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

What are the quotient q(x)q(x) and remainder r(x)r(x) when dividing x3+4x2+x+6x^3+4x^2+x+6 by x+3x+3? Then write x3+4x2+x+6x+3=q(x)+r(x)x+3\frac{x^3+4x^2+x+6}{x+3}=q(x)+\frac{r(x)}{x+3} with deg⁡(r)<1\deg(r)<1.​

x2+x−2+12x+3x^2+x-2+\frac{12}{x+3}
x2+x−2+0x+3x^2+x-2+\frac{0}{x+3}
x2+x+2+12x+3x^2+x+2+\frac{12}{x+3}
x2+7x+6+0x+3x^2+7x+6+\frac{0}{x+3}
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Algebra Quiz: Rewriting Rational Expressions

Practice Rewriting Rational Expressions in Algebra 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.

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

What are the quotient q(x)q(x) and remainder r(x)r(x) when dividing x3+4x2+x+6x^3+4x^2+x+6 by x+3x+3? Then write x3+4x2+x+6x+3=q(x)+r(x)x+3\frac{x^3+4x^2+x+6}{x+3}=q(x)+\frac{r(x)}{x+3} with deg⁡(r)<1\deg(r)<1.​

  1. x2+x−2+12x+3x^2+x-2+\frac{12}{x+3} (correct answer)
  2. x2+x−2+0x+3x^2+x-2+\frac{0}{x+3}
  3. x2+x+2+12x+3x^2+x+2+\frac{12}{x+3}
  4. x2+7x+6+0x+3x^2+7x+6+\frac{0}{x+3}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x) can be rewritten as q(x) + r(x)/b(x), where q(x) is the quotient (the polynomial part), and r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x) must be less than the degree of the divisor b(x)—just like in numerical division where the remainder must be less than the divisor! Dividing (x³ + 4x² + x + 6) by (x + 3) using long division: (1) Divide leading terms: x³/x = x² (first term of quotient). (2) Multiply: x²(x + 3) = x³ + 3x². (3) Subtract from dividend: (x³ + 4x² + x + 6) - (x³ + 3x²) = x² + x + 6. (4) Divide again: x²/x = x (second term of quotient). (5) Multiply: x(x + 3) = x² + 3x. (6) Subtract: (x² + x + 6) - (x² + 3x) = -2x + 6. (7) Divide once more: -2x/x = -2 (third term of quotient). (8) Multiply: -2(x + 3) = -2x - 6. (9) Subtract: (-2x + 6) - (-2x - 6) = 12. Result: quotient is x² + x - 2, remainder is 12, giving x² + x - 2 + 12/(x + 3). Choice A correctly shows x² + x - 2 + 12/(x + 3) where the quotient is a degree 2 polynomial and remainder 12 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has remainder 0, which can't be right because multiplying back doesn't give the original: (x² + x - 2)(x + 3) + 0 = x³ + 3x² + x² + 3x - 2x - 6 = x³ + 4x² + x - 6 ≠ x³ + 4x² + x + 6. The quotient and remainder must satisfy: quotient × divisor + remainder = original numerator. This verification always catches division errors! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. It's exactly like numerical long division! Practice with simpler examples first, then work up to more complex ones.

Question 2

What are the quotient q(x)q(x) and remainder r(x)r(x) when dividing x3−4x2+x+6x^3-4x^2+x+6 by x−3x-3? Then write x3−4x2+x+6x−3=q(x)+r(x)x−3\frac{x^3-4x^2+x+6}{x-3}=q(x)+\frac{r(x)}{x-3} with deg⁡(r)<1\deg(r)<1.

  1. x2−x+2+0x−3x^2-x+2+\frac{0}{x-3}
  2. x2−7x+22+72x−3x^2-7x+22+\frac{72}{x-3}
  3. x2−x−2+0x−3x^2-x-2+\frac{0}{x-3} (correct answer)
  4. x2−x−2+6x−3x^2-x-2+\frac{6}{x-3}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x) can be rewritten as q(x) + r(x)/b(x), where q(x) is the quotient (the polynomial part), and r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x) must be less than the degree of the divisor b(x)—just like in numerical division where the remainder must be less than the divisor! Before dividing, check if numerator factors: x3−4x2+x+6x^3 - 4x^2 + x + 6 factors as (x−3)(x2−x−2)(x - 3)(x^2 - x - 2) since it divides evenly (remainder 0), so the expression simplifies to x2−x−2x^2 - x - 2 with no remainder term! Choice A correctly shows x2−x−2+0x−3x^2 - x - 2 + \frac{0}{x-3} where the quotient is quadratic and remainder has degree less than 1 (it's 0), giving the proper rewritten form. Choice B has a remainder of 6, but since it divides evenly, remainder should be 0—always check if factoring first makes the division unnecessary by plugging in the root (x=3) to see if it's zero! Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? If yes, simplify first—it might eliminate the division entirely! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). If you don't, there's an error somewhere in your division.

Question 3

Rewrite 2x2+7x+1x+3\frac{2x^2+7x+1}{x+3} in the form q(x)+r(x)x+3q(x)+\frac{r(x)}{x+3} (quotient plus remainder over divisor), where deg⁡(r)<1\deg(r)<1.​

  1. 2x+1+−2x+32x+1+\frac{-2}{x+3} (correct answer)
  2. 2x+7+−20x+32x+7+\frac{-20}{x+3}
  3. 2x−1+4x+32x-1+\frac{4}{x+3}
  4. 2x+1+2x+32x+1+\frac{2}{x+3}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. For simple cases, inspection can work: if you need (2x² + 7x + 1)/(x + 3) and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try q(x) = 2x + a for some a, multiply (2x + a)(x + 3), match coefficients with 2x² + 7x + 1, and solve for a and the remainder. This 'educated guess and check' is faster than formal division when it works! For (2x² + 7x + 1)/(x + 3), we suspect quotient is linear: q(x) = 2x + a for some a. Multiplying: (2x + a)(x + 3) = 2x² + 6x + ax + 3a = 2x² + (6 + a)x + 3a. Matching with numerator 2x² + 7x + 1: coefficient of x gives 6 + a = 7, so a = 1. Constant term gives 3a = 3, but we have 1, so remainder is 1 - 3 = -2. Result: (2x + 1) + (-2)/(x + 3). Inspection works when you can guess the quotient form! Choice A correctly shows 2x + 1 + (-2)/(x + 3) where the quotient is linear and remainder -2 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the quotient wrong: when dividing 2x² by x, we get 2x, not something that would lead to 2x + 7. The leading term of the quotient comes from dividing the highest-degree terms of numerator and denominator. This is the first step of polynomial division! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). Let's check: (2x + 1)(x + 3) + (-2) = 2x² + 6x + x + 3 - 2 = 2x² + 7x + 1 ✓. If you don't get the original back, there's an error somewhere in your division. This check works every time and is much faster than redoing the whole division!

Question 4

Rewrite 2x3+x2−4x+5x+1\frac{2x^3+x^2-4x+5}{x+1} in the form q(x)+r(x)x+1q(x)+\frac{r(x)}{x+1} using polynomial division, with deg⁡(r)<deg⁡(x+1)\deg(r)<\deg(x+1).

  1. 2x2−x−3+8x+12x^2-x-3+\frac{8}{x+1} (correct answer)
  2. 2x2−x−3+−8x+12x^2-x-3+\frac{-8}{x+1}
  3. 2x2+x−3+8x+12x^2+x-3+\frac{8}{x+1}
  4. 2x2−x+3+8x+12x^2-x+3+\frac{8}{x+1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x) can be rewritten as q(x) + r(x)/b(x), where q(x) is the quotient (the polynomial part), and r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x) must be less than the degree of the divisor b(x)—just like in numerical division where the remainder must be less than the divisor! Dividing (2x³ + x² - 4x + 5) by (x + 1) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x + 1) = 2x³ + 2x². (3) Subtract from dividend: (2x³ + x² - 4x + 5) - (2x³ + 2x²) = -x² - 4x + 5. (4) Divide again: -x²/x = -x (second term of quotient). (5) Multiply: -x(x + 1) = -x² - x. (6) Subtract: (-x² - 4x + 5) - (-x² - x) = -3x + 5. (7) Divide once more: -3x/x = -3 (third term of quotient). (8) Multiply: -3(x + 1) = -3x - 3. (9) Subtract: (-3x + 5) - (-3x - 3) = 8. Result: quotient is 2x² - x - 3, remainder is 8, giving 2x² - x - 3 + 8/(x + 1). Choice A correctly shows 2x² - x - 3 + 8/(x + 1) where the quotient is a degree-2 polynomial and remainder 8 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong sign on the x term: it shows 2x² + x - 3 instead of 2x² - x - 3. When dividing -x² by x, we get -x, not +x. The signs in polynomial division require careful attention—each subtraction step can introduce sign errors! Verification is your friend: after dividing, multiply your quotient 2x² - x - 3 by the divisor (x + 1) and add your remainder 8. You get (2x² - x - 3)(x + 1) + 8 = 2x³ + 2x² - x² - x - 3x - 3 + 8 = 2x³ + x² - 4x + 5, which matches the original numerator perfectly! This check works every time and catches sign errors quickly.

Question 5

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

  1. x2−x+4+5x−1x^2-x+4+\frac{5}{x-1} (correct answer)
  2. x2−x+4+4x−1x^2-x+4+\frac{4}{x-1}
  3. x2+x+4+5x−1x^2+x+4+\frac{5}{x-1}
  4. x2−x+3+5x−1x^2-x+3+\frac{5}{x-1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. Dividing (x³ - 2x² + 5x + 1) by (x - 1) using long division: (1) Divide leading terms: x³/x = x² (first term of quotient). (2) Multiply: x²(x - 1) = x³ - x². (3) Subtract from dividend: (x³ - 2x² + 5x + 1) - (x³ - x²) = -x² + 5x + 1. (4) Divide again: -x²/x = -x (second term of quotient). (5) Multiply: -x(x - 1) = -x² + x. (6) Subtract: (-x² + 5x + 1) - (-x² + x) = 4x + 1. (7) Divide once more: 4x/x = 4 (third term of quotient). (8) Multiply: 4(x - 1) = 4x - 4. (9) Subtract: (4x + 1) - (4x - 4) = 5. Result: quotient is x² - x + 4, remainder is 5, giving x² - x + 4 + 5/(x - 1). Choice A correctly shows x² - x + 4 + 5/(x - 1) where the quotient is a degree 2 polynomial and remainder 5 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong remainder: the final subtraction gives (4x + 1) - (4x - 4) = 5, not 4. When subtracting (4x - 4) from (4x + 1), we get 4x + 1 - 4x + 4 = 5. Polynomial division has multiple steps where arithmetic errors can creep in—subtraction of polynomials is especially tricky. Double-check each subtraction step carefully! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). Let's check: (x² - x + 4)(x - 1) + 5 = x³ - x² - x² + x + 4x - 4 + 5 = x³ - 2x² + 5x + 1 ✓. If you don't get the original back, there's an error somewhere in your division. This check works every time and is much faster than redoing the whole division!

Question 6

Simplify and rewrite x2−4x−2\frac{x^2-4}{x-2} in the form q(x)+r(x)x−2q(x)+\frac{r(x)}{x-2}, with deg⁡(r)<1\deg(r)<1.

  1. x+2+xx−2x+2+\frac{x}{x-2}
  2. x+2+4x−2x+2+\frac{4}{x-2}
  3. x−2+0x−2x-2+\frac{0}{x-2}
  4. x+2+0x−2x+2+\frac{0}{x-2} (correct answer)
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how 17/5=3+2/517/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x)a(x)/b(x) can be rewritten as q(x)+r(x)/b(x)q(x) + r(x)/b(x), where q(x)q(x) is the quotient (the polynomial part), and r(x)r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x)r(x) must be less than the degree of the divisor b(x)b(x)—just like in numerical division where the remainder must be less than the divisor! Before dividing, check if numerator factors: x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2), so (x−2)(x+2)/(x−2)(x - 2)(x + 2)/(x - 2) simplifies to x+2x + 2 with remainder 0 (for x≠2x ≠ 2). Choice A correctly shows x+2+0/(x−2)x + 2 + 0/(x - 2) where the quotient is linear and remainder has degree less than 1 (it's 0), giving the proper rewritten form. Choice C has remainder 4, but since it factors and cancels, no remainder is needed—always check if factoring first makes the division unnecessary! Verification is your friend: after dividing, multiply your quotient q(x)q(x) by the divisor b(x)b(x) and add your remainder r(x)r(x). You should get back the original numerator a(x)a(x). If you don't, there's an error somewhere in your division.

Question 7

Verify by multiplying back that the following rewriting is correct. Which expression equals x2+6x−1\frac{x^2 + 6}{x - 1} in the form q(x)+r(x)x−1q(x) + \frac{r(x)}{x - 1} with deg⁡(r)<1\deg(r) < 1?

  1. x+1+5x−1x + 1 + \frac{5}{x - 1}
  2. x+1+7xx−1x + 1 + \frac{7x}{x - 1}
  3. x+1+7x−1x + 1 + \frac{7}{x - 1} (correct answer)
  4. x−1+7x−1x - 1 + \frac{7}{x - 1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how 175=3+25\frac{17}{5} = 3 + \frac{2}{5} in arithmetic. Rewriting reveals structure: the expression x2+6x−1\frac{x^2 + 6}{x - 1} looks like one complicated fraction, but rewritten as x+1+7x−1x + 1 + \frac{7}{x - 1}, we see it's a linear function plus a small fraction term. For large ∣x∣|x|, the fraction part →0\to 0, so the expression behaves like x+1x + 1. This separation into polynomial + fractional part is very useful for understanding the expression! Dividing (x2+0x+6)(x^2 + 0x + 6) by (x−1)(x - 1): (1) x2/x=xx^2 / x = x. (2) Multiply: x(x−1)=x2−xx(x - 1) = x^2 - x. (3) Subtract: gets x+6x + 6. (4) x/x=1x / x = 1. (5) Multiply: 1(x−1)=x−11(x - 1) = x - 1. (6) Subtract: (x+6)−(x−1)=7(x + 6) - (x - 1) = 7. Result: x+1+7x−1x + 1 + \frac{7}{x - 1}. Choice A correctly shows x+1+7x−1x + 1 + \frac{7}{x - 1} where the quotient is linear and remainder has degree 0, less than 1, giving the proper rewritten form. Choice B has remainder 5, but multiplying back (x+1)(x−1)+5=x2−1+5=x2+4≠x2+6(x + 1)(x - 1) + 5 = x^2 - 1 + 5 = x^2 + 4 \neq x^2 + 6—the quotient and remainder must satisfy quotient ×\times divisor + remainder = original numerator, so this verification catches errors! Polynomial long division steps: (1) Divide leading terms, (2) Multiply entire divisor, (3) Subtract, (4) Repeat until remainder degree < divisor degree. It's exactly like numerical long division!

Question 8

Use polynomial division to rewrite 2x3+3x2−5x+1x−2\frac{2x^3+3x^2-5x+1}{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+7x+9+19x−22x^2+7x+9+\frac{19}{x-2} (correct answer)
  2. 2x2+7x+9+17x−22x^2+7x+9+\frac{17}{x-2}
  3. 2x2+7x−9+19x−22x^2+7x-9+\frac{19}{x-2}
  4. 2x2+3x+1+0x−22x^2+3x+1+\frac{0}{x-2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. Dividing (2x³ + 3x² - 5x + 1) by (x - 2) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x - 2) = 2x³ - 4x². (3) Subtract from dividend: (2x³ + 3x² - 5x + 1) - (2x³ - 4x²) = 7x² - 5x + 1. (4) Divide again: 7x²/x = 7x (second term of quotient). (5) Multiply: 7x(x - 2) = 7x² - 14x. (6) Subtract: (7x² - 5x + 1) - (7x² - 14x) = 9x + 1. (7) Divide once more: 9x/x = 9 (third term of quotient). (8) Multiply: 9(x - 2) = 9x - 18. (9) Subtract: (9x + 1) - (9x - 18) = 19. Result: quotient is 2x² + 7x + 9, remainder is 19, giving 2x² + 7x + 9 + 19/(x - 2). Choice A correctly shows 2x² + 7x + 9 + 19/(x - 2) where the quotient is a degree 2 polynomial and remainder 19 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong remainder: the final subtraction gives (9x + 1) - (9x - 18) = 19, not 17. This is an arithmetic error in the last step of the division process. Polynomial division has multiple steps where arithmetic errors can creep in—subtraction of polynomials is especially tricky. Double-check each subtraction step carefully! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). Let's check: (2x² + 7x + 9)(x - 2) + 19 = 2x³ - 4x² + 7x² - 14x + 9x - 18 + 19 = 2x³ + 3x² - 5x + 1 ✓. If you don't get the original back, there's an error somewhere in your division. This check works every time and is much faster than redoing the whole division!

Question 9

Rewrite x3+4x2+x−6x+2\frac{x^3+4x^2+x-6}{x+2} in the form q(x)+r(x)x+2q(x)+\frac{r(x)}{x+2} using polynomial division, with deg⁡(r)<deg⁡(x+2)\deg(r)<\deg(x+2).

  1. x2+2x−3+0x+2x^2+2x-3+\frac{0}{x+2} (correct answer)
  2. x2+2x+3+0x+2x^2+2x+3+\frac{0}{x+2}
  3. x2+2x−3+6x+2x^2+2x-3+\frac{6}{x+2}
  4. x2+2x+−3x+2x^2+2x+\frac{-3}{x+2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Before dividing, check if numerator factors: attempting to factor x³ + 4x² + x - 6. Let's try grouping or synthetic division with x = -2: (-2)³ + 4(-2)² + (-2) - 6 = -8 + 16 - 2 - 6 = 0. So (x + 2) is a factor! Using synthetic division or factoring, we find x³ + 4x² + x - 6 = (x + 2)(x² + 2x - 3), so the expression simplifies to (x² + 2x - 3) with no remainder! Choice A correctly shows x² + 2x - 3 + 0/(x + 2), which we can write simply as x² + 2x - 3 since the remainder is 0. Choice C shows a remainder of 6, but that can't be right because we've shown the numerator is exactly divisible by (x + 2)—when a polynomial divides evenly, the remainder must be 0. Verification is your friend: after dividing, multiply your quotient x² + 2x - 3 by the divisor (x + 2) and add your remainder 0. You get (x² + 2x - 3)(x + 2) + 0 = x³ + 2x² + 2x² + 4x - 3x - 6 = x³ + 4x² + x - 6, which matches the original numerator perfectly! Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? If yes, simplify first—it might eliminate the division entirely! (2) Is the numerator degree less than denominator degree already? Then q(x) = 0 and r(x) = a(x). (3) If neither, proceed with division. In this case, checking for factors saved us from doing long division!

Question 10

Rewrite x2+6x−1\frac{x^2+6}{x-1} in the form q(x)+r(x)x−1q(x)+\frac{r(x)}{x-1} using polynomial division, where deg⁡(r)<deg⁡(x−1)\deg(r)<\deg(x-1).

  1. x+1+7x−1x+1+\frac{7}{x-1} (correct answer)
  2. x−1+7x−1x-1+\frac{7}{x-1}
  3. x+1+5x−1x+1+\frac{5}{x-1}
  4. x+1x−1+7x+\frac{1}{x-1}+7
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. For simple cases, inspection can work: if you need (x² + 6)/(x - 1) and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try q(x) = x + a for some a, multiply (x + a)(x - 1), match coefficients with x² + 6, and solve for a and the remainder. For (x² + 6)/(x - 1), we suspect quotient is linear: q(x) = x + a for some a. Multiplying: (x + a)(x - 1) = x² - x + ax - a = x² + (a - 1)x - a. Matching with numerator x² + 0x + 6: coefficient of x gives a - 1 = 0, so a = 1. Constant term gives -a = -1, but we need +6, so remainder is 6 - (-1) = 7. Result: (x + 1) + 7/(x - 1). Inspection works when you can guess the quotient form! Choice A correctly shows x + 1 + 7/(x - 1) where the quotient is linear and remainder 7 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C shows remainder 5 instead of 7, which would come from an arithmetic error: if the quotient is x + 1, then (x + 1)(x - 1) = x² - 1, and the remainder would be (x² + 6) - (x² - 1) = 7, not 5. Always double-check the arithmetic when using inspection! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. For this problem: x²/x = x, multiply x(x-1) = x² - x, subtract from x² + 6 to get x + 6, then x/x = 1, multiply 1(x-1) = x - 1, subtract from x + 6 to get 7. Answer: x + 1 + 7/(x-1).

Question 11

Use polynomial division to express x3−2x2+4x−8x−2\frac{x^3-2x^2+4x-8}{x-2} as q(x)+rx−2q(x)+\frac{r}{x-2}, where deg⁡(r)<1\deg(r)<1.

  1. x2+4+0x−2x^2+4+\frac{0}{x-2} (correct answer)
  2. x2−2x+4+−8x−2x^2-2x+4+\frac{-8}{x-2}
  3. x2+4+−8x−2x^2+4+\frac{-8}{x-2}
  4. x2+4+8x−2x^2+4+\frac{8}{x-2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5=3+2/517/5 = 3 + 2/5 in arithmetic. Before dividing, check if numerator factors: attempting to factor x3−2x2+4x−8x^3 - 2x^2 + 4x - 8. We can factor by grouping: x2(x−2)+4(x−2)=(x2+4)(x−2)x^2(x - 2) + 4(x - 2) = (x^2 + 4)(x - 2). Since the denominator is (x−2)(x - 2), we can cancel: x3−2x2+4x−8x−2=(x2+4)(x−2)x−2=x2+4\frac{x^3 - 2x^2 + 4x - 8}{x - 2} = \frac{(x^2 + 4)(x - 2)}{x - 2} = x^2 + 4. The expression simplifies to x2+4x^2 + 4 with no remainder! Choice A correctly shows x2+4+0x−2x^2 + 4 + \frac{0}{x-2}, which we can write simply as x2+4x^2 + 4 since the remainder is 0. Choice B would give a different quotient x2−2x+4x^2 - 2x + 4, but that can't be right because multiplying back: (x2−2x+4)(x−2)+(−8)=x3−2x2−2x2+4x+4x−8−8=x3−4x2+8x−16≠x3−2x2+4x−8(x^2 - 2x + 4)(x - 2) + (-8) = x^3 - 2x^2 - 2x^2 + 4x + 4x - 8 - 8 = x^3 - 4x^2 + 8x - 16 \neq x^3 - 2x^2 + 4x - 8. Always check if factoring first makes the division unnecessary! When the numerator factors with the denominator as one of its factors, the division becomes a simple cancellation, leaving you with just the other factor as your quotient and remainder 0.

Question 12

Use polynomial division to rewrite the rational expression x2+5x+7x+2\frac{x^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. x+3+1x+2x+3+\frac{1}{x+2} (correct answer)
  2. x+2+3x+2x+2+\frac{3}{x+2}
  3. x+3+7x+2x+3+\frac{7}{x+2}
  4. x+4+−1x+2x+4+\frac{-1}{x+2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. Dividing (x² + 5x + 7) by (x + 2) using long division: (1) Divide leading terms: x²/x = x (first term of quotient). (2) Multiply: x(x + 2) = x² + 2x. (3) Subtract from dividend: (x² + 5x + 7) - (x² + 2x) = 3x + 7. (4) Divide again: 3x/x = 3 (second term of quotient). (5) Multiply: 3(x + 2) = 3x + 6. (6) Subtract: (3x + 7) - (3x + 6) = 1. Result: quotient is x + 3, remainder is 1, giving (x + 3) + 1/(x + 2). Choice A correctly shows (x + 3) + 1/(x + 2) where the quotient is linear and remainder has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong quotient coefficient: when we divide 3x by x in step 4, we get 3, not 2, so the quotient should be x + 3, not x + 2. Verification is your friend: after dividing, multiply your quotient (x + 3) by the divisor (x + 2) and add your remainder 1. You get (x + 3)(x + 2) + 1 = x² + 2x + 3x + 6 + 1 = x² + 5x + 7, which matches the original numerator perfectly!

Question 13

Use polynomial division to rewrite x4+2x3−3x2+4x−5x2+1\frac{x^4+2x^3-3x^2+4x-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+2x−4+2x2−1x2+1x^2+2x-4+\frac{2x^2-1}{x^2+1}
  2. x2+2x−4+2x−1x2+1x^2+2x-4+\frac{2x-1}{x^2+1} (correct answer)
  3. x2+2x−3+2x−1x2+1x^2+2x-3+\frac{2x-1}{x^2+1}
  4. x2+2x−4+2x+1x2+1x^2+2x-4+\frac{2x+1}{x^2+1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. For x4+2x3−3x2+4x−5x2+1\frac{x^4 + 2x^3 - 3x^2 + 4x - 5}{x^2 + 1}, dividing: (1) x4/x2=x2x^4 / x^2 = x^2. (2) Multiply: x2(x2+1)=x4+x2x^2(x^2 + 1) = x^4 + x^2. (3) Subtract: gets 2x3−4x2+4x−52x^3 - 4x^2 + 4x - 5. (4) 2x3/x2=2x2x^3 / x^2 = 2x. (5) Multiply: 2x(x2+1)=2x3+2x2x(x^2 + 1) = 2x^3 + 2x. (6) Subtract: gets −4x2+2x−5-4x^2 + 2x - 5. (7) −4x2/x2=−4-4x^2 / x^2 = -4. (8) Multiply: −4(x2+1)=−4x2−4-4(x^2 + 1) = -4x^2 - 4. (9) Subtract: gets 2x−12x - 1. Result: x2+2x−4+2x−1x2+1x^2 + 2x - 4 + \frac{2x - 1}{x^2 + 1}. Choice A correctly shows x2+2x−4+2x−1x2+1x^2 + 2x - 4 + \frac{2x - 1}{x^2 + 1} where the quotient is quadratic and remainder has degree 1, less than divisor degree 2, giving the proper rewritten form. Choice D's remainder has degree 2 which is NOT less than the divisor's degree 2—if the remainder has degree ≥ divisor degree, we can divide further, so keep dividing until the remainder's degree is strictly less! The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here, remainder 2x−12x - 1 has degree 1 < 2, so we're good. For Algebra 1, stick to divisors of degree 1 or simple degree 2 like x2+1x^2 + 1, and know that technology can handle messier cases.

Question 14

Rewrite 2x3−x2+4x−3x+2\frac{2x^3-x^2+4x-3}{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. 2x2−5x+14+−31x+22x^2-5x+14+\frac{-31}{x+2} (correct answer)
  2. 2x2−5x+14+−15x+22x^2-5x+14+\frac{-15}{x+2}
  3. 2x2+x+2+1x+22x^2+x+2+\frac{1}{x+2}
  4. 2x2−5x−14+−31x+22x^2-5x-14+\frac{-31}{x+2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When you can't divide anymore (when what's left has smaller degree than the divisor), that leftover is your remainder, and what you've built up is your quotient. Dividing (2x³ - x² + 4x - 3) by (x + 2) using long division: (1) Divide leading terms: 2x³/x = 2x² (first term of quotient). (2) Multiply: 2x²(x + 2) = 2x³ + 4x². (3) Subtract from dividend: (2x³ - x² + 4x - 3) - (2x³ + 4x²) = -5x² + 4x - 3. (4) Divide again: -5x²/x = -5x (second term of quotient). (5) Multiply: -5x(x + 2) = -5x² - 10x. (6) Subtract: (-5x² + 4x - 3) - (-5x² - 10x) = 14x - 3. (7) Divide once more: 14x/x = 14 (third term of quotient). (8) Multiply: 14(x + 2) = 14x + 28. (9) Subtract: (14x - 3) - (14x + 28) = -31. Result: quotient is 2x² - 5x + 14, remainder is -31, giving 2x² - 5x + 14 + (-31)/(x + 2). Choice A correctly shows 2x² - 5x + 14 + (-31)/(x + 2) where the quotient is a degree 2 polynomial and remainder -31 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the wrong remainder: the final subtraction gives (14x - 3) - (14x + 28) = -31, not -15. When subtracting (14x + 28) from (14x - 3), we get 14x - 3 - 14x - 28 = -31. Polynomial division has multiple steps where arithmetic errors can creep in—subtraction of polynomials is especially tricky. Double-check each subtraction step carefully! Verification is your friend: after dividing, multiply your quotient q(x) by the divisor b(x) and add your remainder r(x). You should get back the original numerator a(x). Let's check: (2x² - 5x + 14)(x + 2) + (-31) = 2x³ + 4x² - 5x² - 10x + 14x + 28 - 31 = 2x³ - x² + 4x - 3 ✓. If you don't get the original back, there's an error somewhere in your division. This check works every time and is much faster than redoing the whole division!

Question 15

Express x4−1x2−1\frac{x^4-1}{x^2-1} in the form q(x)+r(x)x2−1q(x)+\frac{r(x)}{x^2-1} using polynomial division, where deg⁡(r)<2\deg(r)<2.

  1. x2+1+0x2−1x^2+1+\frac{0}{x^2-1} (correct answer)
  2. x2−1+0x2−1x^2-1+\frac{0}{x^2-1}
  3. x2+1x2−1x^2+\frac{1}{x^2-1}
  4. x2+1+1x2−1x^2+1+\frac{1}{x^2-1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Before dividing, check if numerator factors: x⁴ - 1 = (x²)² - 1² = (x² - 1)(x² + 1) using difference of squares. Since the denominator is (x² - 1), we can cancel: (x⁴ - 1)/(x² - 1) = (x² - 1)(x² + 1)/(x² - 1) = x² + 1. The expression simplifies to x² + 1 with no remainder! Choice A correctly shows x² + 1 + 0/(x² - 1), which we can write simply as x² + 1 since adding 0 doesn't change the value. Choice B would give x² - 1, but that can't be right because multiplying back doesn't give the original: (x² - 1)(x² - 1) = x⁴ - 2x² + 1 ≠ x⁴ - 1. Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? If yes, simplify first—it might eliminate the division entirely! This problem demonstrates that factoring first can turn a potentially complex division into a simple cancellation.

Question 16

Rewrite x2+5x+7x+2\frac{x^2+5x+7}{x+2} in the form q(x)+rx+2q(x)+\frac{r}{x+2} using polynomial division (quotient + remainder/divisor), with remainder degree less than the divisor degree.

  1. x+3+1x+2x+3+\frac{1}{x+2} (correct answer)
  2. x+3+7x+2x+3+\frac{7}{x+2}
  3. x+1+1x+2x+1+\frac{1}{x+2}
  4. x+2+3x+2x+2+\frac{3}{x+2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. For simple cases, inspection can work: if you need (x² + 5x + 7)/(x + 2) and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try q(x) = x + a for some a, multiply (x + a)(x + 2), match coefficients with x² + 5x + 7, and solve for a and the remainder. For (x² + 5x + 7)/(x + 2), we suspect quotient is linear: q(x) = x + a for some a. Multiplying: (x + a)(x + 2) = x² + (2 + a)x + 2a. Matching with numerator x² + 5x + 7: coefficient of x gives 2 + a = 5, so a = 3. Constant term gives 2a = 6, but we have 7, so remainder is 7 - 6 = 1. Result: (x + 3) + 1/(x + 2). Choice B correctly shows x + 3 + 1/(x + 2) where the quotient is linear and remainder 1 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong remainder: 7 instead of 1, which might come from forgetting to subtract 2a = 6 from the constant term 7—the remainder is what's left after accounting for the quotient times divisor! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. Practice with (x² + 5x + 7)/(x + 2): divide x² by x → x, multiply x(x+2) = x² + 2x, subtract to get 3x + 7, divide 3x by x → 3, multiply 3(x+2) = 3x + 6, subtract to get remainder 1. Answer: x + 3 + 1/(x+2).

Question 17

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

  1. x2−2+0x2−1x^2-2+\frac{0}{x^2-1} (correct answer)
  2. x2−2+2xx2−1x^2-2+\frac{2x}{x^2-1}
  3. x2−1+−1x2−1x^2-1+\frac{-1}{x^2-1}
  4. x2−2+1x2−1x^2-2+\frac{1}{x^2-1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Polynomial long division works exactly like the long division you learned in elementary school, just with polynomials instead of numbers: divide the leading terms, multiply back, subtract, bring down the next term, repeat. When dividing by a degree-2 polynomial like x² - 1, we're looking for how many times x² - 1 goes into our numerator. Dividing (x⁴ - 3x² + 2) by (x² - 1) using long division: (1) Divide leading terms: x⁴/x² = x² (first term of quotient). (2) Multiply: x²(x² - 1) = x⁴ - x². (3) Subtract from dividend: (x⁴ - 3x² + 2) - (x⁴ - x²) = -2x² + 2. (4) Divide again: -2x²/x² = -2 (second term of quotient). (5) Multiply: -2(x² - 1) = -2x² + 2. (6) Subtract: (-2x² + 2) - (-2x² + 2) = 0. Result: quotient is x² - 2, remainder is 0, giving (x² - 2) + 0/(x² - 1). Choice A correctly shows x² - 2 + 0/(x² - 1) where the quotient is degree 2 and remainder 0 has degree less than the divisor's degree 2, giving the proper rewritten form. Choice B shows a remainder of 2x, but that has degree 1, and we need the remainder to have degree less than 2 (the degree of x² - 1). If the remainder has degree ≥ divisor degree, we can divide further! Keep dividing until the remainder's degree is strictly less than the divisor's. The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': in our case, the remainder 0 has degree -∞ (or we say it has no degree), which is certainly less than the divisor's degree 2, so we're done. This zero remainder also tells us that x⁴ - 3x² + 2 is exactly divisible by x² - 1, which we could verify by factoring: x⁴ - 3x² + 2 = (x² - 1)(x² - 2).

Question 18

For large ∣x∣|x|, the rational expression x2+6x+1x+3\frac{x^2+6x+1}{x+3} can be written as q(x)+r(x)x+3q(x)+\frac{r(x)}{x+3}, and the quotient q(x)q(x) dominates. Which rewrite is correct (with deg⁡(r)<deg⁡(x+3)\deg(r)<\deg(x+3))?

  1. x+3+−8x+3x+3+\frac{-8}{x+3} (correct answer)
  2. x+3+8x+3x+3+\frac{8}{x+3}
  3. x+6+−17x+3x+6+\frac{-17}{x+3}
  4. x+3+−8xx+3x+3+\frac{-8x}{x+3}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. Rewriting reveals structure: the expression (x² + 6x + 1)/(x + 3) looks like one complicated fraction, but rewritten as q(x) + r(x)/(x + 3), we see it's a polynomial q(x) plus a small fraction term. For large |x|, the fraction part → 0, so the expression behaves like q(x). This separation into polynomial + fractional part is very useful for understanding the expression! Dividing (x² + 6x + 1) by (x + 3) using long division: (1) Divide leading terms: x²/x = x (first term of quotient). (2) Multiply: x(x + 3) = x² + 3x. (3) Subtract from dividend: (x² + 6x + 1) - (x² + 3x) = 3x + 1. (4) Divide again: 3x/x = 3 (second term of quotient). (5) Multiply: 3(x + 3) = 3x + 9. (6) Subtract: (3x + 1) - (3x + 9) = 1 - 9 = -8. Result: quotient is x + 3, remainder is -8, giving (x + 3) + (-8)/(x + 3). Choice A correctly shows (x + 3) + (-8)/(x + 3) where the quotient is linear and remainder has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice D incorrectly shows remainder as -8x (degree 1) instead of -8 (degree 0), violating the requirement that remainder degree < divisor degree. If the remainder had degree 1, we could divide further! The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': the remainder -8 has degree 0, which is less than the divisor's degree 1, so we're done. For large |x|, the term -8/(x+3) → 0, so the expression behaves like the linear function x + 3, which is why understanding this rewrite helps analyze the function's behavior.

Question 19

Use polynomial division to express 3x2−2x+4x−1\frac{3x^2-2x+4}{x-1} as q(x)+r(x)x−1q(x)+\frac{r(x)}{x-1}, where deg⁡(r)<deg⁡(x−1)\deg(r)<\deg(x-1).

  1. 3x+1+5x−13x+1+\frac{5}{x-1} (correct answer)
  2. 3x−1+5x−13x-1+\frac{5}{x-1}
  3. 3x+1+3x−13x+1+\frac{3}{x-1}
  4. 3x+x+5x−13x+\frac{x+5}{x-1}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression (fraction of polynomials) in the form quotient + remainder/divisor, just like how 17/5 = 3 + 2/5 in arithmetic. The division algorithm for polynomials says that any rational expression a(x)/b(x) can be rewritten as q(x) + r(x)/b(x), where q(x) is the quotient (the polynomial part), and r(x) is the remainder (what's left over in the numerator). The crucial requirement: the degree of the remainder r(x) must be less than the degree of the divisor b(x)—just like in numerical division where the remainder must be less than the divisor! Dividing (3x² - 2x + 4) by (x - 1) using long division: (1) Divide leading terms: 3x²/x = 3x (first term of quotient). (2) Multiply: 3x(x - 1) = 3x² - 3x. (3) Subtract from dividend: (3x² - 2x + 4) - (3x² - 3x) = x + 4. (4) Divide again: x/x = 1 (second term of quotient). (5) Multiply: 1(x - 1) = x - 1. (6) Subtract: (x + 4) - (x - 1) = 5. Result: quotient is 3x + 1, remainder is 5, giving (3x + 1) + 5/(x - 1). Choice A correctly shows 3x + 1 + 5/(x - 1) where the quotient is linear and remainder 5 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice B has the quotient wrong: it shows 3x - 1 instead of 3x + 1. When we get to the second division step with x + 4 remaining and divide by x to get 1, that +1 is added to our quotient, not subtracted. The signs in polynomial division can be tricky—always double-check! Polynomial long division steps: (1) Divide leading terms to get first term of quotient, (2) Multiply entire divisor by that term, (3) Subtract from dividend, (4) Repeat with what remains until remainder degree < divisor degree. It's exactly like numerical long division! Practice with (x² + 5x + 7)/(x + 2): divide x² by x → x, multiply x(x+2) = x² + 2x, subtract to get 3x + 7, divide 3x by x → 3, multiply 3(x+2) = 3x + 6, subtract to get remainder 1. Answer: x + 3 + 1/(x+2).

Question 20

Use polynomial division to express 2x3+x2−8x+7x+2\frac{2x^3+x^2-8x+7}{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. 2x2−3x−2−11x+22x^2-3x-2-\frac{11}{x+2}
  2. 2x2−3x+2+11x+22x^2-3x+2+\frac{11}{x+2}
  3. 2x2−3x−2+11x+22x^2-3x-2+\frac{11}{x+2} (correct answer)
  4. 2x2−3x−2+11xx+22x^2-3x-2+\frac{11x}{x+2}
Explanation: This question tests your understanding of polynomial division—rewriting a rational expression like (2x3+x2−8x+7)/(x+2)(2x^3 + x^2 - 8x + 7)/(x + 2) in the form quotient + remainder/divisor, just like dividing numbers where 17/5=3+2/517/5 = 3 + 2/5. The division algorithm for polynomials says that any rational expression a(x)/b(x)a(x)/b(x) can be rewritten as q(x)+r(x)/b(x)q(x) + r(x)/b(x), where q(x)q(x) is the quotient, and r(x)r(x) is the remainder with degree less than the degree of b(x)b(x)—here, since b(x)=x+2b(x) = x + 2 is degree 1, r(x)r(x) must be a constant. Dividing (2x3+x2−8x+7)(2x^3 + x^2 - 8x + 7) by (x+2)(x + 2): (1) 2x3/x=2x22x^3/x = 2x^2. (2) Multiply: 2x2(x+2)=2x3+4x22x^2(x + 2) = 2x^3 + 4x^2. (3) Subtract: gets −3x2−8x+7-3x^2 - 8x + 7 (after bringing down). (4) −3x2/x=−3x-3x^2/x = -3x. (5) Multiply: −3x(x+2)=−3x2−6x-3x(x + 2) = -3x^2 - 6x. (6) Subtract: gets −2x+7-2x + 7. (7) −2x/x=−2-2x/x = -2. (8) Multiply: −2(x+2)=−2x−4-2(x + 2) = -2x - 4. (9) Subtract: gets 11. Result: 2x2−3x−2+11/(x+2)2x^2 - 3x - 2 + 11/(x + 2). Choice A correctly shows 2x2−3x−2+11/(x+2)2x^2 - 3x - 2 + 11/(x + 2) where the quotient is quadratic and remainder has degree 0, less than 1, giving the proper rewritten form. Choice C has quotient 2x2−3x+22x^2 - 3x + 2, but when dividing −2x-2x by xx, we get −2-2, not +2+2— the leading term of the quotient comes from dividing the highest-degree terms, and signs matter in each step! Before starting polynomial division, always check: (1) Can I factor the numerator and cancel with the denominator? (2) If numerator degree < denominator, it's already in form with q(x)=0q(x)=0. This pre-check saves work!