Operating with Rational Expressions - Algebra 1

Perform the indicated operation and simplify (the product is a rational expression, showing closure under multiplication): $\frac{3x}{x^2-4}\cdot\frac{x-2}{6}$
Answer: $\frac{x}{2(x+2)}$ This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor first: x²-4 = (x+2)(x-2). So we have [3x/(x+2)(x-2)]·[(x-2)/6]. Before multiplying, we can cancel: (x-2) appears in both a numerator and denominator, so it cancels. We can also simplify 3/6 = 1/2. This leaves us with [x/(x+2)]·[1/2] = x/[2(x+2)]. Choice B correctly performs the multiplication and simplifies to x/[2(x+2)] by factoring and canceling the (x-2) term. Great work! Choice A incorrectly has (x-2) in the denominator instead of (x+2)—remember x²-4 = (x+2)(x-2), not 2(x-2)! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.
Add the rational expressions (polynomials divided by polynomials) and simplify. (The sum is a rational expression because the set is closed under addition.) $\frac{1}{x}+\frac{2}{x+1}$
Answer: $\frac{3x+1}{x(x+1)}$ This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with $1/2 + 1/3$ : we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! The denominators are $x$ and $(x+1)$ , so LCD is $x(x+1)$ ; rewrite as $\frac{1 \cdot (x+1) + 2 \cdot x}{x(x+1)} = \frac{x+1 + 2x}{x(x+1)} = \frac{3x+1}{x(x+1)}$ , which doesn't simplify further. Choice B correctly performs the addition and simplifies to $\frac{3x+1}{x(x+1)}$ by finding the LCD and combining numerators. Great work! For example, choice C might result from adding numerators without the common denominator properly, like just 1+2=3 over something, but always build up to the LCD first—it's a key step, and you're doing great by practicing it. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like $1/6 + 1/4$ : LCD = 12, rewrite as $2/12 + 3/12 = 5/12$ , just with variables! Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). So in $[2/x] + [3/x]$ , you cannot cancel the x's—you can only add numerators because denominators are already the same: $(2+3)/x = 5/x$ .
Add and simplify (remember: rational expressions are closed under addition, so your result should be a rational expression in simplest form): $\frac{1}{x}+\frac{2}{x+1}$
Answer: $\frac{3x+1}{x(x+1)}$ This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For 1/x + 2/(x+1), the denominators are x and (x+1). Since these share no common factors, the LCD is their product: x(x+1). We rewrite: 1/x = (x+1)/[x(x+1)] and 2/(x+1) = 2x/[x(x+1)]. Now we can add: (x+1)/[x(x+1)] + 2x/[x(x+1)] = (x+1+2x)/[x(x+1)] = (3x+1)/[x(x+1)]. Choice D correctly shows (3x+1)/[x(x+1)] as the final answer. Great work! Choice A shows 3/(2x+1)—remember we need to find a common denominator first, not just add the numerators. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6 + 1/4: LCD = 12, rewrite as 2/12 + 3/12 = 5/12, just with variables!
Multiply and simplify the rational expressions (the result is also a rational expression): $\frac{x+3}{x} \cdot \frac{x}{x-2}$
Answer: $\frac{x+3}{x-2}$ This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving $\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d}$ . But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For $\frac{x+3}{x} \cdot \frac{x}{x-2}$ , the x in the numerator of the second fraction cancels with the x in the denominator of the first, leaving $\frac{x+3}{x-2}$ . Choice B correctly performs the multiplication and simplifies to $\frac{x+3}{x-2}$ by canceling the common x factors. For example, choice A forgets to cancel the common x, resulting in an unsimplified expression—remember to always look for cancellations after multiplying! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.
Simplify the rational expression by factoring and canceling common factors: $\frac{x^2-9}{x+3}$
Answer: $x-3$ This question tests your understanding of how to simplify rational expressions—algebraic fractions that work just like regular fractions but with variables. Rational expressions form a system like rational numbers: they're closed under addition, subtraction, multiplication, and division (as long as we don't divide by zero). This means doing any of these operations on rational expressions gives you another rational expression—the result is still a fraction of polynomials! To simplify $\frac{x^2-9}{x+3}$ , factor the numerator as $(x-3)(x+3)$ , then cancel the common $(x+3)$ factor to get $x-3$ . Choice B correctly simplifies to $x-3$ by factoring and canceling the common factor. For instance, choice A doesn't factor or cancel, leaving it unsimplified—always factor to spot cancellations! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.
If $\frac{x-1}{x+3} \cdot \frac{x^2+6x+9}{x^2-1} = \frac{A}{B}$ in lowest terms, what is $A + B$ ?
Answer: $2x + 4$ When you encounter rational expressions being multiplied, your goal is to simplify by factoring and canceling common factors before multiplying across. Let's work through this step-by-step. First, factor each polynomial completely: - $x^2 + 6x + 9 = (x + 3)^2$ (perfect square trinomial) - $x^2 - 1 = (x + 1)(x - 1)$ (difference of squares) Now substitute these factorizations: $\frac{x-1}{x+3} \cdot \frac{(x+3)^2}{(x+1)(x-1)}$ Cancel common factors: The $(x-1)$ terms cancel, and one $(x+3)$ from the numerator cancels with the $(x+3)$ in the denominator: $\frac{\cancel{x-1}}{\cancel{x+3}} \cdot \frac{(x+3)^{\cancel{2}}}{\cancel{(x-1)}(x+1)} = \frac{x+3}{x+1}$ So $A = x + 3$ and $B = x + 1$ , making $A + B = (x + 3) + (x + 1) = 2x + 4$ . Looking at the wrong answers: Choice B ( $2x - 1$ ) likely comes from sign errors when adding the constants. Choice C ( $x + 1$ ) might result from forgetting to add both terms and just taking one of them. Choice D ( $2x + 1$ ) could come from incorrectly combining the constants as $3 + 1 = 4$ but then making an arithmetic mistake. Study tip: Always factor completely before multiplying rational expressions, and double-check your arithmetic when combining like terms. The factoring patterns (perfect squares and difference of squares) appear frequently on algebra exams.
Divide and simplify. (Because rational expressions are closed under division by a nonzero rational expression, the result is rational.) $\frac{x^2-1}{x+1} \div \frac{x-1}{x}$
Answer: $x$ This question tests your understanding of how to divide rational expressions—algebraic fractions that work just like regular fractions but with variables. Dividing rational expressions uses the 'keep, change, flip' rule from numeric fractions: to divide by a fraction, multiply by its reciprocal. So $(a/b) \div (c/d) = (a/b) \cdot (d/c)$ . Flip the second fraction (reciprocal), change \div to \times, then multiply! To divide, rewrite as $\frac{x^2-1}{x+1} \cdot \frac{x}{x-1}$ , factor $ $x^2$ -1 = (x-1)(x+1) $, so $\frac{(x-1)(x+1) \cdot x}{(x+1) \cdot (x-1)}$ , then cancel (x+1) and (x-1), leaving $ x $. Choice C correctly performs the division and simplifies to x by factoring and canceling common factors after flipping. Great work! For example, choice A is the expression before dividing, but always remember to flip the divisor first—it's a crucial step, and you're getting the hang of it! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: $\frac{x^2-4}{x} \cdot \frac{x}{x+2}$ looks messy, but factor $ $x^2$ -4 $ to get $\frac{(x+2)(x-2)}{x} \cdot \frac{x}{x+2}$ , cancel the (x+2) and x, leaving just $ x-2 $. So much cleaner!
Divide and simplify (the quotient is a rational expression): $\frac{x^2-1}{x-1}\div\frac{x+1}{x-1}$
Answer: $x-1$ This question tests your understanding of how to divide rational expressions—algebraic fractions that work just like regular fractions but with variables. Dividing rational expressions uses the 'keep, change, flip' rule from numeric fractions: to divide by a fraction, multiply by its reciprocal. So (a/b) ÷ (c/d) = (a/b)·(d/c). Flip the second fraction (reciprocal), change ÷ to ×, then multiply! Let's apply this: (x²-1)/(x-1) ÷ (x+1)/(x-1) becomes (x²-1)/(x-1) · (x-1)/(x+1). First factor x²-1 = (x+1)(x-1), giving us [(x+1)(x-1)/(x-1)] · [(x-1)/(x+1)]. The (x-1) cancels from numerator and denominator in the first fraction, and then (x+1) and (x-1) cancel between the two fractions, leaving just (x-1). Choice C correctly divides and simplifies to x-1 by using the reciprocal and canceling all common factors. Great work! Choice A shows the original dividend, choice B incorrectly inverts the result, and choice D gives x+1 instead of x-1. Common mistake: forgetting to flip the second fraction when dividing. Remember, division by a fraction is multiplication by its reciprocal—this rule works exactly the same whether you're dividing 2/3 by 1/4 or dividing complex rational expressions!
Divide and simplify (use the reciprocal of the second rational expression): $\frac{x-1}{2}\div\frac{x-1}{4}$
Answer: $2$ This question tests your understanding of how to divide rational expressions—algebraic fractions that work just like regular fractions but with variables. Dividing rational expressions uses the 'keep, change, flip' rule from numeric fractions: to divide by a fraction, multiply by its reciprocal. So (a/b) ÷ (c/d) = (a/b)·(d/c). Flip the second fraction (reciprocal), change ÷ to ×, then multiply! For (x-1)/2 ÷ (x-1)/4, we flip the second fraction to get 4/(x-1) and multiply: (x-1)/2 · 4/(x-1). Multiplying gives us 4(x-1)/[2(x-1)]. The (x-1) factors cancel, leaving 4/2 = 2. Choice B correctly shows 2 as the simplified result. Great work! Choice A shows 1/2, which would be the result if you forgot to flip—remember, dividing by a fraction means multiplying by its reciprocal. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.
Add the rational expressions and simplify. (Because rational expressions are closed under addition, the sum is also a rational expression.) $\frac{2}{x-3}+\frac{5}{x-3}$
Answer: $\frac{7}{x-3}$ This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! Since the denominators are already the same (x-3), simply add the numerators: (2 + 5)/(x-3) = 7/(x-3), and it doesn't simplify further. Choice A correctly performs the addition and simplifies to 7/(x-3) by combining numerators over the common denominator. Great work! For example, choice C might come from subtracting 3 from the denominator or something similar, but keep the denominator as is when adding—only the numerators combine, just like 1/5 + 2/5 = 3/5, not changing the 5. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6 + 1/4: LCD = 12, rewrite as 2/12 + 3/12 = 5/12, just with variables! Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). So in [2/x] + [3/x], you cannot cancel the x's—you can only add numerators because denominators are already the same: (2+3)/x = 5/x.
Add and simplify (the sum is a rational expression, showing closure under addition): $\frac{1}{x-1}+\frac{2}{x+1}$
Answer: $\frac{3x-1}{x^2-1}$ This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For 1/(x-1) + 2/(x+1), the denominators are (x-1) and (x+1). Since these share no common factors, the LCD is their product: (x-1)(x+1) = x²-1. Now rewrite each fraction: 1/(x-1) = (x+1)/[(x-1)(x+1)] = (x+1)/(x²-1) and 2/(x+1) = 2(x-1)/[(x+1)(x-1)] = (2x-2)/(x²-1). Adding the numerators: (x+1) + (2x-2) = x+1+2x-2 = 3x-1, so our answer is (3x-1)/(x²-1). Choice C correctly performs the addition and simplifies to (3x-1)/(x²-1) by finding the LCD and adding numerators. Great work! Choice B shows (3x+1) in the numerator—be careful with signs when distributing and combining like terms! Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added).
Divide and simplify (the result is a rational expression). Use the reciprocal to divide: $\frac{x^2-4}{x+1}\div\frac{x-2}{x+1}$
Answer: $x+2$ This question tests your understanding of how to divide rational expressions—algebraic fractions that work just like regular fractions but with variables. Dividing rational expressions uses the 'keep, change, flip' rule from numeric fractions: to divide by a fraction, multiply by its reciprocal. So (a/b) ÷ (c/d) = (a/b)·(d/c). Flip the second fraction (reciprocal), change ÷ to ×, then multiply! Let's apply this: (x²-4)/(x+1) ÷ (x-2)/(x+1) becomes (x²-4)/(x+1) · (x+1)/(x-2). Now factor x²-4 = (x+2)(x-2), giving us [(x+2)(x-2)/(x+1)] · [(x+1)/(x-2)]. The (x+1) cancels from numerator and denominator, and the (x-2) cancels from numerator and denominator, leaving just (x+2). Choice B correctly divides and simplifies to x+2 by using the reciprocal and canceling common factors. Great work! Choice A incorrectly keeps the original form without dividing, while choices C and D show various errors in the division process. Common mistake: forgetting to flip the second fraction when dividing. Remember, division by a fraction is multiplication by its reciprocal—this rule works exactly the same whether you're dividing 2/3 by 1/4 or dividing complex rational expressions!
What is $\frac{2x}{x^2 - 1} - \frac{1}{x - 1}$ in simplest form?
Answer: $\frac{1}{x + 1}$ When you encounter rational expressions being subtracted, the key is finding a common denominator and then simplifying. Notice that the first denominator $x^2 - 1$ can be factored as $(x-1)(x+1)$ , which will help us work with both fractions. To subtract these fractions, we need a common denominator. Since $x^2 - 1 = (x-1)(x+1)$ , we can rewrite the expression as: $\frac{2x}{(x-1)(x+1)} - \frac{1}{x-1}$ To get a common denominator, multiply the second fraction by $\frac{x+1}{x+1}$ : $\frac{2x}{(x-1)(x+1)} - \frac{1 \cdot (x+1)}{(x-1)(x+1)} = \frac{2x}{(x-1)(x+1)} - \frac{x+1}{(x-1)(x+1)}$ Now subtract the numerators: $\frac{2x - (x+1)}{(x-1)(x+1)} = \frac{2x - x - 1}{(x-1)(x+1)} = \frac{x - 1}{(x-1)(x+1)}$ Finally, cancel the common factor $(x-1)$ from numerator and denominator: $\frac{x-1}{(x-1)(x+1)} = \frac{1}{x+1}$ Choice A gives $\frac{2x-1}{x^2-1}$ , which would result from incorrectly combining numerators without finding a proper common denominator. Choice B gives $\frac{x+1}{x^2-1}$ , which comes from an error in the subtraction step. Choice C gives $\frac{x-1}{x+1}$ , which is what you'd get if you forgot to cancel the common factor. Strategy tip: Always factor denominators first when working with rational expressions, and remember that subtraction requires distributing the negative sign through parentheses in the numerator.
Perform the indicated operation and simplify (the product is a rational expression): $\frac{3}{x}\cdot\frac{x^2}{x+2}$
Answer: $\frac{3x}{x+2}$ This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's multiply: (3/x) · (x²/(x+2)) = (3·x²)/(x·(x+2)) = 3x²/[x(x+2)]. Now we can simplify by canceling one factor of x from numerator and denominator: 3x²/[x(x+2)] = 3x/(x+2). Choice C correctly multiplies and simplifies to 3x/(x+2) by multiplying numerators and denominators, then canceling the common factor of x. Great work! Choice A shows the unsimplified form, choice B cancels too many x's (you can only cancel one!), and choice D has the wrong denominator. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: [(x²-4)/x]·[x/(x+2)] looks messy, but factor x²-4 to get [(x+2)(x-2)/x]·[x/(x+2)], cancel the (x+2) and x, leaving just (x-2). So much cleaner!
Add and simplify (the sum is a rational expression). Find the LCD: $\frac{1}{x-2}+\frac{1}{x+2}$
Answer: $\frac{2x}{x^2-4}$ This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For 1/(x-2) + 1/(x+2), the LCD is (x-2)(x+2) = x²-4. Rewrite the first fraction: 1/(x-2) = (x+2)/[(x-2)(x+2)] = (x+2)/(x²-4). Rewrite the second fraction: 1/(x+2) = (x-2)/[(x-2)(x+2)] = (x-2)/(x²-4). Now add: (x+2)/(x²-4) + (x-2)/(x²-4) = [(x+2)+(x-2)]/(x²-4) = 2x/(x²-4). Choice B correctly adds the fractions and simplifies to 2x/(x²-4) by finding the LCD and combining numerators. Great work! Choice A incorrectly gives just 2 in the numerator, choice C shows the factored form of the denominator (which is equivalent), and choice D has the wrong numerator. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6 + 1/4: LCD = 12, rewrite as 2/12 + 3/12 = 5/12, just with variables!
What is the simplified form of the rational expression (a rational expression remains rational after simplification, showing closure under equivalent transformations)? $\frac{x^2-5x+6}{x^2-9}$
Answer: $\frac{x-2}{x+3}$ This question tests your understanding of how to simplify rational expressions—algebraic fractions that work just like regular fractions but with variables. Rational expressions form a system like rational numbers: they're closed under addition, subtraction, multiplication, and division (as long as we don't divide by zero). This means doing any of these operations on rational expressions gives you another rational expression—the result is still a fraction of polynomials! To simplify (x²-5x+6)/(x²-9), we need to factor both numerator and denominator. The numerator x²-5x+6 factors as (x-2)(x-3), and the denominator x²-9 is a difference of squares: (x+3)(x-3). So we have [(x-2)(x-3)]/[(x+3)(x-3)]. Now we can cancel the common factor (x-3) from numerator and denominator, leaving us with (x-2)/(x+3). Choice A correctly simplifies to (x-2)/(x+3) by factoring and canceling the common factor (x-3). Great work! Choice D shows the original expression without canceling—remember, always simplify by canceling common factors! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Even when just simplifying a single fraction, factoring is your best friend!
Multiply and simplify (factor first, then cancel common factors): $\frac{x^2-1}{x^2+3x+2}\cdot\frac{x+2}{x-1}$
Answer: $\frac{x+1}{x}$ This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor everything: (x²-1)/(x²+3x+2) = [(x+1)(x-1)]/[(x+1)(x+2)] and we're multiplying by (x+2)/(x-1). Now we can see common factors: (x+1) and (x-1) each appear once in a numerator and denominator, as does (x+2). After canceling: [(x+1)(x-1)]/[(x+1)(x+2)] · (x+2)/(x-1) = (x+1)/x. Choice B correctly performs the multiplication and simplifies to (x+1)/x by factoring and canceling all common factors. Great work! Choice A has an extra +2 in the denominator, C and D show different incorrect simplifications. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Notice how factoring x²+3x+2 as (x+1)(x+2) revealed the (x+1) factor that could cancel—without factoring first, you'd miss this!
Perform the indicated operation and simplify. (Rational expressions are closed under division, so the quotient is a rational expression.) Divide: $\frac{x^2-4}{x^2-1}\div \frac{x-2}{x-1}$
Answer: $\frac{x+2}{x+1}$ This question tests your understanding of how to divide rational expressions—algebraic fractions that work just like regular fractions but with variables. Dividing rational expressions uses the 'keep, change, flip' rule from numeric fractions: to divide by a fraction, multiply by its reciprocal. So $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ . Flip the second fraction (reciprocal), change ÷ to ×, then multiply! To divide, rewrite as $\frac{x^2 - 4}{x^2 - 1} \cdot \frac{x-1}{x-2}$ , factor $ $x^2$ - 4 = (x-2)(x+2) $, $ $x^2$ - 1 = (x-1)(x+1) $, so $\frac{(x-2)(x+2) \cdot (x-1)}{(x-1)(x+1) \cdot (x-2)}$ , then cancel (x-2) and (x-1), leaving $\frac{x+2}{x+1}$ . Choice A correctly performs the division and simplifies to $\frac{x+2}{x+1}$ by factoring and canceling common factors after flipping the divisor. Great work! For instance, choice B is just the expression before performing the division, but remember to apply the reciprocal rule first—it's like with numbers, $\frac{1}{2} \div \frac{1}{3} = \frac{1}{2} \times 3 = \frac{3}{2}$ , not leaving it as is. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: $\frac{x^2 - 4}{x} \cdot \frac{x}{x+2}$ looks messy, but factor $ $x^2$ - 4 $ to get $\frac{(x+2)(x-2)}{x} \cdot \frac{x}{x+2}$ , cancel the (x+2) and x, leaving just $ x-2 $. So much cleaner!
Multiply and simplify: $\frac{x^2-1}{x^2-4}\cdot\frac{x+2}{x+1}$
Answer: $\frac{x-1}{x-2}$ This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving $(\frac{a}{b}) \cdot (\frac{c}{d}) = \frac{ac}{bd}$ . But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For $(\frac{x^2-1}{x^2-4} \cdot \frac{x+2}{x+1})$ , factor to $(\frac{(x-1)(x+1)}{(x-2)(x+2)} \cdot \frac{x+2}{x+1})$ , then cancel $(x+1)$ and $(x+2)$ to leave $(\frac{x-1}{x-2})$ . Choice A correctly multiplies and simplifies to $(\frac{x-1}{x-2})$ by factoring and canceling common factors. For example, choice C doesn't factor or cancel fully—always factor differences of squares! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.
Which expression is equivalent to $\frac{x^2 + 5x + 6}{x^2 + 3x + 2} \div \frac{x + 3}{x + 1}$ ?
Answer: $\frac{x + 2}{x + 1}$ First, factor the numerator and denominator: $x^2 + 5x + 6 = (x+2)(x+3)$ and $x^2 + 3x + 2 = (x+1)(x+2)$ . Division by a fraction means multiplication by its reciprocal: $\frac{(x+2)(x+3)}{(x+1)(x+2)} \cdot \frac{x+1}{x+3} = \frac{(x+2)(x+3)(x+1)}{(x+1)(x+2)(x+3)} = \frac{x+2}{x+1}$ after canceling common factors. Choice A fails to cancel all common factors. Choice C inverts the final result. Choice D represents an intermediate step before full simplification.

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