Operating with Rational Expressions - Algebra 2

What is the simplified form of $\frac{\frac{x^2 - 1}{x + 2}}{\frac{x - 1}{x^2 + 4x + 4}}$ ?
Answer: $(x + 1)(x + 2)$ To divide by a fraction, multiply by its reciprocal: $\frac{x^2 - 1}{x + 2} \cdot \frac{x^2 + 4x + 4}{x - 1}$ . Factor: $\frac{(x + 1)(x - 1)}{x + 2} \cdot \frac{(x + 2)^2}{x - 1}$ . Cancel common factors: $\frac{(x + 1)(x - 1)(x + 2)^2}{(x + 2)(x - 1)} = (x + 1)(x + 2)$ . Choice A incorrectly shows the denominator as 1. Choice B incorrectly cancels an additional factor. Choice D fails to complete the division properly.
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.)
Answer: $\frac{x+2}{x-1}$ 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)!
Subtract and simplify to lowest terms: $\frac{x+1}{x^2-4}-\frac{3}{x-2}$ (Assume $x \neq 2, -2$ .)
Answer: $\frac{-2x-5}{x^2-4}$ 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 $x^2-4=(x-2)(x+2)$ 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 $\frac{2}{x - 1} + \frac{3}{x - 1}$ , 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)!
What is the LCD (least common denominator) for the rational expressions $\displaystyle \frac{3}{x^2-4}$ and $\displaystyle \frac{2}{x^2+4x+4}$ ? (Factoring first is recommended.)
Answer: $(x-2)(x+2)^2$ 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 find the LCD of 3/(x² - 4) and 2/(x² + 4x + 4), factor the denominators: x² - 4 = (x - 2)(x + 2) and x² + 4x + 4 = (x + 2)². The LCD must contain all factors with their highest powers: (x - 2) appears once, and (x + 2) appears squared. Therefore, LCD = (x - 2)(x + 2)². Choice C correctly identifies this LCD in factored form. Choice A incorrectly gives just (x - 2)(x + 2), missing that (x + 2) appears squared in the second denominator—the LCD must include the highest power of each factor. The rational expression operation hierarchy: For addition/subtraction, factor denominators completely, then the LCD is the product of all distinct factors, each raised to its highest power appearing in any denominator. This systematic approach ensures you can rewrite all fractions with the same denominator!
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.)
Answer: $\frac{x+3}{x+2}$ 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!
Subtract and simplify to lowest terms. Be careful to distribute the subtraction across the entire numerator when using a common denominator. Subtract: $\frac{4}{x+3}-\frac{1}{x-3}$ (Assume $x\neq \pm 3$ .)
Answer: $\frac{3x-15}{x^2-9}$ 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: for subtraction, find the LCD, rewrite with common denominator, then subtract numerators. To subtract 4/(x + 3) - 1/(x - 3), we need a common denominator. The LCD is (x + 3)(x - 3) = x² - 9. Rewrite each fraction: 4/(x + 3) = 4(x - 3)/[(x + 3)(x - 3)] = (4x - 12)/(x² - 9) and 1/(x - 3) = 1(x + 3)/[(x + 3)(x - 3)] = (x + 3)/(x² - 9). Now subtract: [(4x - 12) - (x + 3)]/(x² - 9) = [4x - 12 - x - 3]/(x² - 9) = [3x - 15]/(x² - 9). Choice C correctly finds the LCD, rewrites both fractions, and carefully distributes the negative sign when subtracting to get (3x - 15)/(x² - 9). Choice A has the wrong sign, B has an arithmetic error, and D oversimplified. Common mistake: forgetting to distribute the negative sign across the entire second numerator. When subtracting (x + 3), you must subtract both terms: -x - 3!
Subtract and simplify to lowest terms. Subtract: $\dfrac{3}{x^2-4}-\dfrac{1}{x-2}$ (assume denominators are nonzero).
Answer: $\dfrac{-x+1}{(x-2)(x+2)}$ 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, use LCD (x - 2)(x + 2) = x² - 4; the first is 3/(x² - 4), second is 1/(x - 2) = (x + 2)/(x² - 4); subtract numerators 3 - (x + 2) = 1 - x, giving (1 - x)/(x² - 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!
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).
Answer: $\dfrac{x^2+3x-2}{x^2-1}$ 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)!
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: $\displaystyle \frac{x^2+5x+6}{x^2-1}\div\frac{x+2}{x-1}$
Answer: $\displaystyle \frac{x+3}{x+1}$ 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!
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.)
Answer: $\frac{x+2}{x-1}$ 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!
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 $x \neq 2, -2$ .)
Answer: $\frac{5x+2}{x^2-4}$ 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, find LCD $(x-2)(x+2)$ ; 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!
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: $\displaystyle \frac{2}{x-3}+\frac{5}{x+2}$
Answer: $\displaystyle \frac{7x-11}{x^2-x-6}$ 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: for addition, find the LCD, rewrite with common denominator, then add numerators. To add 2/(x - 3) + 5/(x + 2), we need LCD = (x - 3)(x + 2). Rewrite each fraction: 2/(x - 3) = 2(x + 2)/[(x - 3)(x + 2)] = (2x + 4)/[(x - 3)(x + 2)] and 5/(x + 2) = 5(x - 3)/[(x - 3)(x + 2)] = (5x - 15)/[(x - 3)(x + 2)]. Now add numerators: (2x + 4 + 5x - 15)/[(x - 3)(x + 2)] = (7x - 11)/[(x - 3)(x + 2)]. Choice C correctly shows this result, which equals (7x - 11)/(x² - x - 6) since (x - 3)(x + 2) = x² - x - 6. Choice B incorrectly adds as (7x + 4)—a sign error when distributing 5(x - 3) = 5x - 15, not 5x + 15. The rational expression operation hierarchy: Addition and subtraction are harder—find LCD as the product of distinct factors, rewrite each fraction with LCD, then add/subtract numerators. Always expand and combine like terms carefully!
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).
Answer: $\dfrac{-2x-5}{(x-2)(x+2)}$ 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)!
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 $x\neq 2,-2$ .)
Answer: $\frac{5x+2}{x^2-4}$ 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, find LCD $(x-2)(x+2)$ ; 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!
What is the LCD (least common denominator) for the rational expressions below? Finding the LCD is the key step for adding/subtracting rational expressions. Find the LCD for: $\frac{3}{x^2-4}\quad\text{and}\quad\frac{2}{x^2+4x+4}$
Answer: $(x-2)(x+2)^2$ 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: for addition/subtraction, find the LCD as the LCM of denominator factors. To find the LCD for 3/(x² - 4) and 2/(x² + 4x + 4), we must factor both denominators. Factor x² - 4 = (x - 2)(x + 2) and x² + 4x + 4 = (x + 2)². The first denominator has factors (x - 2) and (x + 2), while the second has (x + 2)². For the LCD, we need the highest power of each factor that appears: (x - 2) appears once, and (x + 2) appears squared. Therefore, LCD = (x - 2)(x + 2)². Choice C correctly identifies all factors and uses the highest power of each to form the LCD. Choice A incorrectly writes the factors without proper grouping, B misses the squared factor, and D incorrectly squares (x - 2) instead of (x + 2). The rational expression operation hierarchy: Finding the LCD is crucial for addition/subtraction—factor all denominators completely, then take the product of the highest powers of all distinct factors. This ensures both fractions can be rewritten with the same denominator!
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.)
Answer: $(x-2)(x+2)^2$ 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!
Subtract and simplify. Rational expressions behave like rational numbers: find an LCD, subtract the numerators, and simplify. $\displaystyle \frac{x+1}{x^2-4}-\frac{3}{x-2}$
Answer: $\displaystyle \frac{-2x-5}{(x-2)(x+2)}$ 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! Factor $x^2$ - 4 = (x - 2)(x + 2) for LCD (x - 2)(x + 2), rewrite as \frac{x + 1 - 3(x + 2)}{(x - 2)(x + 2)} = \frac{x + 1 - 3x - 6}{(x - 2)(x + 2)} = \frac{-2x - 5}{(x - 2)(x + 2)}. Choice B correctly subtracts and simplifies using the LCD, giving \frac{-2x - 5}{(x - 2)(x + 2)}. Choice A fails by likely miscalculating the numerator as -2x + 7, perhaps from adding instead of subtracting or sign errors. 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)!
Subtract and simplify to lowest terms. This uses the same rules as subtracting fractions and shows closure under subtraction. Subtract: $\displaystyle \frac{x+1}{x^2-4}-\frac{3}{x-2}$
Answer: $\displaystyle \frac{-2x-5}{(x-2)(x+2)}$ 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 subtract (x + 1)/(x² - 4) - 3/(x - 2), first factor x² - 4 = (x - 2)(x + 2). Now we have (x + 1)/[(x - 2)(x + 2)] - 3/(x - 2). The LCD is (x - 2)(x + 2), so rewrite the second fraction: 3/(x - 2) = 3(x + 2)/[(x - 2)(x + 2)] = (3x + 6)/[(x - 2)(x + 2)]. Now subtract: (x + 1 - 3x - 6)/[(x - 2)(x + 2)] = (-2x - 5)/[(x - 2)(x + 2)]. Choice B correctly shows this result in factored form, while Choice C shows the equivalent form with x² - 4 in the denominator. Choice D incorrectly gets (x - 8) in the numerator—likely from a sign error when distributing the negative: -(3x + 6) = -3x - 6, not -3x + 6. Common mistake: trying to cancel before getting common denominator. You can only cancel FACTORS (things multiplied), never TERMS (things added). Know when canceling is valid!
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 $x \neq -3,-1,1$ .)
Answer: $\frac{x+1}{x+3}$ 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 reciprocal: factor $x^2 - 1 = (x-1)(x+1)$ , $x^2 + 4x + 3 = (x+1)(x+3)$ , 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)!
Divide and simplify. State is implicit: rational expressions are closed under division by a nonzero rational expression. $\displaystyle \frac{x^2-1}{x}\div\frac{x+1}{x-1}$
Answer: $\displaystyle \frac{(x-1)^2}{x}$ 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)!

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