Algebra 2 Quiz: Rewriting Rational Expressions
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Rewriting Rational ExpressionsQuestion 1 of 20

Use the polynomial division algorithm to write x4−3x2+5x2+1\frac{x^4-3x^2+5}{x^2+1} in the form q(x)+r(x)x2+1q(x)+\frac{r(x)}{x^2+1}, where deg⁡(r)<2\deg(r)<2.

x2−4+9x2+1x^2-4+\dfrac{9}{x^2+1}
x2−4+9xx2+1x^2-4+\dfrac{9x}{x^2+1}
x2−3+8x2+1x^2-3+\dfrac{8}{x^2+1}
x2−4+9x+1x^2-4+\dfrac{9}{x+1}
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Algebra 2 Quiz

Algebra 2 Quiz: Rewriting Rational Expressions

Practice Rewriting Rational Expressions in Algebra 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Rewriting Rational Expressions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Use the polynomial division algorithm to write x4−3x2+5x2+1\frac{x^4-3x^2+5}{x^2+1} in the form q(x)+r(x)x2+1q(x)+\frac{r(x)}{x^2+1}, where deg⁡(r)<2\deg(r)<2.

  1. x2−4+9x2+1x^2-4+\dfrac{9}{x^2+1} (correct answer)
  2. x2−4+9xx2+1x^2-4+\dfrac{9x}{x^2+1}
  3. x2−3+8x2+1x^2-3+\dfrac{8}{x^2+1}
  4. x2−4+9x+1x^2-4+\dfrac{9}{x+1}
Explanation: 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 (x⁴ - 3x² + 5) by (x² + 1): First, x⁴ ÷ x² = x², multiply back: x²(x² + 1) = x⁴ + x², subtract from x⁴ - 3x² + 5 to get -4x² + 5. Next, -4x² ÷ x² = -4, multiply back: -4(x² + 1) = -4x² - 4, subtract to get 9. Choice A correctly shows quotient x² - 4 and remainder 9, giving us x² - 4 + 9/(x² + 1). Choice B incorrectly has 9x as remainder instead of 9, which would have degree 1—but that's still valid since deg(9x) = 1 < 2 = deg(x² + 1), though the division shows remainder is just 9. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here x⁴ - 3x² + 5 doesn't factor nicely with x² + 1, so division is needed. Since we're dividing by x² + 1 (degree 2), the remainder can have degree at most 1, and here it's degree 0 (the constant 9).

Question 2

Divide x3−1x^3-1 by x−1x-1 and express x3−1x−1\frac{x^3-1}{x-1} in the form q(x)+rx−1q(x)+\frac{r}{x-1}, where deg⁡(r)<deg⁡(x−1)\deg(r)<\deg(x-1).

  1. x2+x+1+0x−1x^2+x+1+\dfrac{0}{x-1} (correct answer)
  2. x2−x+1+0x−1x^2-x+1+\dfrac{0}{x-1}
  3. x2+x+1+1x−1x^2+x+1+\dfrac{1}{x-1}
  4. x2+x+1x−1x^2+x+\dfrac{1}{x-1}
Explanation: 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!

Question 3

Find the quotient q(x)q(x) and remainder r(x)r(x) when dividing x3+4x2−2x+7x^3 + 4x^2 - 2x + 7 by x+2x + 2, and write x3+4x2−2x+7x+2=q(x)+r(x)x+2\frac{x^3 + 4x^2 - 2x + 7}{x + 2} = q(x) + \frac{r(x)}{x+2} with deg⁡(r)<1\deg(r) < 1.

  1. x2+2x−6+19x+2x^2 + 2x - 6 + \dfrac{19}{x + 2} (correct answer)
  2. x2+2x−6+−19x+2x^2 + 2x - 6 + \dfrac{-19}{x + 2}
  3. x2+6x+10+19x+2x^2 + 6x + 10 + \dfrac{19}{x + 2}
  4. x2+2x+19xx+2x^2 + 2x + \dfrac{19x}{x + 2}
Explanation: 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 (x³ + 4x² - 2x + 7) by (x + 2): First, x³ ÷ x = x², multiply (x + 2) by x² to get x³ + 2x², subtract: (x³ + 4x² - 2x + 7) - (x³ + 2x²) = 2x² - 2x + 7. Next, 2x² ÷ x = 2x, multiply (x + 2) by 2x to get 2x² + 4x, subtract: (2x² - 2x + 7) - (2x² + 4x) = -6x + 7. Finally, -6x ÷ x = -6, multiply (x + 2) by -6 to get -6x - 12, subtract: (-6x + 7) - (-6x - 12) = 19. Choice A correctly gives quotient x² + 2x - 6 with remainder 19, where deg(19) = 0 < deg(x + 2) = 1. Choice B has the wrong sign on the remainder (-19), while Choice C has completely different quotient coefficients. Before dividing, always check: can you factor the numerator and cancel with the denominator? If not, proceed systematically with long division, being careful with signs when subtracting!

Question 4

Express 3x3−2x2+4x+1\frac{3x^3-2x^2+4}{x+1} as q(x)+r(x)x+1q(x)+\frac{r(x)}{x+1} using polynomial division, with deg⁡(r)<1\deg(r)<1.

  1. 3x2−3x+1+3x+13x^2-3x+1+\dfrac{3}{x+1}
  2. 3x2−5x+5+−1x+13x^2-5x+5+\dfrac{-1}{x+1} (correct answer)
  3. 3x2−5x+−1x+13x^2-5x+\dfrac{-1}{x+1}
  4. 3x2−5x+5+1x+13x^2-5x+5+\dfrac{1}{x+1}
Explanation: 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).

Question 5

Use polynomial division to rewrite the rational expression 2x3+x2−5x+3x−2\frac{2x^3 + x^2 - 5x + 3}{x - 2} in the form q(x)+r(x)x−2q(x) + \frac{r(x)}{x-2}, where deg⁡(r)<deg⁡(x−2)\deg(r) < \deg(x-2).

  1. 2x2+5x+5+13x−22x^2 + 5x + 5 + \dfrac{13}{x - 2} (correct answer)
  2. 2x2+5x+5+7x−22x^2 + 5x + 5 + \dfrac{7}{x - 2}
  3. 2x2+x−5+13x−22x^2 + x - 5 + \dfrac{13}{x - 2}
  4. 2x2+5x+13xx−22x^2 + 5x + \dfrac{13x}{x - 2}
Explanation: 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 + x² - 5x + 3) by (x - 2): First, 2x³ ÷ x = 2x², so multiply (x - 2) by 2x² to get 2x³ - 4x², then subtract: (2x³ + x² - 5x + 3) - (2x³ - 4x²) = 5x² - 5x + 3. Next, 5x² ÷ x = 5x, multiply (x - 2) by 5x to get 5x² - 10x, subtract: (5x² - 5x + 3) - (5x² - 10x) = 5x + 3. Finally, 5x ÷ x = 5, multiply (x - 2) by 5 to get 5x - 10, subtract: (5x + 3) - (5x - 10) = 13. Choice A correctly shows the quotient 2x² + 5x + 5 with remainder 13, satisfying deg(13) = 0 < deg(x - 2) = 1. Choice D incorrectly leaves 13x in the remainder instead of just 13, violating the degree requirement. Before dividing, always check: can you factor the numerator and cancel with the denominator? If not, proceed with long division systematically, stopping when the remainder's degree drops below the divisor's degree—here, once we get the constant 13, we're done!

Question 6

Use polynomial long division (or inspection) to rewrite the rational expression 2x3+3x2−5x+7x−2\dfrac{2x^3+3x^2-5x+7}{x-2} in the form q(x)+r(x)x−2q(x)+\dfrac{r(x)}{x-2}, where deg⁡(r)<deg⁡(x−2)\deg(r)<\deg(x-2).

  1. 2x2−x−7+21x−22x^2-x-7+\dfrac{21}{x-2}
  2. 2x2+7x+9x+25x−22x^2+7x+\dfrac{9x+25}{x-2}
  3. 2x2+7x+9+25x−22x^2+7x+9+\dfrac{25}{x-2} (correct answer)
  4. 2x2+7x+9+5x−22x^2+7x+9+\dfrac{5}{x-2}
Explanation: 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).

Question 7

Use polynomial division to rewrite the rational expression 2x3+3x2−5x+7x+2\frac{2x^3+3x^2-5x+7}{x+2} in the form q(x)+r(x)x+2q(x)+\frac{r(x)}{x+2}, where deg⁡(r)<deg⁡(x+2)\deg(r)<\deg(x+2).

  1. 2x2−x−3+13x+22x^2-x-3+\dfrac{13}{x+2} (correct answer)
  2. 2x2−x−3+1x+22x^2-x-3+\dfrac{1}{x+2}
  3. 2x2−x+1+13x+22x^2-x+1+\dfrac{13}{x+2}
  4. 2x2−x−3+13xx+22x^2-x-3+\dfrac{13x}{x+2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/517/5 as 3+2/53 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x)a(x)/b(x) can be written as q(x)+r(x)b(x)q(x) + \frac{r(x)}{b(x)}, where q(x)q(x) is the quotient (polynomial part) and r(x)r(x) is the remainder with degree strictly less than the divisor's degree. To divide 2x3+3x2−5x+72x^3 + 3x^2 - 5x + 7 by x+2x + 2, start by dividing 2x32x^3 by xx to get 2x22x^2, multiply by the divisor to get 2x3+4x22x^3 + 4x^2, subtract to yield −x2−5x-x^2 - 5x, then continue to get −x-x and −3-3, resulting in a remainder of 13. Choice A correctly divides to get quotient 2x2−x−32x^2 - x - 3 and remainder 13, with deg⁡(r)<deg⁡(x+2)\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⁡(remainder)<deg⁡(divisor)\deg(\text{remainder}) < \deg(\text{divisor}), just like stopping in numerical division when the remainder is smaller than the divisor.

Question 8

Find the quotient q(x)q(x) and remainder r(x)r(x) when dividing x3+4x2−2x+7x^3+4x^2-2x+7 by x+2x+2, and write x3+4x2−2x+7x+2=q(x)+rx+2\frac{x^3+4x^2-2x+7}{x+2}=q(x)+\frac{r}{x+2} with deg⁡(r)<deg⁡(x+2)\deg(r)<\deg(x+2).

  1. x2+2x−6+19x+2x^2+2x-6+\dfrac{19}{x+2} (correct answer)
  2. x2+2x−6−19x+2x^2+2x-6-\dfrac{19}{x+2}
  3. x2+6x−14+35x+2x^2+6x-14+\dfrac{35}{x+2}
  4. x2+2x+6+−5x+2x^2+2x+6+\dfrac{-5}{x+2}
Explanation: 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!

Question 9

Use polynomial division to rewrite the rational expression 2x3+3x2−5x+6x−2\frac{2x^3+3x^2-5x+6}{x-2} in the form q(x)+r(x)x−2q(x)+\frac{r(x)}{x-2}, where deg⁡(r)<deg⁡(x−2)\deg(r)<\deg(x-2) (the polynomial division algorithm).

  1. 2x2+7x+9+24x−22x^2+7x+9+\dfrac{24}{x-2} (correct answer)
  2. 2x2+7x+9+12x−22x^2+7x+9+\dfrac{12}{x-2}
  3. 2x2+7x+9+24x+22x^2+7x+9+\dfrac{24}{x+2}
  4. 2x2+7x+24xx−22x^2+7x+\dfrac{24x}{x-2}
Explanation: 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!

Question 10

Express 2x2+7x−3x+3\frac{2x^2+7x-3}{x+3} as q(x)+r(x)x+3q(x)+\frac{r(x)}{x+3} using the polynomial division algorithm, where deg⁡(r)<deg⁡(x+3)\deg(r)<\deg(x+3).

  1. 2x+1+−6x+32x+1+\dfrac{-6}{x+3} (correct answer)
  2. 2x+1+6x+32x+1+\dfrac{6}{x+3}
  3. 2x−1+0x+32x-1+\dfrac{0}{x+3}
  4. 2x+1+−6xx+32x+1+\dfrac{-6x}{x+3}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/517/5 as 3+2/53 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x)a(x)/b(x) can be written as q(x)+r(x)/b(x)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 2x2+7x−32x^2 + 7x - 3 by x+3x + 3, divide 2x22x^2 by xx to get 2x2x, multiply by (x+3)(x + 3) to get 2x2+6x2x^2 + 6x, subtract to get x−3x - 3, then divide xx by xx to get 1, multiply to get x+3x + 3, and subtract to yield -6 as remainder. Choice A correctly divides to get quotient 2x+12x + 1 and remainder -6 with deg⁡(r)<deg⁡(x+3)\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⁡\deg of remainder less than deg⁡\deg of divisor) tells you when to stop dividing: if dividing by (x+3)(x + 3) (degree 1), remainder must be degree 0 (constant).

Question 11

Rewrite 3x2−2x+4x−1\frac{3x^2-2x+4}{x-1} in the form q(x)+r(x)x−1q(x)+\frac{r(x)}{x-1}, where deg⁡(r)<deg⁡(x−1)\deg(r)<\deg(x-1). (You may use inspection or long division.)

  1. 3x+1+5x−13x+1+\dfrac{5}{x-1} (correct answer)
  2. 3x+2+4x−13x+2+\dfrac{4}{x-1}
  3. 3x+1+5xx−13x+1+\dfrac{5x}{x-1}
  4. 3x−1+5x−13x-1+\dfrac{5}{x-1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/517/5 as 3+2/53 + 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 3x2−2x+43x^2 - 2x + 4 divided by x−1x - 1, division gives 3x+13x + 1 with remainder 5, verifiable by inspection or synthetic division. Choice A correctly provides the quotient 3x+13x + 1 and remainder 5, with deg⁡(r)<1\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.

Question 12

Use polynomial division to express 2x3−x2+0x+3x2+1\frac{2x^3-x^2+0x+3}{x^2+1} in the form q(x)+r(x)x2+1q(x)+\frac{r(x)}{x^2+1}, where deg⁡(r)<2\deg(r)<2.

  1. 2x−1+−2x+4x2+12x-1+\dfrac{-2x+4}{x^2+1} (correct answer)
  2. 2x−1+−2x+4x2−12x-1+\dfrac{-2x+4}{x^2-1}
  3. 2x+1+−2x+4x2+12x+1+\dfrac{-2x+4}{x^2+1}
  4. 2x−1+−2x2+4x2+12x-1+\dfrac{-2x^2+4}{x^2+1}
Explanation: 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 2x³ - x² + 3 divided by x² + 1, the process results in 2x - 1 with remainder -2x + 4. Choice A correctly identifies this quotient and remainder, with deg(r) < 2. Choice D has an invalid remainder degree, equaling the divisor's. The degree requirement ensures you stop once the remainder is linear or constant for a quadratic divisor.

Question 13

Use polynomial long division to write x4−3x2+5x2+1\frac{x^4 - 3x^2 + 5}{x^2 + 1} in the form q(x)+r(x)x2+1q(x) + \frac{r(x)}{x^2+1}, where deg⁡(r)<2\deg(r) < 2.

  1. x2−4+9x2+1x^2 - 4 + \dfrac{9}{x^2 + 1} (correct answer)
  2. x2−4+9xx2+1x^2 - 4 + \dfrac{9x}{x^2 + 1}
  3. x2−3+8x2+1x^2 - 3 + \dfrac{8}{x^2 + 1}
  4. x2−4+5x2+1x^2 - 4 + \dfrac{5}{x^2 + 1}
Explanation: 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. When dividing (x⁴ - 3x² + 5) by (x² + 1), we treat it like dividing by a quadratic: First, x⁴ ÷ x² = x², multiply (x² + 1) by x² to get x⁴ + x², subtract: (x⁴ - 3x² + 5) - (x⁴ + x²) = -4x² + 5. Next, -4x² ÷ x² = -4, multiply (x² + 1) by -4 to get -4x² - 4, subtract: (-4x² + 5) - (-4x² - 4) = 9. Choice A correctly shows quotient x² - 4 with remainder 9, where deg(9) = 0 < deg(x² + 1) = 2. Choice B incorrectly has remainder 9x instead of 9, violating the degree constraint since deg(9x) = 1 is not less than 2 but still acceptable—however, the division process yields 9, not 9x. Before dividing, always check the degrees: since we're dividing by a degree-2 polynomial, our remainder can have degree at most 1. Here we got degree 0 (the constant 9), which satisfies the requirement perfectly!

Question 14

Use polynomial long division (or inspection) to rewrite the rational expression 2x3+x2−5x+3x−2\frac{2x^3+x^2-5x+3}{x-2} in the form q(x)+r(x)x−2q(x)+\frac{r(x)}{x-2}, where deg⁡(r)<deg⁡(x−2)\deg(r)<\deg(x-2).

  1. 2x2+5x+5+13x−22x^2+5x+5+\dfrac{13}{x-2} (correct answer)
  2. 2x2+5x+5+13x+22x^2+5x+5+\dfrac{13}{x+2}
  3. 2x2+5x+13x−22x^2+5x+\dfrac{13}{x-2}
  4. 2x2+5x+5+13xx−22x^2+5x+5+\dfrac{13x}{x-2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 175\frac{17}{5} as 3+253 + \frac{2}{5}. The division algorithm for polynomials says any rational expression a(x)b(x)\frac{a(x)}{b(x)} can be written as q(x)+r(x)b(x)q(x) + \frac{r(x)}{b(x)}, where q(x)q(x) is the quotient (polynomial part) and r(x)r(x) is the remainder with degree strictly less than the divisor's degree. To divide 2x3+x2−5x+32x^3 + x^2 - 5x + 3 by x−2x - 2, start by dividing 2x32x^3 by xx to get 2x22x^2, multiply by (x−2)(x - 2) to get 2x3−4x22x^3 - 4x^2, subtract to yield 5x2−5x5x^2 - 5x, then continue to get 5x5x next, and finally 55, resulting in a remainder of 1313. Choice A correctly divides to get quotient 2x2+5x+52x^2 + 5x + 5 and remainder 1313 with deg⁡(r)<deg⁡(x−2)\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⁡\deg of remainder less than deg⁡\deg of divisor) tells you when to stop dividing: if dividing by (x−2)(x - 2) (deg⁡\deg 1), remainder must be deg⁡\deg 0 (constant).

Question 15

Use polynomial long division to write x4−3x2+5x2+1\frac{x^4-3x^2+5}{x^2+1} in the form q(x)+r(x)x2+1q(x)+\frac{r(x)}{x^2+1}, where deg⁡(r)<2\deg(r)<2.

  1. x2−4+9x2+1x^2-4+\dfrac{9}{x^2+1} (correct answer)
  2. x2−4+9xx2+1x^2-4+\dfrac{9x}{x^2+1}
  3. x2−3+8x2+1x^2-3+\dfrac{8}{x^2+1}
  4. x2−4+9x+9x2+1x^2-4+\dfrac{9x+9}{x^2+1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/517/5 as 3+2/53 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x)a(x)/b(x) can be written as q(x)+r(x)/b(x)q(x) + r(x)/b(x), where q(x)q(x) is the quotient (polynomial part) and r(x)r(x) is the remainder with degree strictly less than the divisor's degree. For x4−3x2+5x^4 - 3x^2 + 5 divided by x2+1x^2 + 1, divide leading terms x4/x2=x2x^4 / x^2 = x^2, multiply x2x^2 by x2+1x^2 + 1 for x4+x2x^4 + x^2, subtract to get −4x2+5-4x^2 + 5, then −4x2/x2=−4-4x^2 / x^2 = -4, multiply −4-4 by x2+1x^2 + 1 for −4x2−4-4x^2 - 4, subtract to get 9, yielding quotient x2−4x^2 - 4 and remainder 9. Choice A correctly divides to get quotient x2−4x^2 - 4 and remainder 9 with deg⁡(r)<2\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 (x2+1)(x^2 + 1) (degree 2), remainder can be degree 1 or 0 (like 3x+53x + 5 or just 7).

Question 16

Use polynomial division to write x4−2x3+3x2+4x2+1\dfrac{x^4-2x^3+3x^2+4}{x^2+1} in the form q(x)+r(x)x2+1q(x)+\dfrac{r(x)}{x^2+1}, where deg⁡(r)<2\deg(r)<2.

  1. x2−2x+2+2x2+1x^2-2x+2+\dfrac{2}{x^2+1}
  2. x2−2x+1+2x+3x2+1x^2-2x+1+\dfrac{2x+3}{x^2+1}
  3. x2−2x+2+2x+2x2+1x^2-2x+2+\dfrac{2x+2}{x^2+1} (correct answer)
  4. x2−2x+3+2x+1x2+1x^2-2x+3+\dfrac{2x+1}{x^2+1}
Explanation: 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. When dividing by x² + 1, we divide x⁴ - 2x³ + 3x² + 4 by x² + 1: First, x⁴ ÷ x² = x², multiply back: x²(x² + 1) = x⁴ + x², subtract: (x⁴ - 2x³ + 3x² + 4) - (x⁴ + x²) = -2x³ + 2x² + 4. Next, -2x³ ÷ x² = -2x, multiply back: -2x(x² + 1) = -2x³ - 2x, subtract: (-2x³ + 2x² + 4) - (-2x³ - 2x) = 2x² + 2x + 4. Then, 2x² ÷ x² = 2, multiply back: 2(x² + 1) = 2x² + 2, subtract: (2x² + 2x + 4) - (2x² + 2) = 2x + 2. Since deg(2x + 2) = 1 < deg(x² + 1) = 2, we stop here. Choice A correctly gives quotient x² - 2x + 2 and remainder 2x + 2, satisfying the degree requirement. Choice C incorrectly has remainder 2 (missing the 2x term), likely from stopping the division too early. 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 (like 2x + 2 or just 7). Once your remainder's degree drops below the divisor's, you're done—continuing would be like dividing 2 by 5 further after getting 0 remainder 2.

Question 17

Use polynomial division to rewrite x3+2x2−5x+3x−1\frac{x^3+2x^2-5x+3}{x-1} in the form q(x)+r(x)x−1q(x)+\frac{r(x)}{x-1}, where deg⁡(r)<deg⁡(x−1)\deg(r)<\deg(x-1).

  1. x2+3x−2+1x−1x^2+3x-2+\dfrac{1}{x-1} (correct answer)
  2. x2+3x+1x−1x^2+3x+\dfrac{1}{x-1}
  3. x2+3x−2+−1x−1x^2+3x-2+\dfrac{-1}{x-1}
  4. x2+2x−5+3x−1x^2+2x-5+\dfrac{3}{x-1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 175\frac{17}{5} as 3+253 + \frac{2}{5}. The division algorithm for polynomials says any rational expression a(x)/b(x)a(x)/b(x) can be written as q(x)+r(x)b(x)q(x) + \frac{r(x)}{b(x)}, where q(x)q(x) is the quotient (polynomial part) and r(x)r(x) is the remainder with degree strictly less than the divisor's degree. Using synthetic division for x3+2x2−5x+3x^3 + 2x^2 - 5x + 3 by x−1x - 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 x2+3x−2x^2 + 3x - 2 and remainder 11 with deg⁡(r)<deg⁡(x−1)\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⁡\deg of remainder less than deg⁡\deg of divisor) tells you when to stop dividing: if dividing by (x−1)(x - 1) (degree 1), remainder must be degree 0 (constant).

Question 18

Rewrite x2−4x−2\frac{x^2-4}{x-2} in the form q(x)+rx−2q(x)+\frac{r}{x-2} (you may simplify first), with deg⁡(r)<deg⁡(x−2)\deg(r)<\deg(x-2).

  1. x+2+0x−2x+2+\dfrac{0}{x-2} (correct answer)
  2. x−2+0x−2x-2+\dfrac{0}{x-2}
  3. x+2+4x−2x+2+\dfrac{4}{x-2}
  4. x+2+xx−2x+2+\dfrac{x}{x-2}
Explanation: 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!

Question 19

Use polynomial division (or simplify first if you notice a pattern) to express x2−4x−2\frac{x^2-4}{x-2} in the form q(x)+r(x)x−2q(x)+\frac{r(x)}{x-2}, where deg⁡(r)<1\deg(r)<1.

  1. x+2+0x−2x+2+\dfrac{0}{x-2} (correct answer)
  2. x−2+0x−2x-2+\dfrac{0}{x-2}
  3. x+2+4x−2x+2+\dfrac{4}{x-2}
  4. x+2+xx−2x+2+\dfrac{x}{x-2}
Explanation: 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? Notice that x² - 4 = (x + 2)(x - 2), so (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) = x + 2 after canceling the common factor (x - 2). This gives us x + 2 with remainder 0, or x + 2 + 0/(x - 2). Choice A correctly shows this result: x + 2 + 0/(x - 2), which simplifies to just x + 2 since adding 0 doesn't change the value. Choice C incorrectly suggests remainder 4, but when you multiply (x + 2)(x - 2) you get x² - 4 exactly, with no remainder. This pre-check saves enormous time! When you can factor and cancel, the division becomes trivial—no long division needed. The remainder is 0 (degree undefined or -∞), which certainly satisfies deg(r) < 1.

Question 20

Use polynomial long division to express x4+1x−1\frac{x^4 + 1}{x - 1} in the form q(x)+r(x)x−1q(x) + \frac{r(x)}{x-1}, where deg⁡(r)<1\deg(r) < 1.

  1. x3+x2+x+1+2x−1x^3 + x^2 + x + 1 + \dfrac{2}{x - 1} (correct answer)
  2. x3+x2+x+1+0x−1x^3 + x^2 + x + 1 + \dfrac{0}{x - 1}
  3. x3−x2+x−1+2x−1x^3 - x^2 + x - 1 + \dfrac{2}{x - 1}
  4. x3+x2+x+2x−1x^3 + x^2 + x + \dfrac{2}{x - 1}
Explanation: 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 (x⁴ + 1) by (x - 1): First, x⁴ ÷ x = x³, multiply (x - 1) by x³ to get x⁴ - x³, subtract: (x⁴ + 1) - (x⁴ - x³) = x³ + 1. Next, x³ ÷ x = x², multiply by x² to get x³ - x², subtract: (x³ + 1) - (x³ - x²) = x² + 1. Then x² ÷ x = x, multiply by x to get x² - x, subtract: (x² + 1) - (x² - x) = x + 1. Finally, x ÷ x = 1, multiply by 1 to get x - 1, subtract: (x + 1) - (x - 1) = 2. Choice A correctly shows quotient x³ + x² + x + 1 with remainder 2, where deg(2) = 0 < deg(x - 1) = 1. Choice B incorrectly claims remainder 0, which would mean x - 1 divides x⁴ + 1 evenly (false: 1⁴ + 1 = 2 ≠ 0). The degree requirement tells you when to stop: once you get a constant remainder (here, 2), you cannot divide further by (x - 1)!