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Algebra Help: Rewriting Rational Expressions

Review real example questions for Rewriting Rational Expressions in Algebra.

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When $\frac{x^3 - 2x^2 + 4x - 8}{x - 2}$ is simplified, the result can be written in the form $p(x) + \frac{k}{x - 2}$ where $p(x)$ is a polynomial and $k$ is a constant. What is the value of $k$ ?

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Question 1

When $\frac{x^3 - 2x^2 + 4x - 8}{x - 2}$ is simplified, the result can be written in the form $p(x) + \frac{k}{x - 2}$ where $p(x)$ is a polynomial and $k$ is a constant. What is the value of $k$ ?

$-8$
$0$ (correct answer)
$4$
$8$

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 2

Two students are rewriting $\frac{2x^3 - x^2 + 3x - 7}{x^2 + 2}$ . Student A gets $2x - 1 + \frac{-x - 5}{x^2 + 2}$ and Student B gets $2x - 1 + \frac{x + 5}{x^2 + 2}$ . Which statement is correct?

Only Student A is correct because the remainder must have a negative leading coefficient
Only Student B is correct because the constant term in the remainder must be positive
Both students are incorrect because the quotient should be $2x + 1$ , not $2x - 1$
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 3

Rewrite $\frac{4x^2+1}{2x-1}$ in the form $q(x)+\frac{r}{2x-1}$ using polynomial division, with $\deg(r)<1$ .

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

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 $(4x^2 + 1)/(2x - 1)$ and the numerator degree is just 1 more than denominator degree, the quotient will be linear. Dividing $(4x^2 + 1)$ by $(2x - 1)$ using long division: (1) Divide leading terms: $4x^2/(2x) = 2x$ (first term of quotient). (2) Multiply: $2x(2x - 1) = 4x^2 - 2x$ . (3) Subtract: $(4x^2 + 0x + 1) - (4x^2 - 2x) = 2x + 1$ . (4) Divide again: $2x/(2x) = 1$ . (5) Multiply: $1(2x - 1) = 2x - 1$ . (6) Subtract: $(2x + 1) - (2x - 1) = 2$ . Result: quotient is $2x + 1$ , remainder is 2, giving $(2x + 1) + 2/(2x - 1)$ . Choice A correctly shows $2x + 1 + 2/(2x - 1)$ where the quotient is linear and remainder 2 has degree 0 (less than divisor degree 1), giving the proper rewritten form. Choice C has the wrong sign on the remainder: -2 instead of 2, which is a common error—remember that $(2x + 1) - (2x - 1) = 2x + 1 - 2x + 1 = 2$ , not -2! Verification is your friend: after dividing, multiply your quotient $2x + 1$ by the divisor $(2x - 1)$ and add your remainder 2. You get $(2x + 1)(2x - 1) + 2 = 4x^2 - 2x + 2x - 1 + 2 = 4x^2 + 1$ , which matches the original numerator perfectly!

Question 4

Rewrite $\frac{x^2+5x+7}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$ (quotient plus remainder over divisor), with $\deg(r)<\deg(x+2)$ .

$x+3+\frac{1}{x+2}$ (correct answer)
$x+2+\frac{3}{x+2}$
$x+3+\frac{2}{x+2}$
$x+1+\frac{5}{x+2}$

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. This 'educated guess and check' is faster than formal division when it works! 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). Inspection works when you can guess the quotient form! Choice A 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 D has an arithmetic error: it shows x + 1 + 5/(x + 2), but if we verify by multiplying back: (x + 1)(x + 2) + 5 = x² + 3x + 2 + 5 = x² + 3x + 7 ≠ x² + 5x + 7. The quotient and remainder must satisfy: quotient × divisor + remainder = original numerator. This verification always catches division errors! 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), so it's just 0 + a(x)/b(x) = a(x)/b(x) (already in required form). (3) If neither, proceed with division. This pre-check saves work!

Question 5

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

$2x^2+7x+9+\frac{19}{x-2}$ (correct answer)
$2x^2+7x-9+\frac{19}{x-2}$
$2x^2+7x+9+\frac{17}{x-2}$
$2x^2+7x+\frac{9x+19}{x-2}$

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 sign in the quotient (shows +7x - 9 instead of +7x + 9), which would come from an arithmetic error in the division process—polynomial division has multiple steps where arithmetic errors can creep in, especially in the subtraction steps. Verification is your friend: after dividing, multiply your quotient 2x² + 7x + 9 by the divisor (x - 2) and add your remainder 19. You get (2x² + 7x + 9)(x - 2) + 19 = 2x³ - 4x² + 7x² - 14x + 9x - 18 + 19 = 2x³ + 3x² - 5x + 1, which matches the original numerator perfectly! This check works every time and is much faster than redoing the whole division.

Question 6

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

$x+2+\frac{x}{x-2}$
$x+2+\frac{4}{x-2}$
$x-2+\frac{0}{x-2}$
$x+2+\frac{0}{x-2}$ (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

Rewrite the rational expression $\frac{x^3+2x^2-1}{x+1}$ in the form $q(x)+\frac{r(x)}{x+1}$ using polynomial long division (or another valid method). Make sure $\deg(r)<\deg(x+1)$ .

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

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 8

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

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

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 9

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

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

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 10

Use polynomial division to rewrite $\frac{x^3-x^2-4x+4}{x-2}$ as $q(x)+\frac{r(x)}{x-2}$ , with $\deg(r)<\deg(x-2)$ .

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

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!