to errors. Why distractors fail: Option A is wrong because adding $0$-coefficient terms does not change the degree. Option B is wrong because placeholders have no effect on the value of the remainder. Option D is wrong because inserting zeros does not factor the polynomial.

Answer: $-8x^3 + 6x^2$

Perform the first division step: Divide $6x^4$ by $2x^2$ to get $3x^2$. Multiply step: $3x^2(2x^2 + x - 1) = 6x^4 + 3x^3 - 3x^2$. Subtract step: $(6x^4 - 5x^3 + 3x^2) - (6x^4 + 3x^3 - 3x^2) = 0x^4 - 8x^3 + 6x^2$. So the partial remainder is $-8x^3 + 6x^2$ before bringing down the next terms. Why distractors fail: Option A prematurely includes terms that haven't been brought down yet (or has extra terms). Option C doesn't perform the subtraction correctly. Option D has incorrect coefficient computation.

Answer: The quotient will be $0$ and the remainder is $3x^2 + 2x - 1$

Degree comparison before dividing: The dividend has degree 2 and the divisor has degree 3. Since the dividend's degree is already less than the divisor's degree, the stopping condition is immediately met. Why the correct answer works: Option A is correct: when the dividend has a lower degree than the divisor, the quotient is $0$ and the entire dividend becomes the remainder. This is analogous to $\frac{3}{7} = 0$ remainder $3$ in integer division. Why distractors fail: Option B is wrong—the division is not undefined; it simply gives quotient $0$. Option C is wrong—you do not rearrange which polynomial is dividend vs. divisor. Option D is wrong—you never artificially raise the degree of the dividend.

Answer: $k = 5$

Apply the Factor Theorem: If $x + 2$ divides $P(x)$ evenly, then $P(-2) = 0$. Set up the equation: $P(-2) = (-2)^3 + k(-2)^2 + (-2) - 10 = -8 + 4k - 2 - 10 = 4k - 20$. Solve for k: Setting $4k - 20 = 0$ gives $4k = 20$, so $k = 5$. Verification: With $k = 5$: $P(x) = x^3 + 5x^2 + x - 10$. Check: $P(-2) = -8 + 20 - 2 - 10 = 0$. ✓ Dividing: $(x+2)(x^2 + 3x - 5) = x^3 + 3x^2 - 5x + 2x^2 + 6x - 10 = x^3 + 5x^2 + x - 10$. ✓ Why distractors fail: Option A ($k = 4$): $P(-2) = -8 + 16 - 2 - 10 = -4 \neq 0$. Option C ($k = -3$): $P(-2) = -8 - 12 - 2 - 10 = -32 \neq 0$. Option D ($k = 2$): $P(-2) = -8 + 8 - 2 - 10 = -12 \neq 0$.