Practice multiplying and dividing rational expressions with 24 problems, four worked examples, and a printable answer key. Each solution includes the variable restrictions.
Published September 9, 2026 · Last updated September 9, 2026 · Aligned to CCSS HSA.APR.D.7 and HSA.SSE.A.2
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Part of a nine guide unit. See all the rational expressions and rational functions worksheets, listed in a recommended order.
24 problems in four parts, with the complete answer key on the back pages. Every answer shows the simplified expression and the full restriction list, not just the final fraction. Print single sided if you want students to work without the key in view.
Both files are free for classroom and home use. If you post them, please link back to this page rather than rehosting the PDF, so students always land on the current version.
A rational expression is a fraction whose numerator and denominator are polynomials. Multiplying two of them follows the same rule you learned for numerical fractions: multiply the numerators, multiply the denominators. Dividing follows the same rule you learned for numerical fractions too: flip the second fraction and multiply.
So the arithmetic is not the hard part. What makes this an Algebra 2 topic is that the numerators and denominators factor, that the factors cancel, and that every denominator you started with places a condition on x that survives the cancelling. A student who multiplies correctly but reports no restrictions has answered only part of the question.
AB · CD = ACBD AB ÷ CD = AB · DC = ADBC
In the division rule, B, C and D all have to be nonzero. That third condition, C ≠ 0, is easy to miss, because C is written as a numerator and numerators are usually allowed to be zero.
The routine I teach has five steps and never changes. Factor first, restrict second, cancel third. Doing it in that order is what keeps the restrictions from disappearing along with the factors.
Factor every numerator and every denominator completely. Pull out the greatest common factor first, then look for a difference of squares, then a trinomial pattern, then grouping.
Write down the restrictions before you cancel anything. Set every denominator equal to zero and solve. In a division problem, also set the numerator of the divisor equal to zero.
Cancel factors that appear in both a numerator and a denominator. Factors only. A factor is something being multiplied by everything else in that expression.
Multiply what is left across. Leave the result in factored form unless the problem asks otherwise, since factored form is easier to check.
Report the answer together with the restriction list from step 2. The list does not shrink when factors cancel.
Simplify and state the restrictions: 4x + 12x2 − 4 · x2 − 5x + 66x + 18
Step 1, factor. The first numerator has a common factor of 4. The first denominator is a difference of squares. The second numerator is a trinomial with roots 2 and 3. The second denominator has a common factor of 6.
4(x + 3)(x − 2)(x + 2) · (x − 2)(x − 3)6(x + 3)
Step 2, restrictions. The denominators are (x − 2)(x + 2) and 6(x + 3). Setting each factor to zero gives x ≠ 2, x ≠ −2, and x ≠ −3. This is a multiplication problem, so there is no divisor numerator to check.
Step 3, cancel. The factor (x + 3) appears in the first numerator and the second denominator. The factor (x − 2) appears in the second numerator and the first denominator. Both cancel. The constants reduce, since 4/6 = 2/3.
Steps 4 and 5, multiply and report.
2(x − 3)3(x + 2), where x ≠ 2, x ≠ −2, x ≠ −3
The final expression only has (x + 2) in its denominator, so looking at the answer alone you would guess the only restriction is x ≠ −2. But x = 2 and x = −3 both made an original denominator zero, which means the original expression was never defined there. Cancelling a factor does not repair a hole in the domain, it only hides it. This is a common place to lose credit, and it is why step 2 comes before step 3.
Division adds exactly one wrinkle. Before you flip, the second fraction's numerator is sitting in the position of a divisor, and you cannot divide by zero. So that numerator generates a restriction even though it is a numerator.
Simplify and state the restrictions: x2 + x − 6x2 − 9 ÷ x2 − 4x2 + 6x + 9
Step 1, factor everything, including the divisor.
(x + 3)(x − 2)(x − 3)(x + 3) ÷ (x − 2)(x + 2)(x + 3)2
Step 2, restrictions, from three places.
Nothing in the original problem shows (x − 2) or (x + 2) underneath a fraction bar. Both come from the divisor, so record them from the original expression, before flipping or cancelling.
Step 3, flip and cancel.
(x + 3)(x − 2)(x − 3)(x + 3) · (x + 3)2(x − 2)(x + 2)
One (x + 3) cancels from the left fraction, and (x − 2) cancels across. What remains is (x + 3)2 on top and (x − 3)(x + 2) underneath.
(x + 3)2(x − 3)(x + 2), where x ≠ 3, x ≠ −3, x ≠ 2, x ≠ −2
Four restrictions, and only two of them are visible in the answer.
Write the restrictions in a table before cancelling. There are only four places a restriction can come from, and two of them depend on whether the problem is a multiplication or a division.
| Where a restriction comes from | In a multiplication problem | In a division problem |
|---|---|---|
| Denominator of the first fraction | Yes, always | Yes, always |
| Denominator of the second fraction | Yes, always | Yes, always, even though it moves to the top when you flip |
| Numerator of the second fraction | No | Yes, because it is the divisor |
| Numerator of the first fraction | No | No |
2x2 − 8x2 + 5x + 6 · x2 − x − 64x − 8
Factored, this is
2(x − 2)(x + 2)(x + 2)(x + 3) · (x − 3)(x + 2)4(x − 2)
Ledger, filled in before cancelling: the denominators are (x + 2)(x + 3) and 4(x − 2), so x ≠ −2, x ≠ −3, x ≠ 2. Now cancel one (x + 2) and the (x − 2), and reduce 2/4 to 1/2.
(x − 3)(x + 2)2(x + 3), where x ≠ −2, x ≠ −3, x ≠ 2
Here x = 2 is a restriction with nothing left to point at. The factor (x − 2) is gone from both the numerator and the denominator of the answer. Graphically it is a hole in the curve at x = 2, which shows up in the answer only if the restriction was recorded before the factor cancelled. There is also a hole at x = −2, since the (x + 2) that remains on top is a different copy from the one that cancelled underneath.
The expressions (3 − x) and (x − 3) are not the same, so they do not cancel to 1. They are negatives of each other, so the quotient is −1.
3 − xx − 3 = −1(x − 3)x − 3 = −1, where x ≠ 3
The move is always the same. Factor −1 out of the binomial that is written backwards, which flips both signs inside, and then the factors match.
25 − x2x2 − 2x − 15 · x + 3x
The first numerator is a difference of squares written in the reverse order, so 25 − x2 = (5 − x)(5 + x). The first denominator factors as (x − 5)(x + 3). The restrictions come from those two denominators: x ≠ 5, x ≠ −3, x ≠ 0.
Now rewrite (5 − x) as −1(x − 5):
−1(x − 5)(x + 5)(x − 5)(x + 3) · x + 3x = −x + 5x, where x ≠ 5, x ≠ −3, x ≠ 0
Skipping the −1 step gives the answer the wrong sign. A quick check is to substitute a legal number into both the original and the answer. At x = 1 the original is (24/−16)(4/1) = −6 and the answer is −(6/1) = −6. They agree, so the sign is right.
This tool builds a fresh multiplication or division problem every time you press the button, then walks the solution one step at a time so you can try the step yourself before you look. Every problem it makes factors cleanly over the integers, and the restriction list it produces is complete, including the restrictions that disappear when factors cancel.
Problems are generated in your browser, so nothing is sent anywhere and you can run it offline once the page has loaded.
These are the six errors I mark most often on this unit, in rough order of how much credit they cost.
A student sees (x + 3)/3 and crosses out both threes. If that happens in the first line, nothing after it can be correct. A factor is multiplied by everything else in the expression, a term is added to it. In (x + 3) the 3 is added, so it is welded to the x and cannot leave on its own. My test for this is to read the expression out loud: if I have to say the word "plus" between the piece I want to cancel and its neighbor, I cannot cancel it.
This one is easy to miss, because the algebra looks finished. Restrictions written at the end describe a different expression than the one the question asked about. Record them from the original expression, before flipping or cancelling.
In a division problem, the numerator of the second fraction generates restrictions too. Students accept this quickly when I write the arithmetic version next to it: 6 ÷ 0 is undefined, and the zero there was a numerator before we flipped it. Once the fraction is flipped, that numerator sits in a denominator and looks obvious, but by then the cancelling has usually started. Record it from the original expression instead.
When (4 − x) and (x − 4) are in the same problem, students either cancel them to 1 or leave them alone. The correct answer is −1. I tell students to circle any binomial where the variable is written second, because in a factored expression that is nearly always a sign that a −1 needs to come out.
Flipping first is legal, but it makes the restriction from the divisor harder to see, and it tempts students to flip the first fraction by mistake. Once everything is in factors, the flip is a visual move rather than an algebraic one.
A student cancels the obvious pair, writes the answer, and never notices that the denominator still factors. The check is the same one I use in the polynomial unit: pick a value of x that is not on the restriction list, then evaluate both the original expression and your answer. If the two numbers do not match, something in the middle went wrong. Usually x = 1 or x = 10 works, whichever is legal and easier to compute.
This is the same content as the printable cheat sheet, laid out here so it is readable on a phone and so you can check a rule without opening a PDF. The download button repeats at the bottom of this section.
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AB · CD = ACBD
AB ÷ CD = AB · DC
Flip the second fraction only. Never the first, never both.
| Source | Multiply | Divide |
|---|---|---|
| Denominator of the first fraction | Yes | Yes |
| Denominator of the second fraction | Yes | Yes |
| Numerator of the second fraction | No | Yes, it is the divisor |
| Numerator of the first fraction | No | No |
Restrictions never disappear when factors cancel. A cancelled factor leaves a hole in the graph, not a clean domain. Write the list from the original expression, before flipping or cancelling.
a − xx − a = −1, x ≠ a
Factor −1 out of the binomial written backwards, then the factors match. Circle any binomial where the variable comes second.
Legal: a factor, which is multiplied by everything else. (x + 3)(x − 1) / (x + 3) = x − 1
Not legal: a term, which is added. (x + 3)/3 does not become x, and (x2 + 4)/x2 does not become 4.
Read it out loud. If you say the word "plus" between the piece and its neighbor, it stays.
Pick a value of x that is not on your restriction list, usually x = 1 or x = 10. Evaluate the original expression and your answer. If the two numbers disagree, the error is between them, and you can find it before you hand in the page.
Download the cheat sheet (PDF, one page)
These are the same 24 problems that appear in the printable worksheet, in the same order, so a student can work on paper and check here, or work here and print the version with room to show work. Simplify each expression completely and state every restriction on the variable. Answers for each part are behind the button under that part.
The worksheet targets two high school standards. HSA.APR.D.7 asks students to understand that rational expressions form a system analogous to the rational numbers, closed under multiplication and under division by a nonzero rational expression. HSA.SSE.A.2 covers the structural seeing that lets a student recognize x2 − 4 as a difference of squares in the first place. Part D also touches the difference of cubes and factoring by grouping, which are usually taught earlier in the polynomials unit.
If you want a second explanation in a different voice, these are the outside sources I point students toward:
This page sits inside a longer sequence. Simplifying a single rational expression comes first, and adding and subtracting them comes after, since those need a common denominator and this topic does not.
We can work through a few problems together to see which steps need practice. Students who struggle here are often stuck on factoring rather than on the fractions themselves, and the two need different practice.
I tutor Algebra I, Algebra II, Precalculus and AP science in person in Fremont, Newark and Union City, and online elsewhere. Sessions happen at your table, so there is no drive at 4 pm.
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