Algebra Quiz: Rewriting Rational Expressions

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.

Question 1

1. $x^2+x-1+\frac{0}{x+1}$ (correct answer)
2. $x^2+x-1+\frac{1}{x+1}$
3. $x^2+x+1+\frac{-2}{x+1}$
4. $x^2+2x-1+\frac{0}{x+1}$

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. Dividing (x³ + 2x² - 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: (x³ + 2x² - 1) - (x³ + x²) = x² - 1. (4) Divide: x²/x = x. (5) Multiply: x(x + 1) = x² + x. (6) Subtract: (x² - 1) - (x² + x) = -x - 1. (7) Divide: -x/x = -1. (8) Multiply: -1(x + 1) = -x - 1. (9) Subtract: (-x - 1) - (-x - 1) = 0. Result: quotient is x² + x - 1, remainder is 0. Choice A correctly shows x² + x - 1 + 0/(x + 1) where the quotient is quadratic and remainder is 0 (degree -∞, less than divisor degree 1), giving the proper rewritten form. Choice B incorrectly adds a remainder of 1 when our division came out evenly with remainder 0—when the division is exact, there's no fractional part! Verification is your friend: after dividing, multiply your quotient (x² + x - 1) by the divisor (x + 1) and add your remainder 0. You get (x² + x - 1)(x + 1) + 0 = x³ + x² + x² + x - x - 1 = x³ + 2x² - 1, which matches the original numerator perfectly! When remainder is 0, the rational expression simplifies to just the quotient polynomial.

Question 2

1. $2x+1+\frac{-2}{x+3}$ (correct answer)
2. $2x+7+\frac{-20}{x+3}$
3. $2x-1+\frac{4}{x+3}$
4. $2x+1+\frac{2}{x+3}$

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 3

1. $x^2+2x-3+\frac{0}{x+2}$ (correct answer)
2. $x^2+2x+3+\frac{0}{x+2}$
3. $x^2+2x-3+\frac{6}{x+2}$
4. $x^2+2x+\frac{-3}{x+2}$

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 4

1. $x^2+x-2+\frac{0}{x-2}$ (correct answer)
2. $x^2-x-2+\frac{0}{x-2}$
3. $x^2+x-2+\frac{4}{x-2}$
4. $x^2+x+2+\frac{0}{x-2}$

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. Before dividing, check if numerator factors: let's test if x = 2 is a root of x³ - x² - 4x + 4. Substituting: 2³ - 2² - 4(2) + 4 = 8 - 4 - 8 + 4 = 0. Yes! So (x - 2) is a factor. We can use synthetic division or factor by grouping to find the other factor. Dividing (x³ - x² - 4x + 4) by (x - 2) using synthetic division with 2: Bring down 1, multiply by 2 to get 2, add to -1 to get 1. Multiply 1 by 2 to get 2, add to -4 to get -2. Multiply -2 by 2 to get -4, add to 4 to get 0. The quotient is x² + x - 2 with remainder 0. We can verify by factoring: x³ - x² - 4x + 4 = x²(x - 1) - 4(x - 1) = (x² - 4)(x - 1) = (x - 2)(x + 2)(x - 1). Wait, that gives three linear factors, not matching our division. Let me redo the division carefully: (1) x³/x = x², multiply x²(x - 2) = x³ - 2x², subtract to get x² - 4x + 4. (2) x²/x = x, multiply x(x - 2) = x² - 2x, subtract to get -2x + 4. (3) -2x/x = -2, multiply -2(x - 2) = -2x + 4, subtract to get 0. Quotient is x² + x - 2, remainder 0. Choice A correctly shows x² + x - 2 + 0/(x - 2) where the quotient is quadratic and remainder 0 indicates exact division. Choice B has a sign error in the quotient, showing x² - x - 2 instead of x² + x - 2. When we divided x² by x in step 2, we got +x, not -x, because we were dividing the positive x² term that remained after the first subtraction. Verification is your friend: (x² + x - 2)(x - 2) = x³ - 2x² + x² - 2x - 2x + 4 = x³ - x² - 4x + 4 ✓. The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here our remainder is 0, which has no degree (or degree -∞), certainly less than the divisor's degree 1. A zero remainder means the division is exact—the divisor is a factor of the dividend!

Question 5

1. $2x^2+7x+9+\frac{19}{x-2}$ (correct answer)
2. $2x^2+7x+9+\frac{17}{x-2}$
3. $2x^2+7x-9+\frac{19}{x-2}$
4. $2x^2+3x+1+\frac{0}{x-2}$

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 6

1. $x+2+\frac{x}{x-2}$
2. $x+2+\frac{4}{x-2}$
3. $x-2+\frac{0}{x-2}$
4. $x+2+\frac{0}{x-2}$ (correct answer)

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how $17/5 = 3 + 2/5$ 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: $x^2 - 4 = (x - 2)(x + 2)$, so $(x - 2)(x + 2)/(x - 2)$ simplifies to $x + 2$ with remainder 0 (for $x ≠ 2$). Choice A correctly shows $x + 2 + 0/(x - 2)$ 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 $q(x)$ 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 7

1. $x^2+4+\frac{0}{x-2}$ (correct answer)
2. $x^2-2x+4+\frac{-8}{x-2}$
3. $x^2+4+\frac{-8}{x-2}$
4. $x^2+4+\frac{8}{x-2}$

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 $17/5 = 3 + 2/5$ in arithmetic. Before dividing, check if numerator factors: attempting to factor $x^3 - 2x^2 + 4x - 8$. We can factor by grouping: $x^2(x - 2) + 4(x - 2) = (x^2 + 4)(x - 2)$. Since the denominator is $(x - 2)$, we can cancel: $\frac{x^3 - 2x^2 + 4x - 8}{x - 2} = \frac{(x^2 + 4)(x - 2)}{x - 2} = x^2 + 4$. The expression simplifies to $x^2 + 4$ with no remainder! Choice A correctly shows $x^2 + 4 + \frac{0}{x-2}$, which we can write simply as $x^2 + 4$ since the remainder is 0. Choice B would give a different quotient $x^2 - 2x + 4$, but that can't be right because multiplying back: $(x^2 - 2x + 4)(x - 2) + (-8) = x^3 - 2x^2 - 2x^2 + 4x + 4x - 8 - 8 = x^3 - 4x^2 + 8x - 16 \neq x^3 - 2x^2 + 4x - 8$. 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 8

1. $x+3+\frac{1}{x+2}$ (correct answer)
2. $x+2+\frac{3}{x+2}$
3. $x+3+\frac{7}{x+2}$
4. $x+4+\frac{-1}{x+2}$

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 9

1. $q(x)=x^2-x-1, r(x)=-8$ (correct answer)
2. $q(x)=x^2-7x+2, r(x)=-11$
3. $q(x)=x^2-x-1, r(x)=8$
4. $q(x)=x^2-x, r(x)=-8$

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 $(x^3 - 4x^2 + 2x - 5)$ by $(x - 3)$ using long division: (1) Divide leading terms: $x^3/x = x^2$ (first term of quotient). (2) Multiply: $x^2(x - 3) = x^3 - 3x^2$. (3) Subtract: $(x^3 - 4x^2 + 2x - 5) - (x^3 - 3x^2) = -x^2 + 2x - 5$. (4) Divide again: $-x^2/x = -x$. (5) Multiply: $-x(x - 3) = -x^2 + 3x$. (6) Subtract: $(-x^2 + 2x - 5) - (-x^2 + 3x) = -x - 5$. (7) Divide once more: $-x/x = -1$. (8) Multiply: $-1(x - 3) = -x + 3$. (9) Subtract: $(-x - 5) - (-x + 3) = -8$. Result: quotient is $x^2 - x - 1$, remainder is $-8$. Choice A correctly shows $q(x) = x^2 - x - 1$ and $r(x) = -8$, where the quotient is degree 2 and remainder $-8$ has degree 0 (less than divisor degree 1), giving the proper division result. Choice B has the wrong sign on the remainder: $+8$ instead of $-8$, which is a common error in the final subtraction step—remember that $(-x - 5) - (-x + 3) = -8$, not $+8$! Verification is your friend: after dividing, multiply your quotient $x^2 - x - 1$ by the divisor $(x - 3)$ and add your remainder $-8$. You get $(x^2 - x - 1)(x - 3) + (-8) = x^3 - 3x^2 - x^2 + 3x - x + 3 - 8 = x^3 - 4x^2 + 2x - 5$, which matches the original numerator perfectly!

Question 10

1. $x^2-2+\frac{0}{x^2-1}$ (correct answer)
2. $x^2-2+\frac{2x}{x^2-1}$
3. $x^2-1+\frac{-1}{x^2-1}$
4. $x^2-2+\frac{1}{x^2-1}$

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 11

1. $3x-11+\frac{37}{x+3}$ (correct answer)
2. $3x+11+\frac{37}{x+3}$
3. $3x-11+\frac{-37}{x+3}$
4. $3x-8+\frac{28}{x+3}$

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. 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 (3x² - 2x + 4) by (x + 3) using long division: (1) Divide leading terms: 3x²/x = 3x (first term of quotient). (2) Multiply: 3x(x + 3) = 3x² + 9x. (3) Subtract from dividend: (3x² - 2x + 4) - (3x² + 9x) = -11x + 4. (4) Divide again: -11x/x = -11 (second term of quotient). (5) Multiply: -11(x + 3) = -11x - 33. (6) Subtract: (-11x + 4) - (-11x - 33) = 4 - (-33) = 37. Result: quotient is 3x - 11, remainder is 37, giving (3x - 11) + 37/(x + 3). Choice A correctly shows (3x - 11) + 37/(x + 3) 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 sign in the quotient: when we subtract in step 3, we get -11x, not +11x, so the second term of the quotient is -11, making it 3x - 11, not 3x + 11. Verification is your friend: after dividing, multiply your quotient (3x - 11) by the divisor (x + 3) and add your remainder 37. You get (3x - 11)(x + 3) + 37 = 3x² + 9x - 11x - 33 + 37 = 3x² - 2x + 4, which matches the original numerator perfectly!

Question 12

1. $-8$
2. $0$ (correct answer)
3. $4$
4. $8$

Explanation: We can factor the numerator by grouping: $x^3 - 2x^2 + 4x - 8 = x^2(x - 2) + 4(x - 2) = (x - 2)(x^2 + 4)$. Therefore, $\frac{x^3 - 2x^2 + 4x - 8}{x - 2} = \frac{(x - 2)(x^2 + 4)}{x - 2} = x^2 + 4$ (for $x \neq 2$). This means $p(x) = x^2 + 4$ and $k = 0$. Alternatively, using synthetic division or polynomial long division would yield quotient $x^2 + 4$ and remainder $0$. Students might incorrectly think $k = -8$ (the constant term), $k = 4$ (coefficient of $x$), or $k = 8$ (absolute value of constant term).

Question 13

1. $12$
2. $18$
3. $30$ (correct answer)
4. $36$

Explanation: From the given form, we have $\frac{P(x)}{x^2 + 4} = x^2 - 4 + \frac{3x + 12}{x^2 + 4}$. To find $P(x)$, multiply both sides by $x^2 + 4$: $P(x) = (x^2 - 4)(x^2 + 4) + (3x + 12)$. Using the difference of squares: $(x^2 - 4)(x^2 + 4) = x^4 - 16$. Therefore, $P(x) = x^4 - 16 + 3x + 12 = x^4 + 3x - 4$. Now, $P(2) = 2^4 + 3(2) - 4 = 16 + 6 - 4 = 18$. Wait, let me recalculate: $P(x) = x^4 + 3x - 4$, so $P(2) = 16 + 6 - 4 = 18$. Actually, let me verify by substitution: at $x = 2$, we should have $\frac{P(2)}{8} = 4 - 4 + \frac{6 + 12}{8} = 0 + \frac{18}{8} = \frac{18}{8}$. So $P(2) = 18 \cdot \frac{8}{8} = 18$. But this gives choice B, not C. Let me recalculate $P(x)$ more carefully: $P(x) = (x^2 - 4)(x^2 + 4) + 3x + 12 = x^4 - 16 + 3x + 12 = x^4 + 3x - 4$. Then $P(2) = 16 + 6 - 4 = 18$. Since the correct calculation gives 18 but the marked answer is C (30), there's an error in the original answer key.

Question 14

1. Perform polynomial long division from scratch and compare quotients and remainders
2. Substitute $x = 0$ into both expressions and check if they yield the same result
3. Multiply $(2x + 5)(x^2 + 3) + (-9x - 14)$ and compare to the original numerator (correct answer)
4. Check whether the degree of $-9x - 14$ is less than the degree of $x^2 + 3$

Explanation: To verify the polynomial division, we use the division algorithm: if $\frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)}$, then $a(x) = q(x) \cdot b(x) + r(x)$. So we need $2x^3 + 5x^2 - 3x + 1 = (2x + 5)(x^2 + 3) + (-9x - 14)$. Expanding: $(2x + 5)(x^2 + 3) = 2x^3 + 6x + 5x^2 + 15 = 2x^3 + 5x^2 + 6x + 15$. Adding the remainder: $2x^3 + 5x^2 + 6x + 15 + (-9x - 14) = 2x^3 + 5x^2 - 3x + 1$, which matches the original numerator. Choice A is inefficient since we're verifying, not discovering. Choice B only checks one value, not the entire identity. Choice D only verifies a necessary but not sufficient condition.

Question 15

1. Only Student A is correct because the remainder must have a negative leading coefficient
2. Only Student B is correct because the constant term in the remainder must be positive
3. Both students are incorrect because the quotient should be $2x + 1$, not $2x - 1$
4. Only Student A is correct because polynomial division gives a unique quotient and remainder (correct answer)

Explanation: Polynomial division has a unique result. Let's verify: $(2x - 1)(x^2 + 2) + (-x - 5) = 2x^3 + 4x - x^2 - 2 - x - 5 = 2x^3 - x^2 + 3x - 7$ ✓. For Student B: $(2x - 1)(x^2 + 2) + (x + 5) = 2x^3 + 4x - x^2 - 2 + x + 5 = 2x^3 - x^2 + 5x + 3$ ≠ original numerator. Therefore, only Student A is correct. Choice A is wrong because there's no rule about negative leading coefficients. Choice B is wrong because there's no rule about positive constant terms. Choice C is wrong because the quotient $2x - 1$ is correct, as verified by the division algorithm.

Question 16

1. $3x - 10 + \frac{4}{x + 3}$
2. $3x + 8 + \frac{40}{x + 3}$
3. $3x - 10 + \frac{40x}{x + 3}$
4. $3x - 10 + \frac{40}{x + 3}$ (correct answer)

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression in the form quotient + remainder/divisor, just like how $17/5 = 3 + 2/5$ in arithmetic. For simple cases, inspection can work: if you need $(3x^2 - x + 10)/(x + 3)$ and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Try $q(x) = 3x + a$ for some a, multiply $(3x + a)(x + 3)$, match coefficients with $3x^2 - x + 10$, and solve for a and the remainder. This 'educated guess and check' is faster than formal division when it works! For $(3x^2 - x + 10)/(x + 3)$, we suspect quotient is linear: $q(x) = 3x + a$. Multiplying: $(3x + a)(x + 3) = 3x^2 + (9 + a)x + 3a$. Matching with $3x^2 - x + 10$: coefficient of x gives $9 + a = -1$, so a = -10. Constant term gives $3a = -30$, but we have +10, so remainder is $10 - (-30) = 40$. Result: $3x - 10 + \frac{40}{x + 3}$. Choice A correctly shows $3x - 10 + \frac{40}{x + 3}$ where the quotient is linear and remainder has degree 0, less than 1, giving the proper rewritten form. Choice B has remainder 4 instead of 40, likely from an arithmetic error in the constant term—polynomial division has multiple steps where errors can creep in, so double-check each subtraction! 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 $a(x)$. This check works every time and is much faster than redoing the whole division!

Question 17

1. $x^2-x+2+\frac{0}{x-3}$
2. $x^2-7x+22+\frac{72}{x-3}$
3. $x^2-x-2+\frac{0}{x-3}$ (correct answer)
4. $x^2-x-2+\frac{6}{x-3}$

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: $x^3 - 4x^2 + x + 6$ factors as $(x - 3)(x^2 - x - 2)$ since it divides evenly (remainder 0), so the expression simplifies to $x^2 - x - 2$ with no remainder term! Choice A correctly shows $x^2 - x - 2 + \frac{0}{x-3}$ 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 18

1. $2x+\frac{11}{2}+\frac{9/2}{2x-1}$ (correct answer)
2. $2x+\frac{11}{2}+\frac{9}{2x-1}$
3. $2x+\frac{9}{2}+\frac{11/2}{2x-1}$
4. $2x+\frac{11}{2}-\frac{9/2}{2x-1}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression like $\frac{4x^2 + 9x - 1}{2x - 1}$ in the form quotient + remainder/divisor, just like dividing numbers where $17/5 = 3 + 2/5$. 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 $4x^2 + 9x - 1$ by $2x - 1$: (1) $4x^2 / (2x) = 2x$. (2) Multiply: $2x(2x - 1) = 4x^2 - 2x$. (3) Subtract: gets $11x - 1$. (4) $11x / (2x) = 11/2$. (5) Multiply: $(11/2)(2x - 1) = 11x - 11/2$. (6) Subtract: gets $-1 + 11/2 = 9/2$. Result: $2x + 11/2 + \frac{9/2}{2x - 1}$. Choice A correctly shows $2x + 11/2 + \frac{9/2}{2x - 1}$ where the quotient is linear (with fractions, that's okay) and remainder has degree 0, less than 1, giving the proper rewritten form. Choice D has a negative remainder term, but the subtraction leads to positive $9/2$—remember to handle fractions carefully in subtraction, as arithmetic errors are tricky here! 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). This check works every time and is much faster than redoing the whole division!

Question 19

1. $x+3+\frac{1}{x+2}$ (correct answer)
2. $x+3-\frac{1}{x+2}$
3. $x+2+\frac{3}{x+2}$
4. $x+3+\frac{x+1}{x+2}$

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. 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 $(x^2 + 5x + 7)/(x + 2)$, dividing: (1) $x^2/x = x$. (2) Multiply: $x(x + 2) = x^2 + 2x$. (3) Subtract: $(x^2 + 5x + 7) - (x^2 + 2x) = 3x + 7$. (4) $3x/x = 3$. (5) Multiply: $3(x + 2) = 3x + 6$. (6) Subtract: $(3x + 7) - (3x + 6) = 1$. Result: $x + 3 + 1/(x + 2)$. 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 a negative remainder, but the subtraction gives positive 1—arithmetic errors in subtraction are common, so verify by multiplying back! Practice with $(x^2 + 5x + 7)/(x + 2)$: divide $x^2$ by $x$ → $x$, multiply $x(x+2) = x^2 + 2x$, subtract to get $3x + 7$, divide $3x$ by $x$ → $3$, multiply $3(x+2) = 3x + 6$, subtract to get remainder $1$. The degree requirement (degree of remainder < degree of divisor) is what makes the division 'done': here, remainder $1$ has degree 0 < 1, so we're good—if it had an x-term, you'd divide further!

Question 20

1. $x^2+2x-4+\frac{2x^2-1}{x^2+1}$
2. $x^2+2x-4+\frac{2x-1}{x^2+1}$ (correct answer)
3. $x^2+2x-3+\frac{2x-1}{x^2+1}$
4. $x^2+2x-4+\frac{2x+1}{x^2+1}$

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 $\frac{x^4 + 2x^3 - 3x^2 + 4x - 5}{x^2 + 1}$, dividing: (1) $x^4 / x^2 = x^2$. (2) Multiply: $x^2(x^2 + 1) = x^4 + x^2$. (3) Subtract: gets $2x^3 - 4x^2 + 4x - 5$. (4) $2x^3 / x^2 = 2x$. (5) Multiply: $2x(x^2 + 1) = 2x^3 + 2x$. (6) Subtract: gets $-4x^2 + 2x - 5$. (7) $-4x^2 / x^2 = -4$. (8) Multiply: $-4(x^2 + 1) = -4x^2 - 4$. (9) Subtract: gets $2x - 1$. Result: $x^2 + 2x - 4 + \frac{2x - 1}{x^2 + 1}$. Choice A correctly shows $x^2 + 2x - 4 + \frac{2x - 1}{x^2 + 1}$ 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 $2x - 1$ has degree 1 < 2, so we're good. For Algebra 1, stick to divisors of degree 1 or simple degree 2 like $x^2 + 1$, and know that technology can handle messier cases.