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. Dividing (2x⁴ - 3x³ + x² + 4) by (x² + 2): First, 2x⁴ ÷ x² = 2x², multiply back: 2x²(x² + 2) = 2x⁴ + 4x², subtract to get -3x³ - 3x² + 4. Next, -3x³ ÷ x² = -3x, multiply back: -3x(x² + 2) = -3x³ - 6x, subtract to get -3x² + 6x + 4. Finally, -3x² ÷ x² = -3, multiply back: -3(x² + 2) = -3x² - 6, subtract to get 6x + 10. Choice A correctly shows quotient 2x² - 3x - 3 and remainder 6x + 10, giving us 2x² - 3x - 3 + (6x + 10)/(x² + 2). Choice C incorrectly suggests remainder 6x² + 10, but that would have degree 2, violating the requirement that deg(r) < 2 = deg(x² + 2). The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: since x² + 2 has degree 2, remainder must have degree at most 1. Our remainder 6x + 10 has degree 1, which is perfect!

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. Let's divide 2x³ + 3x² - 5x + 7 by x - 2: First, 2x³ ÷ x = 2x², multiply back: 2x²(x - 2) = 2x³ - 4x², subtract: (2x³ + 3x² - 5x + 7) - (2x³ - 4x²) = 7x² - 5x + 7. Next, 7x² ÷ x = 7x, multiply back: 7x(x - 2) = 7x² - 14x, subtract: (7x² - 5x + 7) - (7x² - 14x) = 9x + 7. Finally, 9x ÷ x = 9, multiply back: 9(x - 2) = 9x - 18, subtract: (9x + 7) - (9x - 18) = 25. Choice A correctly gives quotient 2x² + 7x + 9 and remainder 25, with deg(25) = 0 < deg(x - 2) = 1. Choice B has the wrong remainder (5 instead of 25), likely from an arithmetic error in the final subtraction. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here, no obvious factorization exists, so long division is necessary. The degree requirement tells you when to stop: since we're dividing by (x - 2) which has degree 1, we stop when the remainder is a constant (degree 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 $q(x) + \frac{r(x)}{b(x)}$, 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 $2x^3 + 3x^2 - 5x + 7$ by $x + 2$, start by dividing $2x^3$ by $x$ to get $2x^2$, multiply by the divisor to get $2x^3 + 4x^2$, subtract to yield $-x^2 - 5x$, then continue to get $-x$ and $-3$, resulting in a remainder of 13. Choice A correctly divides to get quotient $2x^2 - x - 3$ and remainder 13, with $\deg(r) < \deg(x + 2)$. A common mistake, like in choice B, is miscalculating the remainder as 1 instead of 13 due to an arithmetic error in the final subtraction. Before dividing, always check if you can factor the numerator and cancel with the denominator, but here no common factors exist, so long division is necessary. The degree requirement ensures the remainder is proper, stopping when $\deg(\text{remainder}) < \deg(\text{divisor})$, just like stopping in numerical division when the remainder is smaller than the divisor.

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. Let's divide (2x³ + 3x² - 5x + 6) by (x - 2): First, 2x³ ÷ x = 2x², multiply back: 2x²(x - 2) = 2x³ - 4x², subtract to get 7x² - 5x + 6. Next, 7x² ÷ x = 7x, multiply back: 7x(x - 2) = 7x² - 14x, subtract to get 9x + 6. Finally, 9x ÷ x = 9, multiply back: 9(x - 2) = 9x - 18, subtract to get 24. Choice A correctly shows quotient 2x² + 7x + 9 and remainder 24, giving us 2x² + 7x + 9 + 24/(x - 2). Choice B has the wrong remainder (12 instead of 24), likely from an arithmetic error in the final subtraction step. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here we can't simplify first, so long division is necessary. The degree requirement tells you when to stop: since we're dividing by (x - 2) which has degree 1, our remainder must have degree 0 (a constant), which 24 is!

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. 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. Let's divide (x³ + 2x² - 5x + 3) by (x - 1): First, x³/x = x², multiply (x - 1) by x² to get x³ - x², subtract to get 3x² - 5x + 3. Next, 3x²/x = 3x, multiply (x - 1) by 3x to get 3x² - 3x, subtract to get -2x + 3. Then, -2x/x = -2, multiply (x - 1) by -2 to get -2x + 2, subtract to get 1. Since 1 has degree 0 < degree 1 of (x - 1), we stop! Choice A correctly gives quotient x² + 3x - 2 and remainder 1, written as x² + 3x - 2 + 1/(x - 1). Choice C has the wrong sign on the remainder fraction, while B and D have incorrect quotients from calculation errors. The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: once your remainder (1) has degree 0, which is less than the divisor's degree 1, you're done—stop dividing!

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. Let's divide (x³ + 4x² - 2x + 7) by (x + 2): First, x³/x = x², multiply (x + 2) by x² to get x³ + 2x², subtract to get 2x² - 2x + 7. Next, 2x²/x = 2x, multiply (x + 2) by 2x to get 2x² + 4x, subtract to get -6x + 7. Then, -6x/x = -6, multiply (x + 2) by -6 to get -6x - 12, subtract to get 19. Since 19 has degree 0 < degree 1 of (x + 2), we stop! Choice A correctly gives quotient x² + 2x - 6 and remainder 19, written as x² + 2x - 6 + 19/(x + 2). Choice B has the wrong sign on the remainder fraction, C has an incorrect quotient from arithmetic errors, and D has both wrong quotient and remainder. The degree requirement tells you when to stop: once your remainder (19) has degree 0, which is less than the divisor's degree 1, you're done—continuing would be like dividing 2 by 5 further after getting 0 remainder 2!

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. Before dividing, let's check if we can factor: x³ - 1 = (x - 1)(x² + x + 1) by the difference of cubes formula! So (x³ - 1)/(x - 1) = (x - 1)(x² + x + 1)/(x - 1) = x² + x + 1 after canceling. This means the quotient is x² + x + 1 with remainder 0. Choice A correctly gives x² + x + 1 + 0/(x - 1), which simplifies to just x² + x + 1. Choice B has the wrong middle coefficient, C adds an unnecessary remainder of 1, and D is missing the constant term in the quotient. Before dividing, always check: can you factor the numerator and cancel with the denominator? Recognizing x³ - 1 as a difference of cubes that factors as (x - 1)(x² + x + 1) gives you the answer instantly without any division work!

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. For $$3x^2 - 2x + 4$$ divided by $$x - 1$$, division gives $3x + 1$ with remainder 5, verifiable by inspection or synthetic division. Choice A correctly provides the quotient $3x + 1$ and remainder 5, with $\deg(r) < 1$. Choice B might result from a sign error in subtraction, yielding -1 instead of +1. The degree requirement tells you to stop when the remainder is a constant for a linear divisor, preventing unnecessary further steps.

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).