Algebra 2 Quiz: Rewriting Rational Expressions

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Rewriting Rational ExpressionsQuestion 1 of 20

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

2x+1+\dfrac{-6}{x+3}

2x+1+\dfrac{6}{x+3}

2x-1+\dfrac{0}{x+3}

2x+1+\dfrac{-6x}{x+3}

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Algebra 2 Quiz

Algebra 2 Quiz: Rewriting Rational Expressions

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

What this quiz covers

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

How to use this quiz

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

All questions

Question 1

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

$2x+1+\dfrac{-6}{x+3}$ (correct answer)
$2x+1+\dfrac{6}{x+3}$
$2x-1+\dfrac{0}{x+3}$
$2x+1+\dfrac{-6x}{x+3}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting

17/5

3 + 2/5

. The division algorithm for polynomials says any rational expression

a(x)/b(x)

can be written as

q(x) + r(x)/b(x)

, where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. To divide

2x^2 + 7x - 3

x + 3

, divide

2x^2

x

to get

2x

, multiply by

(x + 3)

to get

2x^2 + 6x

, subtract to get

x - 3

, then divide

x

x

to get 1, multiply to get

x + 3

, and subtract to yield -6 as remainder. Choice A correctly divides to get quotient

2x + 1

and remainder -6 with

\deg(r) < \deg(x + 3)

. Choice B might come from a sign error in the final subtraction, turning -6 into +6. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (

\deg

of remainder less than

\deg

of divisor) tells you when to stop dividing: if dividing by

(x + 3)

(degree 1), remainder must be degree 0 (constant).

Question 2

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

$2x^2+5x+5+\dfrac{13}{x-2}$ (correct answer)
$2x^2+5x+5+\dfrac{13}{x+2}$
$2x^2+5x+\dfrac{13}{x-2}$
$2x^2+5x+5+\dfrac{13x}{x-2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting

\frac{17}{5}

3 + \frac{2}{5}

. The division algorithm for polynomials says any rational expression

\frac{a(x)}{b(x)}

can be written as

q(x) + \frac{r(x)}{b(x)}

, where

q(x)

is the quotient (polynomial part) and

r(x)

is the remainder with degree strictly less than the divisor's degree. To divide

2x^3 + x^2 - 5x + 3

x - 2

, start by dividing

2x^3

x

to get

2x^2

, multiply by

(x - 2)

to get

2x^3 - 4x^2

, subtract to yield

5x^2 - 5x

, then continue to get

5x

next, and finally

5

, resulting in a remainder of

13

. Choice A correctly divides to get quotient

2x^2 + 5x + 5

and remainder

13

with

\deg(r) < \deg(x - 2)

. Choice C forgets the constant term in the quotient and the full remainder, likely from stopping too early in the division. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (

\deg

of remainder less than

\deg

of divisor) tells you when to stop dividing: if dividing by

(x - 2)

(

\deg

1), remainder must be

\deg

0 (constant).

Question 3

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

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

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Before dividing, let's check if we can factor: x² - 4 = (x + 2)(x - 2), so (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) = x + 2 after canceling! This means the quotient is x + 2 with remainder 0. Choice A correctly gives x + 2 + 0/(x - 2), which simplifies to just x + 2. Choice B incorrectly has quotient x - 2, C adds an unnecessary remainder of 4, and D includes an x in the remainder which would violate the degree requirement. Before dividing, always check: can you factor the numerator and cancel with the denominator? If (x² - 4)/(x - 2) factors to (x + 2)(x - 2)/(x - 2) = x + 2 (no remainder!), that's instant. Only do long division if simplification doesn't eliminate the division entirely. This pre-check saves enormous time!

Question 4

Use polynomial long division to rewrite $\frac{4x^3-x^2+2x-1}{x+1}$ in the form $q(x)+\frac{r(x)}{x+1}$ , where $\deg(r)<1$ .

$4x^2-5x+7+\dfrac{-8}{x+1}$ (correct answer)
$4x^2-5x+7+\dfrac{8}{x+1}$
$4x^2+5x+7+\dfrac{-8}{x+1}$
$4x^2-5x+\dfrac{-8x}{x+1}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Let's divide (4x³ - x² + 2x - 1) by (x + 1): First, 4x³ ÷ x = 4x², multiply back: 4x²(x + 1) = 4x³ + 4x², subtract to get -5x² + 2x - 1. Next, -5x² ÷ x = -5x, multiply back: -5x(x + 1) = -5x² - 5x, subtract to get 7x - 1. Finally, 7x ÷ x = 7, multiply back: 7(x + 1) = 7x + 7, subtract to get -8. Choice A correctly shows quotient 4x² - 5x + 7 and remainder -8, giving us 4x² - 5x + 7 + (-8)/(x + 1). Choice B has remainder 8 instead of -8, missing the negative sign from the final subtraction: -1 - 7 = -8. Before dividing, always check if you can factor and simplify—here we can't, so long division is necessary. The degree requirement tells you to stop when remainder degree < divisor degree: -8 has degree 0, which is less than 1, so we're done!

Question 5

The rational expression $\frac{x^4 - 3x^3 + 5x - 2}{x^2 - x + 1}$ can be written as $q(x) + \frac{r(x)}{x^2 - x + 1}$ where the degree of $r(x)$ is less than 2. What is the coefficient of $x$ in $r(x)$ ?

$-1$
$2$
$3$
$4$ (correct answer)

Explanation: Performing polynomial long division:

\frac{x^4 - 3x^3 + 5x - 2}{x^2 - x + 1} = x^2 - 2x - 1 + \frac{4x - 3}{x^2 - x + 1}

. The remainder

r(x) = 4x - 3

, so the coefficient of

x

is 4. Choice A comes from sign errors during division. Choice B results from incorrectly computing the final subtraction step. Choice C occurs from arithmetic mistakes in the intermediate steps.

Question 6

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

$x^2 + 2x - 6 + \dfrac{19}{x + 2}$ (correct answer)
$x^2 + 2x - 6 + \dfrac{-19}{x + 2}$
$x^2 + 2x - 6 + \dfrac{19x}{x + 2}$
$x^2 + 4x - 2 + \dfrac{15}{x + 2}$

x^3 + 4x^2 - 2x + 7

x + 2

(root

-2

), bring down 1, multiply by

-2

to get

-2

, add to 4 for 2, multiply for

-4

, add to

-2

for

-6

, multiply for 12, add to 7 for 19 as remainder. Choice A correctly divides to get quotient

x^2 + 2x - 6

and remainder 19 with deg(r) < deg(

x + 2

). Choice B might stem from a sign error in the remainder, flipping 19 to

-19

. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by (

x + 2

) (degree 1), remainder must be degree 0 (constant).

Question 7

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

$2x^2+5x+5+\dfrac{13}{x-2}$ (correct answer)
$2x^2+5x+5+\dfrac{-13}{x-2}$
$2x^2+x-3+\dfrac{9}{x-2}$
$2x^2+5x+\dfrac{13}{x-2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting

17/5

3 + 2/5

. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Using synthetic division with root 2 for

2x^3 + x^2 - 5x + 3

divided by

x - 2

, bring down 2, multiply by 2 for 4, add to 1 for 5, multiply by 2 for 10, add to -5 for 5, multiply by 2 for 10, add to 3 for 13, giving quotient

2x^2 + 5x + 5

and remainder 13. Choice A correctly divides to get quotient

2x^2 + 5x + 5

and remainder 13 with

\deg(r) < 1

. A distractor like choice B might flip the remainder sign from a subtraction error, but positive arithmetic checks out. Before dividing, always check: can you factor the numerator and cancel with the denominator? The degree requirement (

\deg

of remainder less than

\deg

of divisor) tells you when to stop dividing: if dividing by

(x - 2)

(

\deg

1), remainder must be

\deg

0 (constant).

Question 8

Rewrite $\frac{x^3-5x+2}{x-3}$ in the form $q(x)+\frac{r(x)}{x-3}$ using polynomial division, where $\deg(r)<1$ .

$x^2+3x+4+\dfrac{14}{x-3}$ (correct answer)
$x^2+3x+4+\dfrac{-14}{x-3}$
$x^2+3x-4+\dfrac{14}{x-3}$
$x^2+3x+4+\dfrac{14}{x+3}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Let's divide (x³ - 5x + 2) by (x - 3), noting the missing x² term means coefficient 0: First, x³ ÷ x = x², multiply back: x²(x - 3) = x³ - 3x², subtract to get 3x² - 5x + 2. Next, 3x² ÷ x = 3x, multiply back: 3x(x - 3) = 3x² - 9x, subtract to get 4x + 2. Finally, 4x ÷ x = 4, multiply back: 4(x - 3) = 4x - 12, subtract to get 14. Choice A correctly shows quotient x² + 3x + 4 and remainder 14, giving us x² + 3x + 4 + 14/(x - 3). Choice B has remainder -14 instead of 14, likely from computing 2 - (-12) = 2 + 12 = 14 but then incorrectly negating it. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here we can't simplify first, so long division is needed. Since we're dividing by (x - 3) with degree 1, remainder must have degree 0 (constant), which 14 is!

Question 9

When dividing $a(x)$ by $b(x)$ , the division algorithm says $\frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)}$ where $\deg(r) < \deg(b)$ . For $\frac{x^5 + 2x^4 - x^3 + 4}{x^2 - 3x + 1},$ what must be the degree of the remainder polynomial $r(x)$ ?

$\deg(r)=2$
$\deg(r)\le 1$ (correct answer)
$\deg(r)\le 2$
$\deg(r)=3$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. This separation shows the polynomial behavior (q(x)) separately from the fractional part (r(x)/b(x)), which approaches zero as x gets large. In this problem, we're dividing by b(x) = x² - 3x + 1, which has degree 2. The key constraint is deg(r) < deg(b) = 2, meaning the remainder must have degree less than 2. Choice B correctly states deg(r) ≤ 1, which means the remainder can be degree 1 (like 3x + 5) or degree 0 (like 7) or even the zero polynomial. Choice A incorrectly claims deg(r) = 2, but that would violate the fundamental requirement deg(r) < deg(b). The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: if dividing by a degree-2 polynomial, the remainder must be degree 1 or less. This is analogous to dividing integers: when dividing by 10, the remainder must be less than 10 (0 through 9)!

Question 10

Express $\frac{2x^4-3x^3+x^2+4}{x^2+2}$ as $q(x)+\frac{r(x)}{x^2+2}$ using polynomial division, where $\deg(r)<2$ .

$2x^2-3x-3+\dfrac{6x+10}{x^2+2}$ (correct answer)
$2x^2-3x-3+\dfrac{6x+10}{x^2-2}$
$2x^2-3x-3+\dfrac{6x^2+10}{x^2+2}$
$2x^2-3x+3+\dfrac{6x+10}{x^2+2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Dividing (2x⁴ - 3x³ + x² + 4) by (x² + 2): First, 2x⁴ ÷ x² = 2x², multiply back: 2x²(x² + 2) = 2x⁴ + 4x², subtract to get -3x³ - 3x² + 4. Next, -3x³ ÷ x² = -3x, multiply back: -3x(x² + 2) = -3x³ - 6x, subtract to get -3x² + 6x + 4. Finally, -3x² ÷ x² = -3, multiply back: -3(x² + 2) = -3x² - 6, subtract to get 6x + 10. Choice A correctly shows quotient 2x² - 3x - 3 and remainder 6x + 10, giving us 2x² - 3x - 3 + (6x + 10)/(x² + 2). Choice C incorrectly suggests remainder 6x² + 10, but that would have degree 2, violating the requirement that deg(r) < 2 = deg(x² + 2). The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: since x² + 2 has degree 2, remainder must have degree at most 1. Our remainder 6x + 10 has degree 1, which is perfect!

Question 11

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

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

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Let's divide (x³ + 2x² - 5x + 3) by (x - 1): First, x³/x = x², multiply (x - 1) by x² to get x³ - x², subtract to get 3x² - 5x + 3. Next, 3x²/x = 3x, multiply (x - 1) by 3x to get 3x² - 3x, subtract to get -2x + 3. Then, -2x/x = -2, multiply (x - 1) by -2 to get -2x + 2, subtract to get 1. Since 1 has degree 0 < degree 1 of (x - 1), we stop! Choice A correctly gives quotient x² + 3x - 2 and remainder 1, written as x² + 3x - 2 + 1/(x - 1). Choice C has the wrong sign on the remainder fraction, while B and D have incorrect quotients from calculation errors. The degree requirement (deg of remainder less than deg of divisor) tells you when to stop dividing: once your remainder (1) has degree 0, which is less than the divisor's degree 1, you're done—stop dividing!

Question 12

Use polynomial division to write $\frac{x^4+1}{x-1}$ in the form $q(x)+\frac{r(x)}{x-1}$ , where $\deg(r)<\deg(x-1)$ .

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

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. For x⁴ + 1 divided by x - 1, synthetic or long division yields x³ + x² + x + 1 with remainder 2. Choice A correctly captures this, with deg(r) < 1. Choice B incorrectly claims zero remainder. Use the degree constraint to confirm stopping at a constant remainder.

Question 13

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

$2x-1+\dfrac{-2x+4}{x^2+1}$ (correct answer)
$2x-1+\dfrac{-2x+4}{x^2-1}$
$2x+1+\dfrac{-2x+4}{x^2+1}$
$2x-1+\dfrac{-2x^2+4}{x^2+1}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. For 2x³ - x² + 3 divided by x² + 1, the process results in 2x - 1 with remainder -2x + 4. Choice A correctly identifies this quotient and remainder, with deg(r) < 2. Choice D has an invalid remainder degree, equaling the divisor's. The degree requirement ensures you stop once the remainder is linear or constant for a quadratic divisor.

Question 14

When $\frac{2x^4 - 3x^3 + x^2 - x + 5}{x^2 - 2x + 1}$ is written in the form $q(x) + \frac{r(x)}{x^2 - 2x + 1}$ , what is the degree of $q(x) \cdot r(x)$ ?

$2$
$3$ (correct answer)
$4$
$5$

Explanation: The dividend has degree 4 and divisor has degree 2, so

q(x)

has degree

4 - 2 = 2

. Since the divisor has degree 2, the remainder

r(x)

must have degree less than 2, so degree 1. Therefore,

q(x) \cdot r(x)

has degree

2 + 1 = 3

. Note that

x^2 - 2x + 1 = (x-1)^2

, but this doesn't change the degree analysis. Choice A assumes

r(x)

has degree 0. Choice C incorrectly uses the original dividend degree. Choice D adds degrees instead of using polynomial division rules.

Question 15

Rewrite $\frac{x^2+5x+7}{x+2}$ in the form $q(x)+\frac{r(x)}{x+2}$ , where $\deg(r)<\deg(x+2)$ . (Inspection is allowed.)

$x+3+\dfrac{1}{x+2}$ (correct answer)
$x+2+\dfrac{3}{x+2}$
$x+3+\dfrac{1}{x-2}$
$x+3+\dfrac{x+1}{x+2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting

17/5

3 + 2/5

. The division algorithm for polynomials says any rational expression

a(x)/b(x)

can be written as

q(x) + r(x)/b(x)

, where

q(x)

is the quotient (polynomial part) and

r(x)

is the remainder with degree strictly less than the divisor's degree. Inspecting or dividing

x^2 + 5x + 7

x + 2

gives

x + 3

with remainder

1

. Choice A correctly provides this form, meeting the degree condition. Choice B might stem from a miscalculation in the constant term. Before dividing, check for simplification, and remember the remainder must be constant for linear divisors.

Question 16

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

$2x^2-x-7+\dfrac{21}{x-2}$
$2x^2+7x+\dfrac{9x+25}{x-2}$
$2x^2+7x+9+\dfrac{25}{x-2}$ (correct answer)
$2x^2+7x+9+\dfrac{5}{x-2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Let's divide 2x³ + 3x² - 5x + 7 by x - 2: First, 2x³ ÷ x = 2x², multiply back: 2x²(x - 2) = 2x³ - 4x², subtract: (2x³ + 3x² - 5x + 7) - (2x³ - 4x²) = 7x² - 5x + 7. Next, 7x² ÷ x = 7x, multiply back: 7x(x - 2) = 7x² - 14x, subtract: (7x² - 5x + 7) - (7x² - 14x) = 9x + 7. Finally, 9x ÷ x = 9, multiply back: 9(x - 2) = 9x - 18, subtract: (9x + 7) - (9x - 18) = 25. Choice A correctly gives quotient 2x² + 7x + 9 and remainder 25, with deg(25) = 0 < deg(x - 2) = 1. Choice B has the wrong remainder (5 instead of 25), likely from an arithmetic error in the final subtraction. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here, no obvious factorization exists, so long division is necessary. The degree requirement tells you when to stop: since we're dividing by (x - 2) which has degree 1, we stop when the remainder is a constant (degree 0).

Question 17

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

$2x^2-x-3+\dfrac{13}{x+2}$ (correct answer)
$2x^2-x-3+\dfrac{1}{x+2}$
$2x^2-x+1+\dfrac{13}{x+2}$
$2x^2-x-3+\dfrac{13x}{x+2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting

17/5

3 + 2/5

. The division algorithm for polynomials says any rational expression

a(x)/b(x)

can be written as

q(x) + \frac{r(x)}{b(x)}

, where

q(x)

is the quotient (polynomial part) and

r(x)

is the remainder with degree strictly less than the divisor's degree. To divide

2x^3 + 3x^2 - 5x + 7

x + 2

, start by dividing

2x^3

x

to get

2x^2

, multiply by the divisor to get

2x^3 + 4x^2

, subtract to yield

-x^2 - 5x

, then continue to get

-x

and

-3

, resulting in a remainder of 13. Choice A correctly divides to get quotient

2x^2 - x - 3

and remainder 13, with

\deg(r) < \deg(x + 2)

. A common mistake, like in choice B, is miscalculating the remainder as 1 instead of 13 due to an arithmetic error in the final subtraction. Before dividing, always check if you can factor the numerator and cancel with the denominator, but here no common factors exist, so long division is necessary. The degree requirement ensures the remainder is proper, stopping when

\deg(\text{remainder}) < \deg(\text{divisor})

, just like stopping in numerical division when the remainder is smaller than the divisor.

Question 18

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

$2x^2+7x+9+\dfrac{24}{x-2}$ (correct answer)
$2x^2+7x+9+\dfrac{12}{x-2}$
$2x^2+7x+9+\dfrac{24}{x+2}$
$2x^2+7x+\dfrac{24x}{x-2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Let's divide (2x³ + 3x² - 5x + 6) by (x - 2): First, 2x³ ÷ x = 2x², multiply back: 2x²(x - 2) = 2x³ - 4x², subtract to get 7x² - 5x + 6. Next, 7x² ÷ x = 7x, multiply back: 7x(x - 2) = 7x² - 14x, subtract to get 9x + 6. Finally, 9x ÷ x = 9, multiply back: 9(x - 2) = 9x - 18, subtract to get 24. Choice A correctly shows quotient 2x² + 7x + 9 and remainder 24, giving us 2x² + 7x + 9 + 24/(x - 2). Choice B has the wrong remainder (12 instead of 24), likely from an arithmetic error in the final subtraction step. Before dividing, always check: can you factor the numerator and cancel with the denominator? Here we can't simplify first, so long division is necessary. The degree requirement tells you when to stop: since we're dividing by (x - 2) which has degree 1, our remainder must have degree 0 (a constant), which 24 is!

Question 19

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

$x^2+2x-6+\dfrac{19}{x+2}$ (correct answer)
$x^2+2x-6-\dfrac{19}{x+2}$
$x^2+6x-14+\dfrac{35}{x+2}$
$x^2+2x+6+\dfrac{-5}{x+2}$

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. Polynomial long division works exactly like numerical long division: (1) divide the leading terms to get the first term of quotient, (2) multiply the entire divisor by that term, (3) subtract from the dividend, (4) repeat with what remains until the remainder's degree drops below the divisor's degree. Let's divide (x³ + 4x² - 2x + 7) by (x + 2): First, x³/x = x², multiply (x + 2) by x² to get x³ + 2x², subtract to get 2x² - 2x + 7. Next, 2x²/x = 2x, multiply (x + 2) by 2x to get 2x² + 4x, subtract to get -6x + 7. Then, -6x/x = -6, multiply (x + 2) by -6 to get -6x - 12, subtract to get 19. Since 19 has degree 0 < degree 1 of (x + 2), we stop! Choice A correctly gives quotient x² + 2x - 6 and remainder 19, written as x² + 2x - 6 + 19/(x + 2). Choice B has the wrong sign on the remainder fraction, C has an incorrect quotient from arithmetic errors, and D has both wrong quotient and remainder. The degree requirement tells you when to stop: once your remainder (19) has degree 0, which is less than the divisor's degree 1, you're done—continuing would be like dividing 2 by 5 further after getting 0 remainder 2!

Question 20

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

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

Explanation: This question tests your understanding of polynomial division—rewriting a rational expression as a polynomial quotient plus a proper fraction, just like rewriting 17/5 as 3 + 2/5. The division algorithm for polynomials says any rational expression a(x)/b(x) can be written as q(x) + r(x)/b(x), where q(x) is the quotient (polynomial part) and r(x) is the remainder with degree strictly less than the divisor's degree. Before dividing, let's check if we can factor: x³ - 1 = (x - 1)(x² + x + 1) by the difference of cubes formula! So (x³ - 1)/(x - 1) = (x - 1)(x² + x + 1)/(x - 1) = x² + x + 1 after canceling. This means the quotient is x² + x + 1 with remainder 0. Choice A correctly gives x² + x + 1 + 0/(x - 1), which simplifies to just x² + x + 1. Choice B has the wrong middle coefficient, C adds an unnecessary remainder of 1, and D is missing the constant term in the quotient. Before dividing, always check: can you factor the numerator and cancel with the denominator? Recognizing x³ - 1 as a difference of cubes that factors as (x - 1)(x² + x + 1) gives you the answer instantly without any division work!