Algebra Quiz: Operating With Rational Expressions

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.

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.

Question 1

1. $\frac{x-2}{x+3}$ (correct answer)
2. $\frac{x-2}{x-3}$
3. $\frac{x^2-5x+6}{(x-3)(x+3)}$
4. $\frac{x-3}{x-2}$

Explanation: 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!

Question 2

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

Explanation: 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.

Question 3

1. $x+4$
2. $\frac{x-4}{x+4}$
3. $x-4$ (correct answer)
4. $\frac{x^2-16}{x+4}$

Explanation: 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, factor numerator and denominator: $x^2 - 16 = (x-4)(x+4)$, so $\frac{(x-4)(x+4)}{x+4}$, then cancel the common factor $(x+4)$, leaving $x-4$ (assuming $x \neq -4$ to avoid division by zero). Choice B correctly simplifies to $x-4$ by factoring and canceling the common factor. Great work! For example, choice C is the original expression without simplifying, but always check for factoring opportunities—it's like reducing 4/6 to 2/3 by canceling 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. Example: $[(x^2-4)/x] \cdot [x/(x+2)]$ looks messy, but factor $x^2-4$ to get $[(x+2)(x-2)/x] \cdot [x/(x+2)]$, cancel the $(x+2)$ and $x$, leaving just $(x-2)$. So much cleaner!

Question 4

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

Explanation: 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.

Question 5

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

Explanation: 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 work with x/(x+3) · (x²-9)/x. First, factor x²-9 = (x+3)(x-3). So we have: x/(x+3) · [(x+3)(x-3)]/x. Now we can see that x appears in both a numerator and denominator, so it cancels. Also, (x+3) appears in both a numerator and denominator, so it cancels too. After canceling: we're left with just (x-3), which equals x-3. Choice B correctly performs the multiplication and simplifies to x-3 by factoring x²-9 as (x+3)(x-3) and canceling the common factors x and (x+3). Great work! Choice A leaves it as a fraction, C gives the wrong factor, and D has the wrong denominator. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. Notice how factoring x²-9 as a difference of squares revealed both factors we needed to cancel—this is why factoring first is so powerful!

Question 6

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

Explanation: This question tests your understanding of how to subtract 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 $\frac{1}{2} + \frac{1}{3}$: 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 $(x-1)$ and $(x+1)$, so LCD is $(x-1)(x+1) = x^2 - 1$; rewrite as $[3(x+1) - 1(x-1)] / (x^2 - 1) = (3x + 3 - x + 1) / (x^2 - 1) = (2x + 4) / (x^2 - 1)$, which can be left as is or factored but doesn't simplify further. Choice B correctly performs the subtraction and simplifies to $(2x+4)/(x^2-1)$ by finding the LCD and combining numerators. Great work! For example, choice A might come from incorrectly adding instead of subtracting or mishandling numerators, but double-check the signs when subtracting—it's like $\frac{3}{4} - \frac{1}{5} = (15-4)/20 = 11/20$, and practice makes it easier! 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 $\frac{1}{6} + \frac{1}{4}$: 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: $(2+3)/x = 5/x$.

Question 7

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

Explanation: 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 first: (x²-9)/(x²-4) · (x+2)/(x-3) = [(x+3)(x-3)]/[(x+2)(x-2)] · (x+2)/(x-3). Now we can cancel: the (x-3) cancels from numerator and denominator, and the (x+2) cancels from numerator and denominator, leaving us with (x+3)/(x-2). Choice A correctly performs the multiplication and simplifies to (x+3)/(x-2) by factoring and canceling common factors. Great work! Choice B shows the unsimplified form without canceling, while choices C and D have incorrect denominators from factoring errors. 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!

Question 8

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

Explanation: 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 3/x + 2/(x+1), the LCD is x(x+1) since these factors share no common factors. We rewrite: 3/x = 3(x+1)/[x(x+1)] = (3x+3)/[x(x+1)] and 2/(x+1) = 2x/[x(x+1)]. Now we can add: (3x+3)/[x(x+1)] + 2x/[x(x+1)] = (3x+3+2x)/[x(x+1)] = (5x+3)/[x(x+1)]. Choice B correctly adds the fractions and simplifies to (5x+3)/[x(x+1)] by finding the LCD and combining numerators. Great work! Choice A incorrectly adds to get just 5 in the numerator, C has an arithmetic error getting 5x+2, and D shows the original unsimplified expression. 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!

Question 9

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

Explanation: 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! For 3/x · x²/(x+2), we multiply numerators to get 3·x² = 3x² and denominators to get x·(x+2) = x(x+2), giving us 3x²/[x(x+2)]. Notice we can cancel one factor of x from both numerator and denominator: 3x²/[x(x+2)] = 3x/(x+2). Choice A correctly shows 3x/(x+2) as the simplified result. Great work! Choice B shows 3x²/(x+2) without canceling the common factor of x—always simplify by canceling common factors. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. 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!

Question 10

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

Explanation: This question tests your understanding of how to subtract 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 x/(x-1) - 2/(x-1), we're lucky—the denominators are already the same! When denominators match, we simply subtract the numerators: (x-2)/(x-1). No need to find an LCD or rewrite anything. Choice A correctly subtracts the fractions to get (x-2)/(x-1) by subtracting numerators while keeping the common denominator. Great work! Choice B incorrectly adds instead of subtracts, choice C reverses the subtraction order giving (2-x)/(x-1), and choice D incorrectly cancels the denominator. 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.

Question 11

1. $\frac{7}{x}$
2. $\frac{7}{x-3}$ (correct answer)
3. $\frac{7}{x-6}$
4. $\frac{10}{x-3}$

Explanation: 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.

Question 12

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

Explanation: 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! For (x+3)/x · x/(x-2), we multiply numerators to get x(x+3) and denominators to get x(x-2), giving us x(x+3)/[x(x-2)]. Notice we can cancel the common factor x from both numerator and denominator, leaving (x+3)/(x-2). Choice C correctly shows this simplified result of (x+3)/(x-2). Great work! Choice A shows the unsimplified form with x still in the denominator—remember to always cancel 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.

Question 13

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

Explanation: 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!

Question 14

1. $\frac{x+3}{x-2}$ (correct answer)
2. $\frac{x^2-9}{x^2-4}$
3. $\frac{x^2-9}{(x-2)(x-3)}$
4. $\frac{x+3}{x+2}$

Explanation: 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: (x²-9)/(x²-4) = [(x+3)(x-3)]/[(x+2)(x-2)] and we're multiplying by (x+2)/(x-3). Now we can see that (x-3) appears in both a numerator and denominator, as does (x+2), so they cancel: [(x+3)(x-3)]/[(x+2)(x-2)] · (x+2)/(x-3) = (x+3)/(x-2). Choice A correctly performs the multiplication and simplifies to (x+3)/(x-2) by factoring and canceling the common factors (x-3) and (x+2). Great work! Choice B shows the original first fraction without any multiplication, while C and D have errors in their simplification process. 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. Remember, when multiplying fractions, any factor that appears in both a numerator and denominator (anywhere in the problem) can be canceled—it's like reducing 6/8 · 4/3 by noticing the 3 and 4 can cancel across fractions!

Question 15

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

Explanation: 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)/(x²-1) ÷ (x-2)/(x+1) becomes (x²-4)/(x²-1) · (x+1)/(x-2). Now factor: [(x+2)(x-2)]/[(x+1)(x-1)] · (x+1)/(x-2). We can cancel (x-2) from numerator and denominator, and (x+1) from numerator and denominator, leaving us with (x+2)/(x-1). Choice A correctly divides by multiplying by the reciprocal and simplifies to (x+2)/(x-1) by factoring and canceling. Great work! Choice B has the numerator and denominator flipped, C doesn't simplify at all, and D shows the original problem without performing the division. For adding/subtracting: this tip doesn't apply here, but remember division is just multiplication by the reciprocal! Common mistake: forgetting to flip the second fraction when dividing—always remember 'keep, change, flip' means keep the first fraction, change ÷ to ×, and flip the second fraction upside down.

Question 16

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

Explanation: 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: $(2+3)/x = 5/x$.

Question 17

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

Explanation: 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! To multiply, combine: $\frac{3 \cdot x^2}{x \cdot (x+2)} = \frac{3x^2}{x^2 + 2x}$, then cancel a common x from numerator and denominator (factoring x from denominator): $\frac{3x}{x + 2}$. Choice A correctly performs the multiplication and simplifies to $\frac{3x}{x+2}$ by canceling the common x factor. Great work! For example, choice B might be the product before simplifying, but don't forget to cancel common factors afterward—it's like $\frac{2}{3} \cdot \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$, not stopping at $\frac{6}{12}$. 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!

Question 18

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

Explanation: 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 first: x²-9 = (x+3)(x-3) and x²-4 = (x+2)(x-2). So we have [(x+3)(x-3)/(x+2)(x-2)]·[(x+2)/(x-3)]. Now we can cancel: (x-3) cancels from numerator and denominator, and (x+2) cancels from numerator and denominator, leaving us with (x+3)/(x-2). Choice A correctly performs the multiplication and simplifies to (x+3)/(x-2) by factoring and canceling common factors. Great work! Choice B shows the unsimplified product without canceling any common factors—remember to always simplify by canceling! 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.

Question 19

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

Explanation: 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 (x²-4)/(x+1) ÷ (x+2)/(x+1), we flip the second fraction to get (x+1)/(x+2) and multiply: (x²-4)/(x+1) · (x+1)/(x+2). First, factor x²-4 = (x+2)(x-2), giving us [(x+2)(x-2)/(x+1)] · [(x+1)/(x+2)]. Now cancel: (x+1) cancels with (x+1), and (x+2) cancels with (x+2), leaving just (x-2). Choice A correctly shows x-2 as the simplified result. Great work! Choice C shows the original numerator over the second fraction's numerator—remember to flip before multiplying when dividing. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. 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!

Question 20

1. $2x + 4$ (correct answer)
2. $2x - 1$
3. $x + 1$
4. $2x + 1$

Explanation: 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)