Algebra 2 Quiz: Operating With Rational Expressions

20 questions · exam conditions

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Operating With Rational ExpressionsQuestion 1 of 20

Subtract and simplify to lowest terms. (The difference of rational expressions is still a rational expression.)

Subtract: $\dfrac{x+1}{x^2-4}-\dfrac{3}{x-2}$ (assume denominators are nonzero).

\dfrac{-2x-5}{(x-2)(x+2)}

\dfrac{-2x+7}{(x-2)(x+2)}

\dfrac{4x+1}{(x-2)(x+2)}

\dfrac{x-8}{x^2-4}

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Algebra 2 Quiz

Algebra 2 Quiz: Operating With Rational Expressions

Practice Operating With 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 Operating With 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

Subtract and simplify to lowest terms. (The difference of rational expressions is still a rational expression.)

Subtract: $\dfrac{x+1}{x^2-4}-\dfrac{3}{x-2}$ (assume denominators are nonzero).

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! For this subtraction, factor x² - 4 = (x - 2)(x + 2) as the LCD; rewrite the second as 3(x + 2)/((x - 2)(x + 2)); subtract numerators (x + 1) - 3(x + 2) = x + 1 - 3x - 6 = -2x - 5, giving (-2x - 5)/((x - 2)(x + 2)) in lowest terms. Choice A correctly performs the subtraction and simplifies, giving (-2x - 5)/((x - 2)(x + 2)) in lowest terms. A common mistake is switching signs in subtraction, leading to positive terms like in choice C, but remember subtraction means minus the entire numerator. Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 2

Simplify the complex fraction. Because rational expressions are closed under $\div$ (by nonzero), the result is still a rational expression in lowest terms.

$\displaystyle \frac{\frac{x}{x-2}}{\frac{3}{x+1}}$

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! This complex fraction simplifies by multiplying by the reciprocal: \frac{x}{x - 2} cdot \frac{x + 1}{3} = \frac{x(x + 1)}{3(x - 2)}, with no common factors to cancel further. Choice B correctly simplifies the complex fraction to \frac{x(x + 1)}{3(x - 2)}. Choice D fails by perhaps canceling incorrectly across the complex structure, leading to an oversimplified \frac{x}{3}. Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 3

Divide and simplify to lowest terms. This demonstrates closure under division by a nonzero rational expression:

$\frac{x^2-4}{x^2-1}\div\frac{x-2}{x+1}$

(Assume $x$ is any real number that does not make a denominator zero.)

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! To divide, multiply by the reciprocal: factor x²-4=(x-2)(x+2), x²-1=(x-1)(x+1), so \frac{(x-2)(x+2)}{(x-1)(x+1)} cdot \frac{x+1}{x-2}; cancel (x-2) and (x+1) to get \frac{x+2}{x-1}, which is simplified and demonstrates closure under division. Choice A correctly performs the division by multiplying by the reciprocal, factoring, and canceling common factors, giving \frac{x+2}{x-1} in lowest terms. Choice B fails by swapping the numerator and denominator, perhaps from incorrectly handling the reciprocal—always flip only the divisor! Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 4

Simplify. This uses multiplication of rational expressions (factor, cancel, multiply), demonstrating closure under multiplication.

$\displaystyle \frac{x^2+5x+6}{x^2-9}\cdot\frac{x-3}{x+2}$

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. The golden rule: factor everything BEFORE operating! Multiplying (x² - 4)/(x + 3) by (x + 3)/(x + 2) looks messy, but factor x² - 4 = (x + 2)(x - 2) first: [(x + 2)(x - 2)/(x + 3)] · [(x + 3)/(x + 2)]. Now (x + 3) and (x + 2) cancel immediately, leaving just (x - 2). Factoring first prevents working with huge expressions! Factor x^2 + 5x + 6 = (x + 2)(x + 3) and x^2 - 9 = (x - 3)(x + 3), so \frac{(x + 2)(x + 3)}{(x - 3)(x + 3)} cdot \frac{x - 3}{x + 2} cancels to 1. Choice C correctly factors and cancels all common factors, giving 1. Choice A fails by not fully canceling or misfactoring, leaving an unsimplified form. The rational expression operation hierarchy: Multiplication and division are easier—factor everything, cancel common factors, then multiply (or flip and multiply for division). Addition and subtraction are harder—factor denominators, find LCD as the LCM of denominator factors, rewrite each fraction with LCD, then add/subtract numerators. Always simplify at the end by factoring the numerator and canceling any new common factors with the denominator. This systematic approach prevents errors!

Question 5

Perform the operation and simplify to lowest terms (like fraction arithmetic with polynomials).

$\displaystyle \frac{3}{x-1}-\frac{1}{x+1}$

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! To subtract, use LCD (x - 1)(x + 1), rewrite as \frac{3(x + 1) - 1(x - 1)}{(x - 1)(x + 1)} = \frac{3x + 3 - x + 1}{(x - 1)(x + 1)} = \frac{2x + 4}{(x - 1)(x + 1)}. Choice B correctly subtracts and simplifies using the LCD, giving \frac{2x + 4}{x^2 - 1}. Choice C fails by likely miscalculating the numerator signs, such as subtracting incorrectly to get 2x - 4. Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 6

Perform the operation and simplify to lowest terms:

$\frac{2}{x-3}-\frac{1}{x+3}$

(Assume $x\neq 3,-3$ .)

$\frac{1}{x^2-9}$
$\frac{x+9}{x^2-9}$ (correct answer)
$\frac{x+3}{x^2-9}$
$\frac{x-9}{x^2-9}$

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! To subtract, find LCD (x-3)(x+3); \frac{2(x+3) - 1(x-3)}{x^2-9} = \frac{2x+6 - x + 3}{x^2-9} = \frac{x+9}{x^2-9}, which doesn't simplify further. Choice B correctly finds the LCD, subtracts the numerators (distributing the negative), and simplifies, giving \frac{x+9}{x^2-9} in lowest terms. Choice D fails by getting the signs wrong in the numerator, perhaps from subtracting in the wrong order—always distribute the negative carefully! Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 7

Subtract and simplify to lowest terms:

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

(Assume $x\neq 2,-2$ .)

$\frac{x+1}{(x-2)(x+2)}-\frac{3}{(x-2)(x+2)}$
$\frac{x-5}{x^2-4}$
$\frac{-2x+7}{x^2-4}$
$\frac{-2x-5}{x^2-4}$ (correct answer)

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! To subtract, factor x²-4=(x-2)(x+2) for LCD (x-2)(x+2); rewrite second as \frac{3(x+2)}{(x-2)(x+2)}, then \frac{x+1 - (3x+6)}{x^2-4} = \frac{-2x-5}{x^2-4}, which is simplified. Choice B correctly finds the LCD, subtracts the numerators carefully (noting the negative), and simplifies, giving \frac{-2x-5}{x^2-4} in lowest terms. Choice C fails by getting the signs wrong in the numerator, perhaps forgetting to distribute the negative in subtraction—always subtract the entire numerator! Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 8

Divide and simplify. State is implicit: rational expressions are closed under division by a nonzero rational expression.

$\displaystyle \frac{x^2-1}{x}\div\frac{x+1}{x-1}$

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! For this division, rewrite as \frac{x^2 - 1}{x} cdot \frac{x - 1}{x + 1}, factor x^2 - 1 = (x - 1)(x + 1), so \frac{(x - 1)(x + 1)(x - 1)}{x (x + 1)} cancels to \frac{(x - 1)^2}{x}. Choice A correctly divides and simplifies by factoring and canceling, giving \frac{(x - 1)^2}{x}. Choice B fails by perhaps not factoring properly or inverting incorrectly, resulting in the reciprocal. Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 9

Add and simplify to lowest terms. Since rational expressions are closed under addition, the sum is also a rational expression (for allowed $x$ values).

Add: $\displaystyle \frac{2}{x-3}+\frac{5}{x+2}$

$\displaystyle \frac{7}{x^2-x-6}$
$\displaystyle \frac{7x+4}{x^2-x-6}$
$\displaystyle \frac{7x-11}{x^2-x-6}$ (correct answer)
$\displaystyle \frac{7x-11}{(x-3)(x+2)}$

Question 10

Add and simplify to lowest terms. (Rational expressions behave like fractions: find an LCD, combine, and reduce.)

$\frac{3}{x-2}+\frac{2}{x+2}$

(Assume $x\neq 2,-2$ .)

$\frac{5}{x^2-4}$
$\frac{5x-2}{x^2-4}$
$\frac{5x+2}{x^2-4}$ (correct answer)
$\frac{3}{x^2-4}+\frac{2}{x^2-4}$

(x-2)(x+2)

; rewrite as

\frac{3(x+2) + 2(x-2)}{(x-2)(x+2)} = \frac{3x+6 + 2x-4}{x^2-4} = \frac{5x+2}{x^2-4}

, which can't be factored further to cancel, so it's simplified. Choice C correctly finds the LCD, combines the numerators, and simplifies, giving

\frac{5x+2}{x^2-4}

in lowest terms. Choice A fails by incorrectly adding the numerators without distributing, perhaps treating it like same denominators when they're not—always use the LCD for different denominators! The rational expression operation hierarchy: Multiplication and division are easier—factor everything, cancel common factors, then multiply (or flip and multiply for division). Addition and subtraction are harder—factor denominators, find LCD as the LCM of denominator factors, rewrite each fraction with LCD, then add/subtract numerators. Always simplify at the end by factoring the numerator and canceling any new common factors with the denominator. This systematic approach prevents errors!

Question 11

Divide and simplify to lowest terms. This illustrates that rational expressions are closed under division by a nonzero rational expression (like rational numbers).

Divide: $\frac{x^2-4}{x^2-1}\div\frac{x-2}{x+1}$ (Assume all denominators are nonzero and you are not dividing by zero.)

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for division, multiply by the reciprocal (flip the second fraction). The closure property guarantees your result is always another rational expression! To divide (x² - 4)/(x² - 1) by (x - 2)/(x + 1), we multiply by the reciprocal: [(x² - 4)/(x² - 1)] · [(x + 1)/(x - 2)]. Factor everything first: x² - 4 = (x - 2)(x + 2) and x² - 1 = (x - 1)(x + 1). This gives us [(x - 2)(x + 2)]/[(x - 1)(x + 1)] · [(x + 1)/(x - 2)]. Now cancel: (x - 2) cancels from numerator and denominator, and (x + 1) cancels from numerator and denominator, leaving (x + 2)/(x - 1). Choice A correctly performs the division by multiplying by the reciprocal, factors, and cancels to get the simplified result. Choice B incorrectly inverts the final answer, C shows an intermediate step without simplification, and D forgot to cancel the (x + 1) factor. Common mistake: forgetting to flip the second fraction when dividing. Remember: division means multiply by the reciprocal, then factor and cancel as usual!

Question 12

Divide and simplify to lowest terms by multiplying by the reciprocal ("keep, change, flip"). This demonstrates closure under division by a nonzero rational expression.

Divide: $\displaystyle \frac{x^2+5x+6}{x^2-1}\div\frac{x+2}{x-1}$

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. The golden rule: factor everything BEFORE operating! To divide (x² + 5x + 6)/(x² - 1) by (x + 2)/(x - 1), multiply by the reciprocal. First factor: x² + 5x + 6 = (x + 2)(x + 3) and x² - 1 = (x - 1)(x + 1). This gives [(x + 2)(x + 3)]/[(x - 1)(x + 1)] · [(x - 1)/(x + 2)]. Now (x + 2) cancels and (x - 1) cancels, leaving (x + 3)/(x + 1). Choice A correctly shows this simplified result. Choice B incorrectly shows (x + 3)/(x + 2)—perhaps forgetting to flip the second fraction before multiplying. The rational expression operation hierarchy: For division, always flip the second fraction and multiply. Factoring first prevents working with huge expressions and makes cancellation obvious!

Question 13

Perform the operation and simplify to lowest terms. Rational expressions behave like rational numbers: factor, cancel, and simplify.

Simplify: $\frac{x^2+5x+6}{x^2-9}\cdot\frac{x-3}{x+2}$ (Assume denominators are nonzero.)

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. The golden rule: factor everything BEFORE operating! Multiplying (x² + 5x + 6)/(x² - 9) by (x - 3)/(x + 2) requires factoring first. Factor x² + 5x + 6 = (x + 2)(x + 3) and x² - 9 = (x - 3)(x + 3). Now we have [(x + 2)(x + 3)]/[(x - 3)(x + 3)] · [(x - 3)/(x + 2)]. Before multiplying, cancel common factors: (x + 3) cancels from numerator and denominator of the first fraction, (x - 3) cancels between the denominator of the first and numerator of the second, and (x + 2) cancels between the numerator of the first and denominator of the second. After all cancellations, we're left with (x + 3)/(x + 2). Choice D correctly factors all expressions, identifies and cancels all common factors, giving the result in lowest terms. Choice A incorrectly paired factors, B has the wrong denominator, and C reversed the fraction. The rational expression operation hierarchy: For multiplication, factor everything first, then cancel all common factors before multiplying what remains. This systematic approach keeps expressions simple and prevents errors!

Question 14

Perform the operation and simplify to lowest terms. This illustrates the closure property because the result of multiplying two rational expressions is still a rational expression (for allowed $x$ values).

Multiply: $\displaystyle \frac{x^2-9}{x^2-4x+3}\cdot\frac{x-3}{x+1}$

State any restrictions on $x$ from the original expressions.

$\displaystyle \frac{x^2-9}{(x-1)(x+1)}$ , with $x\ne 1,3$
$\displaystyle \frac{x+3}{(x-1)(x+1)}$ , with $x\ne 1,-1,3$ (correct answer)
$\displaystyle \frac{x+3}{(x-1)(x+1)}$ , with $x\ne 1,3$
$\displaystyle \frac{x+3}{x-1}$ , with $x\ne 1,-1,3$

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. The golden rule: factor everything BEFORE operating! Multiplying (x² - 9)/(x² - 4x + 3) by (x - 3)/(x + 1) looks messy, but factor first: x² - 9 = (x + 3)(x - 3), x² - 4x + 3 = (x - 3)(x - 1). Now we have [(x + 3)(x - 3)]/[(x - 3)(x - 1)] · [(x - 3)/(x + 1)]. The (x - 3) factors cancel from the first fraction, leaving [(x + 3)/(x - 1)] · [(x - 3)/(x + 1)] = [(x + 3)(x - 3)]/[(x - 1)(x + 1)]. Choice A correctly shows this result with all restrictions: x ≠ 1, -1 (from denominators) and x ≠ 3 (from the original denominator before canceling). Choice B misses the x ≠ -1 restriction from the final denominator—a common error when tracking restrictions. The rational expression operation hierarchy: Multiplication and division are easier—factor everything, cancel common factors, then multiply. This systematic approach prevents errors and keeps expressions manageable!

Question 15

Subtract and simplify to lowest terms.

Subtract: $\dfrac{3}{x^2-4}-\dfrac{1}{x-2}$ (assume denominators are nonzero).

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

(x - 2)(x + 2) = x^2 - 4

; the first is

\frac{3}{x^2 - 4}

, second is

1/(x - 2) = \frac{x + 2}{x^2 - 4}

; subtract numerators

3 - (x + 2) = 1 - x

, giving

(1 - x)/(x^2 - 4)

or equivalently

(-x + 1)/((x - 2)(x + 2))

in lowest terms. Choice C correctly performs the subtraction and simplifies, giving

(-x + 1)/((x - 2)(x + 2))

in lowest terms. A common mistake is forgetting the negative in subtraction, leading to positive numerators like in choice D, but apply the minus to the entire second numerator. The rational expression operation hierarchy: Multiplication and division are easier—factor everything, cancel common factors, then multiply (or flip and multiply for division). Addition and subtraction are harder—factor denominators, find LCD as the LCM of denominator factors, rewrite each fraction with LCD, then add/subtract numerators. Always simplify at the end by factoring the numerator and canceling any new common factors with the denominator. This systematic approach prevents errors!

Question 16

Add and simplify to lowest terms. (Like fraction arithmetic, you must use an LCD; the result is still a rational expression.)

Add: $\dfrac{x}{x-1}+\dfrac{2}{x+1}$ (assume denominators are nonzero).

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

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. Operating on rational expressions uses the same rules as fraction arithmetic, just with polynomials instead of integers: for multiplication, multiply numerators and denominators (but factor and cancel first to keep things simple!); for division, multiply by the reciprocal (flip the second fraction); for addition/subtraction, find the LCD, rewrite with common denominator, then add/subtract numerators. The closure property guarantees your result is always another rational expression! To add these, use LCD (x - 1)(x + 1) = x² - 1; rewrite as x(x + 1)/(x² - 1) + 2(x - 1)/(x² - 1); combine numerators x² + x + 2x - 2 = x² + 3x - 2, giving (x² + 3x - 2)/(x² - 1) in lowest terms. Choice B correctly finds the LCD and combines, giving (x² + 3x - 2)/(x² - 1) in lowest terms. A common mistake is incorrectly expanding numerators, like forgetting the -2, leading to something like choice C, but verify the combined terms. Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In [2/(x - 1)] + [3/(x - 1)], you can't cancel the (x - 1)—you add the numerators: (2 + 3)/(x - 1) = 5/(x - 1). But in [2/(x - 1)] · [3/(x - 1)], you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 17

Divide and simplify to lowest terms. (Use "multiply by the reciprocal," like fraction division.)

$\frac{x^2-1}{x^2+4x+3}\div\frac{x-1}{x+1}$

(Assume $x \neq -3, -1, 1$ .)

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

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

x^2+4x+3=(x+1)(x+3)

, so

\frac{(x-1)(x+1)}{(x+1)(x+3)} \cdot \frac{x+1}{x-1}

; cancel

(x-1)

and

(x+1)

to get

\frac{x+1}{x+3}

, simplified. Choice A correctly performs the division by factoring, multiplying by the reciprocal, and canceling, giving

\frac{x+1}{x+3}

in lowest terms. Choice B fails by not fully canceling or perhaps forgetting to flip the divisor—always use the reciprocal for division! Common mistake: trying to cancel in addition before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). In

\frac{2}{x - 1} + \frac{3}{x - 1}

, you can't cancel the

(x - 1)

—you add the numerators:

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

. But in

\frac{2}{x - 1} \cdot \frac{3}{x - 1}

, you CAN simplify the result. Know when canceling is valid (multiplication/division after factoring) vs invalid (addition/subtraction without common denominator)!

Question 18

Subtract and simplify to lowest terms. (The difference of rational expressions is still a rational expression.)

Subtract: $\dfrac{x+1}{x^2-4}-\dfrac{3}{x-2}$ (assume denominators are nonzero).

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

Question 19

What is the LCD (least common denominator) for the rational expressions

$\frac{3}{x^2-4} \quad\text{and}\quad \frac{2}{x^2+4x+4}?$

(Factor denominators first.)

$(x-2)(x+2)^2$ (correct answer)
$(x-2)(x+2)(x+2)$
$(x-2)(x+2)$
$(x^2-4)(x^2+4x+4)$

Explanation: This question tests your understanding that rational expressions (fractions with polynomials) form a system just like rational numbers—closed under addition, subtraction, multiplication, and division by nonzero expressions. The golden rule: factor everything BEFORE operating! Multiplying (x² - 4)/(x + 3) by (x + 3)/(x + 2) looks messy, but factor x² - 4 = (x + 2)(x - 2) first: [(x + 2)(x - 2)/(x + 3)] · [(x + 3)/(x + 2)]. Now (x + 3) and (x + 2) cancel immediately, leaving just (x - 2). Factoring first prevents working with huge expressions! To find the LCD, factor denominators: x²-4 = (x-2)(x+2), x²+4x+4 = (x+2)^2; the least common multiple takes the highest powers, so (x-2)(x+2)^2. Choice C correctly factors the denominators and takes the LCM with highest powers, giving (x-2)(x+2)^2. Choice B fails by not using the highest power of (x+2), perhaps overlooking the square—always check exponents in factoring! The rational expression operation hierarchy: Multiplication and division are easier—factor everything, cancel common factors, then multiply (or flip and multiply for division). Addition and subtraction are harder—factor denominators, find LCD as the LCM of denominator factors, rewrite each fraction with LCD, then add/subtract numerators. Always simplify at the end by factoring the numerator and canceling any new common factors with the denominator. This systematic approach prevents errors!

Question 20

Add and simplify to lowest terms. (Rational expressions behave like fractions: find an LCD, combine, and reduce.)

$\frac{3}{x-2}+\frac{2}{x+2}$

(Assume $x \neq 2, -2$ .)

$\frac{5}{x^2-4}$
$\frac{5x-2}{x^2-4}$
$\frac{5x+2}{x^2-4}$ (correct answer)
$\frac{3}{x^2-4}+\frac{2}{x^2-4}$

(x-2)(x+2)

; rewrite as

\frac{3(x+2) + 2(x-2)}{(x-2)(x+2)} = \frac{3x+6 + 2x-4}{x^2-4} = \frac{5x+2}{x^2-4}

, which can't be factored further to cancel, so it's simplified. Choice C correctly finds the LCD, combines the numerators, and simplifies, giving

\frac{5x+2}{x^2-4}