This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. To divide x⁴ + 1 by x² + 1, divide x⁴ by x² to get x², multiply by (x² + 1) to get x⁴ + x², subtract to get -x² + 1, then divide -x² by x² to get -1, multiply to get -x² - 1, and subtract to yield 2 as remainder. Choice A correctly divides to get quotient x² - 1 and remainder 2 with deg(r) < deg(x² + 1). Choice B might come from ignoring the subtraction and assuming no remainder. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by (x² + 1) (degree 2), remainder can be degree 1 or 0.

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting $17/5$ as $3 + 2/5$. The division algorithm for polynomials says any rational expression $a(x)/b(x)$ can be written as $q(x) + r(x)/b(x)$, where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. To divide $2x^2 + 7x - 3$ by $x + 3$, divide $2x^2$ by $x$ to get $2x$, multiply by $(x + 3)$ to get $2x^2 + 6x$, subtract to get $x - 3$, then divide $x$ by $x$ to get 1, multiply to get $x + 3$, and subtract to yield -6 as remainder. Choice A correctly divides to get quotient $2x + 1$ and remainder -6 with $\deg(r) < \deg(x + 3)$. Choice B might come from a sign error in the final subtraction, turning -6 into +6. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement ($\deg$ of remainder less than $\deg$ of divisor) tells you when to stop dividing: if dividing by $(x + 3)$ (degree 1), remainder must be degree 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Using synthetic division for $x^3 + 4x^2 - 2x + 7$ by $x + 2$ (root $-2$), bring down 1, multiply by $-2$ to get $-2$, add to 4 for 2, multiply for $-4$, add to $-2$ for $-6$, multiply for 12, add to 7 for 19 as remainder. Choice A correctly divides to get quotient $x^2 + 2x - 6$ and remainder 19 with deg(r) < deg($x + 2$). Choice B might stem from a sign error in the remainder, flipping 19 to $-19$. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by ($x + 2$) (degree 1), remainder must be degree 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting $\frac{17}{5}$ as $3 + \frac{2}{5}$. The division algorithm for polynomials says any rational expression $\frac{a(x)}{b(x)}$ can be written as $q(x) + \frac{r(x)}{b(x)}$, where $q(x)$ is the quotient (polynomial part) and $r(x)$ is the remainder with degree strictly less than the divisor's degree. To divide $2x^3 + x^2 - 5x + 3$ by $x - 2$, start by dividing $2x^3$ by $x$ to get $2x^2$, multiply by $(x - 2)$ to get $2x^3 - 4x^2$, subtract to yield $5x^2 - 5x$, then continue to get $5x$ next, and finally $5$, resulting in a remainder of $13$. Choice A correctly divides to get quotient $2x^2 + 5x + 5$ and remainder $13$ with $\deg(r) < \deg(x - 2)$. Choice C forgets the constant term in the quotient and the full remainder, likely from stopping too early in the division. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement ($\deg$ of remainder less than $\deg$ of divisor) tells you when to stop dividing: if dividing by $(x - 2)$ ($\deg$ 1), remainder must be $\deg$ 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. For (x³ - 8)/(x - 2), recognize x³ - 8 = (x - 2)(x² + 2x + 4) by difference of cubes, so it divides evenly with quotient x² + 2x + 4 and remainder 0. Choice A correctly identifies quotient x² + 2x + 4 and remainder 0 with deg(r) < deg(x - 2). Choice C adds an incorrect remainder, perhaps from misapplying synthetic division. Before dividing, always check: can you factor the numerator and cancel with the denominator? If (x³ - 8)/(x - 2) factors to x² + 2x + 4 (no remainder!), that's instant. The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by (x - 2) (degree 1), remainder must be degree 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting $\frac{17}{5}$ as $3 + \frac{2}{5}$. The division algorithm for polynomials says any rational expression $a(x)/b(x)$ can be written as $q(x) + \frac{r(x)}{b(x)}$, where $q(x)$ is the quotient (polynomial part) and $r(x)$ is the remainder with degree strictly less than the divisor's degree. Using synthetic division for $x^3 + 2x^2 - 5x + 3$ by $x - 1$ (root 1), bring down 1, multiply by 1 for 1, add to 2 for 3, multiply for 3, add to -5 for -2, multiply for -2, add to 3 for 1 as remainder. Choice A correctly divides to get quotient $x^2 + 3x - 2$ and remainder $1$ with $\deg(r) < \deg(x - 1)$. Choice C probably comes from an arithmetic error in the coefficients during division. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement ($\deg$ of remainder less than $\deg$ of divisor) tells you when to stop dividing: if dividing by $(x - 1)$ (degree 1), remainder must be degree 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. For (x² - 4)/(x - 2), factor numerator as (x - 2)(x + 2), cancel (x - 2) to get x + 2 directly, with remainder 0. Choice A correctly identifies quotient x + 2 and remainder 0 with deg(r) < deg(x - 2). Choice C adds an incorrect remainder, perhaps from partial division without factoring. Before dividing, always check: can you factor the numerator and cancel with the denominator? If (x² - 4)/(x - 2) factors to (x + 2)(x - 2)/(x - 2) = x + 2 (no remainder!), that's instant. The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by (x - 2) (degree 1), remainder must be degree 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting $17/5$ as $3 + 2/5$. The division algorithm for polynomials says any rational expression $a(x)/b(x)$ can be written as $q(x) + r(x)/b(x)$, where $q(x)$ is the quotient (polynomial part) and $r(x)$ is the remainder with degree strictly less than the divisor's degree. To divide $3x^2 - x + 4$ by $x - 2$, divide $3x^2$ by $x$ to get $3x$, multiply by $(x - 2)$ to get $3x^2 - 6x$, subtract to get $5x + 4$, then divide $5x$ by $x$ to get $5$, multiply to get $5x - 10$, and subtract to yield $14$ as remainder. Choice A correctly divides to get quotient $3x + 5$ and remainder $14$ with $\deg(r) < \deg(x - 2)$. Choice B could result from a sign mistake in the final subtraction. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement ($\deg$ of remainder less than $\deg$ of divisor) tells you when to stop dividing: if dividing by $(x - 2)$ ($\deg$ 1), remainder must be $\deg$ 0 (constant).

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting $17/5$ as $3 + 2/5$. The division algorithm for polynomials says any rational expression $a(x)/b(x)$ can be written as $q(x) + r(x)/b(x)$, where $q(x)$ is the quotient (polynomial part) and $r(x)$ is the remainder with degree strictly less than the divisor's degree. To divide $x^4 - 3x^2 + 5$ by $x^2 + 1$, divide $x^4$ by $x^2$ to get $x^2$, multiply by $(x^2 + 1)$ to get $x^4 + x^2$, subtract to get $-4x^2 + 5$, then divide $-4x^2$ by $x^2$ to get $-4$, multiply to get $-4x^2 - 4$, and subtract to yield $9$ as remainder. Choice A correctly divides to get quotient $x^2 - 4$ and remainder $9$ with $\deg(r) < \deg(x^2 + 1)$. Choice C likely results from a miscalculation in the subtraction, getting $-3$ instead of $-4$ and $8$ instead of $9$. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement ($\deg$ of remainder less than $\deg$ of divisor) tells you when to stop dividing: if dividing by $(x^2 + 1)$ (degree 2), remainder can be degree 1 or 0.

This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Using synthetic division with root -1 for 3x³ - 2x² + 4 divided by x + 1 (adding 0x: 3|-2|0|4), bring down 3, multiply by -1 for -3, add to -2 for -5, multiply by -1 for 5, add to 0 for 5, multiply by -1 for -5, add to 4 for -1, giving quotient 3x² - 5x + 5 and remainder -1. Choice A correctly divides to get quotient 3x² - 5x + 5 and remainder -1 with deg(r) < 1. A distractor like choice B might flip the remainder sign from an addition error, but the calculation shows negative. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by (x + 1) (degree 1), remainder must be degree 0 (constant).