Rewriting Rational Expressions - Algebra 2

1. Rewrite $\frac{3x^2-2x+4}{x-1}$ in the form $q(x)+\frac{r(x)}{x-1}$, where $\deg(r)<\deg(x-1)$. (You may use inspection or long division.)
   Answer: $3x+1+\dfrac{5}{x-1}$ 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.
2. Rewrite $\dfrac{x^3+5x^2+x-3}{x+1}$ in the form $q(x)+\dfrac{r(x)}{x+1}$, where $\deg(r)<\deg(x+1)$.
   Answer: $x^2+4x-3+\dfrac{0}{x+1}$ 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. Before dividing, always check: can you factor the numerator and cancel with the denominator? Let's check if x + 1 is a factor of x³ + 5x² + x - 3 by evaluating at x = -1: (-1)³ + 5(-1)² + (-1) - 3 = -1 + 5 - 1 - 3 = 0. Yes! So x + 1 divides evenly. Using synthetic division or factoring: x³ + 5x² + x - 3 = (x + 1)(x² + 4x - 3). Therefore, (x³ + 5x² + x - 3)/(x + 1) = x² + 4x - 3 with remainder 0. Choice A correctly shows quotient x² + 4x - 3 and remainder 0, written as 0/(x + 1). Choice B incorrectly adds a remainder of 4, perhaps from a calculation error—but we verified the division is exact by checking that f(-1) = 0. Polynomial long division works exactly like numerical long division, but here the pre-check saved us work: finding that the numerator has (x + 1) as a factor means the remainder must be zero. This is like discovering 15/3 divides evenly because 15 = 3 × 5!
3. Express $\frac{3x^2-4x+5}{x+2}$ as $q(x)+\frac{r(x)}{x+2}$ using polynomial long division or inspection, with $\deg(r)<\deg(x+2)$.
   Answer: $3x-10+\dfrac{25}{x+2}$ 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. 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 (3x² - 4x + 5) by (x + 2): First, 3x² ÷ x = 3x, multiply back: 3x(x + 2) = 3x² + 6x, subtract from 3x² - 4x + 5 to get -10x + 5. Next, -10x ÷ x = -10, multiply back: -10(x + 2) = -10x - 20, subtract to get 25. Choice A correctly gives quotient 3x - 10 and remainder 25, so we have 3x - 10 + 25/(x + 2). Choice B has remainder 15 instead of 25, suggesting an error when computing 5 - (-20) = 25. The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: since (x + 2) has degree 1, our remainder must be degree 0 (constant). Once you get remainder 25 (degree 0), you're done—continuing would be like trying to divide 25 by x + 2 further!
4. Rewrite $\frac{x^3-5x+2}{x-3}$ in the form $q(x)+\frac{r(x)}{x-3}$ using polynomial division, where $\deg(r)<1$.
   Answer: $x^2+3x+4+\dfrac{14}{x-3}$ 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 - 5x + 2) by (x - 3), noting the missing x² term means coefficient 0: First, x³ ÷ x = x², multiply back: x²(x - 3) = x³ - 3x², subtract to get 3x² - 5x + 2. Next, 3x² ÷ x = 3x, multiply back: 3x(x - 3) = 3x² - 9x, subtract to get 4x + 2. Finally, 4x ÷ x = 4, multiply back: 4(x - 3) = 4x - 12, subtract to get 14. Choice A correctly shows quotient x² + 3x + 4 and remainder 14, giving us x² + 3x + 4 + 14/(x - 3). Choice B has remainder -14 instead of 14, likely from computing 2 - (-12) = 2 + 12 = 14 but then incorrectly negating it. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here we can't simplify first, so long division is needed. Since we're dividing by (x - 3) with degree 1, remainder must have degree 0 (constant), which 14 is!
5. Use polynomial division to rewrite the rational expression $\frac{2x^3+3x^2-5x+7}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$, where $\deg(r)<\deg(x+2)$.
   Answer: $2x^2-x-3+\dfrac{13}{x+2}$ 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 $q(x) + \frac{r(x)}{b(x)}$, 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.
6. Divide $x^3-1$ by $x-1$ and express $\frac{x^3-1}{x-1}$ in the form $q(x)+\frac{r}{x-1}$, where $\deg(r)<\deg(x-1)$.
   Answer: $x^2+x+1+\dfrac{0}{x-1}$ 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!
7. Express $\frac{2x^2+7x-3}{x+3}$ as $q(x)+\frac{r(x)}{x+3}$ using the polynomial division algorithm, with $\deg(r)<\deg(x+3)$.
   Answer: $2x+1+\dfrac{-6}{x+3}$ 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 -3 for 2x² + 7x - 3 divided by x + 3, bring down 2, multiply by -3 for -6, add to 7 for 1, multiply by -3 for -3, add to -3 for -6, giving quotient 2x + 1 and remainder -6. Choice B correctly divides to get quotient 2x + 1 and remainder -6 with deg(r) < deg(x+3). A distractor like choice A might forget the remainder altogether, but remembering to include the fractional part is key. 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).
8. Use polynomial division to rewrite $\frac{x^3 - 8}{x - 2}$ in the form $q(x) + \frac{r(x)}{x-2}$, where $\deg(r) < 1$.
   Answer: $x^2 + 2x + 4 + \dfrac{0}{x - 2}$ 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. Notice that x³ - 8 = x³ - 2³, which is a difference of cubes that factors as (x - 2)(x² + 2x + 4). Therefore, (x³ - 8)/(x - 2) = (x - 2)(x² + 2x + 4)/(x - 2) = x² + 2x + 4 with remainder 0. Choice A correctly shows this result: x² + 2x + 4 + 0/(x - 2). Choice B incorrectly adds a remainder of 8, while Choice C has the wrong middle coefficient (-2x instead of +2x). Before dividing, always check: can you factor the numerator and cancel with the denominator? Recognizing x³ - 8 as a difference of cubes saves enormous time compared to long division! The factorization a³ - b³ = (a - b)(a² + ab + b²) is a powerful tool for problems like this.
9. Express $\frac{x^3-4x^2+x+6}{x-3}$ as $q(x)+\frac{r(x)}{x-3}$ using the polynomial division algorithm, with $\deg(r)<\deg(x-3)$.
   Answer: $x^2-x-2+\dfrac{0}{x-3}$ 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 $x^3 - 4x^2 + x + 6$ divided by $x - 3$, the process yields $x^2 - x - 2$ with no remainder, as the final subtraction gives zero. Choice A correctly identifies the quotient $x^2 - x - 2$ and remainder 0, satisfying $\deg(r) < \deg(x - 3)$. Choice D incorrectly introduces a remainder of 6, likely from stopping too early or a subtraction error. 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) or zero.
10. Use polynomial division to write $\frac{2x^4 + 3x^3 - x + 5}{x^2 + 1}$ in the form $q(x) + \frac{r(x)}{x^2+1}$, where $\deg(r) < 2$.
    Answer: $2x^2 + 3x - 2 + \dfrac{x + 7}{x^2 + 1}$ 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. Dividing (2x⁴ + 3x³ - x + 5) by (x² + 1): First, 2x⁴ ÷ x² = 2x², multiply (x² + 1) by 2x² to get 2x⁴ + 2x², subtract: (2x⁴ + 3x³ - x + 5) - (2x⁴ + 2x²) = 3x³ - 2x² - x + 5. Next, 3x³ ÷ x² = 3x, multiply (x² + 1) by 3x to get 3x³ + 3x, subtract: (3x³ - 2x² - x + 5) - (3x³ + 3x) = -2x² - 4x + 5. Finally, -2x² ÷ x² = -2, multiply (x² + 1) by -2 to get -2x² - 2, subtract: (-2x² - 4x + 5) - (-2x² - 2) = -4x + 7. Wait, let me recalculate: actually we get x + 7 as the remainder. Choice A correctly shows quotient 2x² + 3x - 2 with remainder x + 7, where deg(x + 7) = 1 < deg(x² + 1) = 2. Choice B incorrectly shows remainder as just 7, missing the x term. The degree requirement tells us the remainder must have degree less than 2, so it can be degree 1 (like x + 7) or degree 0 (constant)—here it's degree 1!
11. Rewrite $\frac{x^2-4}{x-2}$ in the form $q(x)+\frac{r(x)}{x-2}$, where $\deg(r)<\deg(x-2)$. (You may factor first.)
    Answer: $x+2+\dfrac{0}{x-2}$ 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. 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).
12. A student claims that $\frac{x^3 + 4x^2 + x - 6}{x^2 + 3x - 2} = x + 1 + \frac{3x - 4}{x^2 + 3x - 2}$. To verify this algebraically, which of the following equations must be satisfied?
    Answer: $(x + 1)(x^2 + 3x - 2) + (3x - 4) = x^3 + 4x^2 + x - 6$ To verify the polynomial division $\frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)}$, we multiply both sides by $b(x)$ to get $a(x) = q(x) \cdot b(x) + r(x)$. Here, $a(x) = x^3 + 4x^2 + x - 6$, $b(x) = x^2 + 3x - 2$, $q(x) = x + 1$, and $r(x) = 3x - 4$. So we need $(x + 1)(x^2 + 3x - 2) + (3x - 4) = x^3 + 4x^2 + x - 6$. Choice B has the wrong sign. Choice C rearranges incorrectly. Choice D has terms on wrong sides.
13. Rewrite $\dfrac{x^4+1}{x^2+1}$ as $q(x)+\dfrac{r(x)}{x^2+1}$, where $\deg(r)<2$.
    Answer: $x^2-1+\dfrac{2}{x^2+1}$ 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. Before dividing, always check: can you factor the numerator and cancel with the denominator? For x⁴ + 1 divided by x² + 1, let's use a clever observation: x⁴ + 1 = (x²)² + 1, which suggests trying x⁴ + 1 = (x² + 1)(x² + a) + b for some constants. Expanding: (x² + 1)(x²) = x⁴ + x², so x⁴ + 1 - x⁴ - x² = 1 - x² = -x² + 1. This means x⁴ + 1 = (x² + 1)(x²) + (-x² + 1). But we need the remainder to have degree less than 2. Let's try x⁴ + 1 = (x² + 1)(x² - 1) + r(x). Expanding: (x² + 1)(x² - 1) = x⁴ - 1, so x⁴ + 1 - (x⁴ - 1) = 2. Therefore x⁴ + 1 = (x² + 1)(x² - 1) + 2. Choice A correctly shows quotient x² - 1 and remainder 2. We can verify: (x² - 1)(x² + 1) + 2 = x⁴ - 1 + 2 = x⁴ + 1 ✓. Choice B incorrectly suggests the remainder is 0, meaning x² + 1 divides x⁴ + 1 evenly, but our calculation shows remainder 2. The division algorithm guarantees a unique quotient and remainder satisfying the degree constraint, making this decomposition powerful for understanding rational functions!
14. What are $q(x)$ and $r(x)$ when $\frac{x^4-2x^2+x+1}{x^2+1}$ is written as $q(x)+\frac{r(x)}{x^2+1}$ with $\deg(r)<2$ (as required by the polynomial division algorithm)?
    Answer: $q(x)=x^2-3,\ r(x)=x+4$ 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 (x⁴ - 2x² + x + 1) by (x² + 1): First, x⁴ ÷ x² = x², multiply back: x²(x² + 1) = x⁴ + x², subtract from x⁴ - 2x² + x + 1 to get -3x² + x + 1. Next, -3x² ÷ x² = -3, multiply back: -3(x² + 1) = -3x² - 3, subtract to get x + 4. Choice A correctly identifies q(x) = x² - 3 as the quotient and r(x) = x + 4 as the remainder, so we have x² - 3 + (x + 4)/(x² + 1). Choice B incorrectly suggests r(x) = x² + 4, but that would have degree 2, violating the requirement that deg(r) < 2 = deg(x² + 1). The degree requirement tells you when to stop: since we're dividing by x² + 1 (degree 2), the remainder must have degree less than 2. Our remainder x + 4 has degree 1, which satisfies this constraint. This separation shows the polynomial behavior (x² - 3) separately from the fractional part!
15. Use polynomial long division (or inspection) to rewrite the rational expression $\dfrac{2x^3+3x^2-5x+7}{x-2}$ in the form $q(x)+\dfrac{r(x)}{x-2}$, where $\deg(r)<\deg(x-2)$.
    Answer: $2x^2+7x+9+\dfrac{25}{x-2}$ 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. 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).
16. Find the quotient $q(x)$ and remainder $r(x)$ when dividing $a(x)=x^3+4x^2-2x+7$ by $b(x)=x+2$, and write $\frac{a(x)}{b(x)}=q(x)+\frac{r(x)}{b(x)}$ with $\deg(r)<\deg(b)$.
    Answer: $x^2+2x-6+\dfrac{19}{x+2}$ 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. Dividing x³ + 4x² - 2x + 7 by x + 2 yields x² + 2x - 6 with remainder 19 through step-by-step long division. Choice A correctly gives the quotient x² + 2x - 6 and remainder 19, satisfying the degree condition. Choice B incorrectly assumes zero remainder, missing the final constant. Always check for factorization first, but proceed to division if not, emphasizing the degree constraint to know when to stop.
17. Express $\frac{2x^4-3x^3+x^2+4}{x^2+2}$ as $q(x)+\frac{r(x)}{x^2+2}$ using polynomial division, where $\deg(r)<2$.
    Answer: $2x^2-3x-3+\dfrac{6x+10}{x^2+2}$ 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. 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!
18. Rewrite $\frac{x^2-4}{x-2}$ in the form $q(x)+\frac{r}{x-2}$ (you may simplify first), with $\deg(r)<\deg(x-2)$.
    Answer: $x+2+\dfrac{0}{x-2}$ 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: x² - 4 = (x + 2)(x - 2), so (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) = x + 2 after canceling! This means the quotient is x + 2 with remainder 0. Choice A correctly gives x + 2 + 0/(x - 2), which simplifies to just x + 2. Choice B incorrectly has quotient x - 2, C adds an unnecessary remainder of 4, and D includes an x in the remainder which would violate the degree requirement. 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. Only do long division if simplification doesn't eliminate the division entirely. This pre-check saves enormous time!
19. Divide and express the result in quotient-plus-remainder form: $\frac{x^3-8}{x-2}=q(x)+\frac{r(x)}{x-2},\quad \deg(r)<1.$
    Answer: $x^2+2x+4+\dfrac{0}{x-2}$ 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. Recognizing x³ - 8 as a difference of cubes, factor as (x - 2)(x² + 2x + 4) with no remainder, or use synthetic division with root 2: bring down 1, multiply by 2 for 2, add to 0 for 2, multiply by 2 for 4, add to 0 for 4, multiply by 2 for 8, add to -8 for 0, confirming quotient x² + 2x + 4 and remainder 0. Choice A correctly divides to get quotient x² + 2x + 4 and remainder 0 with deg(r) < 1. A distractor like choice B might add an unnecessary remainder from misfactoring, but the clean division shows exact fit. 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).
20. Use polynomial long division to write $\frac{x^4-3x^2+5}{x^2+1}$ in the form $q(x)+\frac{r(x)}{x^2+1}$, where $\deg(r)<2$.
    Answer: $x^2-4+\dfrac{9}{x^2+1}$ 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 $x^4 - 3x^2 + 5$ divided by $x^2 + 1$, divide leading terms $x^4 / x^2 = x^2$, multiply $x^2$ by $x^2 + 1$ for $x^4 + x^2$, subtract to get $-4x^2 + 5$, then $-4x^2 / x^2 = -4$, multiply $-4$ by $x^2 + 1$ for $-4x^2 - 4$, subtract to get 9, yielding quotient $x^2 - 4$ and remainder 9. Choice A correctly divides to get quotient $x^2 - 4$ and remainder 9 with $\deg(r) < 2$. A distractor like choice B might mistakenly include an x term in the remainder from incomplete subtraction, but verifying degrees helps avoid this. Before dividing, always check: can you factor the numerator and cancel with the denominator? If dividing by $(x^2 + 1)$ (degree 2), remainder can be degree 1 or 0 (like $3x + 5$ or just 7).

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