This deck covers domain restrictions, simplifying by factoring, multiplying and dividing, the least common denominator, adding and subtracting, complex fractions, and the difference quotient. It targets the biggest error in the course - cancelling added terms instead of factors - along with restrictions lost after a cancellation, the dropped minus sign when subtracting, and a common denominator that is either wrong or needlessly huge.
Subject: College Algebra · 129 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
College Algebra - Deck 03
Fractions, but the pieces are polynomials. Factor first, cancel factors only, and never lose a restriction.
Objectives
By the end of this deck you can:
And one habit above all: you may cancel factors, never terms.
Warm-up
Discussion prompt
Before we open Rational Expressions: without looking back, what was the main idea of Polynomials and Factoring, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers polynomial vocabulary, adding and subtracting polynomials, every way to multiply them, and the special products, then lays out a complete factoring strategy that runs from the GCF through grouping to the cubes. It targets the dropped subtraction sign, the missing middle term when a binomial is squared, trying to factor a sum of squares, and skipping the GCF and then calling what is left unfactorable.
Section
Part 1
Concept
A rational expression is one polynomial divided by another. That is the entire definition.
\[ \frac{P}{Q}, \qquad Q \ne 0 \]
rational expression — A quotient of two polynomials. The word rational comes from ratio: it is a ratio of polynomials, exactly the way a rational number is a ratio of integers.
\[ \frac{3x^{2} - 5x + 1}{x - 4} \]
Counterexample
Discussion prompt
A rational expression is one polynomial divided by another. That is the entire definition.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Picture it
Animation
Shows: Cancel factors, never terms — a rendered Manim animation.
Rendered with Manim.
Takeaway: Cancelling across a plus sign is the single commonest error here.
Concept
Top and bottom must both be polynomials: whole-number exponents, no variable trapped under a radical, no variable stuck in a nested denominator.
\[ \text{rational:}\quad \frac{x+7}{x^{2}-9}, \qquad \frac{5}{x}, \qquad \frac{x^{3}-1}{2} \]
The last one still counts. A polynomial over a constant is a perfectly legal rational expression - the denominator is just a very boring polynomial.
\[ \text{not rational:}\quad \frac{\sqrt{x}+1}{x}, \qquad \frac{x^{-2}+3}{x+1} \]
Analogy
Discussion prompt
Explain Which expressions qualify by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Top and bottom must both be polynomials: whole-number exponents, no variable trapped under a radical, no variable stuck in a nested denominator.
Intuition
Nothing in this deck is a new idea. It is grade-school fraction arithmetic with bigger pieces.
Reduce, multiply straight across, flip to divide, find a common denominator. Same four moves. The only change is that the numbers have been replaced by polynomials.
| With numbers | With polynomials |
|---|---|
| Reduce by dividing out a common factor | Divide out a common polynomial factor |
| Multiply straight across | Multiply straight across |
| Flip the divisor and multiply | Flip the divisor and multiply |
| Find the least common denominator | Find the least common denominator |
There is exactly one genuinely new thing: with numbers you never had to worry about the denominator secretly being zero. Now you do.
Comparison
Comparison matrix
From You already know how to do this: refill the With polynomials column from what you know. The rest of the table is as it appeared.
| With numbers | With polynomials |
|---|---|
| Reduce by dividing out a common factor | Divide out a common polynomial factor |
| Multiply straight across | Multiply straight across |
| Flip the divisor and multiply | Flip the divisor and multiply |
| Find the least common denominator | Find the least common denominator |
Concept
Division asks a question: what number times the bottom gives the top?
\[ \frac{6}{3} = 2 \quad \text{because} \quad 2 \cdot 3 = 6 \]
Now ask that question with a zero on the bottom. What times zero gives six? Nothing does. What times zero gives zero? Everything does.
One question has no answer, the other has infinitely many. Neither is usable, so division by zero is left undefined - not zero, not infinity, just not a number.
Explain it
Discussion prompt
Explain Why a zero denominator is fatal to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Division asks a question: what number times the bottom gives the top?
Picture it
Figure (svg): A number line with open circles punched out at 2 and 3
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
So a rational expression is a machine that happily accepts almost every input - except the handful that zero out a denominator. Those inputs are removed from the domain entirely.
Intuition
So a rational expression is a machine that happily accepts almost every input - except the handful that zero out a denominator. Those inputs are removed from the domain entirely.
\[ \frac{x+1}{(x-2)(x-3)} \]
Figure (svg): A number line with open circles punched out at 2 and 3
domain restriction — A value the variable is not allowed to take because it makes a denominator zero. The expression has no value at all there.
Definition probe
Sort into buckets
Every line below is part of the definition of rational expression or of domain restriction — one or the other, never both. Put each where it belongs.
Ranking
Put in order
Put the moves of Worked example: find the restrictions into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The numerator is allowed to be anything at all, including zero.
Worked example
Find every value the variable is not allowed to take.
\[ \frac{x+3}{x^{2}-5x+6} \]
Look only at the denominator
Why: The numerator is allowed to be anything at all, including zero. Only the bottom can break the expression.
Factor the denominator completely
Why: A product equals zero exactly when one of its factors equals zero, so factoring turns one hard question into two easy ones.
\[ x^{2}-5x+6 = (x-2)(x-3) \]
Set each factor equal to zero and solve
Why: These are precisely the inputs that would make the whole denominator vanish.
\[ x-2 = 0 \;\Rightarrow\; x = 2 \qquad x-3 = 0 \;\Rightarrow\; x = 3 \]
State the answer the way a test asks for it: the excluded values.
\[ x \ne 2, \qquad x \ne 3 \]
Verify by substituting each excluded value into the original denominator
Why: At 2 the denominator is 4 minus 10 plus 6, which is zero. At 3 it is 9 minus 15 plus 6, which is also zero. Both really do break it, and the factoring shows no other value can.
\[ 2^{2}-5(2)+6 = 0 \qquad 3^{2}-5(3)+6 = 0 \]
Picture it
Animation
Shows: Each line of the worked example "find the restrictions", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At 2 the denominator is 4 minus 10 plus 6, which is zero. At 3 it is 9 minus 15 plus 6, which is also zero. Both really do break it, and the factoring shows no other value can.
Concept
Here is the rule that saves you a whole letter grade: find the restrictions first, from the expression exactly as it was handed to you, before you simplify anything.
Cancelling destroys the evidence. Once a factor is gone from the page you can no longer see that it used to be a denominator.
The value is still illegal. The original expression was undefined there, so any form you rewrite it into has to be undefined there too.
Step zero
Discussion prompt
Worked example: a restriction that is about to cancel — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Factor the denominator
Answer:
Worked example
State the domain restrictions.
\[ \frac{x^{2}-9}{x^{2}+x-6} \]
Factor the denominator
Why: Two numbers that multiply to negative six and add to one are three and negative two.
\[ x^{2}+x-6 = (x+3)(x-2) \]
Set each factor equal to zero
Why: Either factor being zero kills the whole product, and a zero denominator is undefined.
\[ x = -3 \quad \text{or} \quad x = 2 \]
Now notice something uncomfortable: the numerator contains that very same first factor.
\[ x^{2}-9 = (x-3)(x+3) \]
Restrict both values anyway
Why: The restriction is a fact about the original denominator, not about what survives the cancelling. Both values stay excluded even though one factor is about to disappear.
\[ x \ne -3, \qquad x \ne 2 \]
Verify by substituting both values into the original
Why: At negative three the expression reads zero over zero, which is undefined - not zero. At two it reads negative five over zero, also undefined. Both genuinely break the original.
\[ x=-3:\ \frac{0}{0} \qquad x=2:\ \frac{-5}{0} \]
Picture it
Animation
Shows: Each line of the worked example "a restriction that is about to cancel", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At negative three the expression reads zero over zero, which is undefined - not zero. At two it reads negative five over zero, also undefined. Both genuinely break the original.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Simplify first, then read the domain off whatever is left.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The simplified form is a plain polynomial with no denominator at all, so it looks completely safe.
Read the restriction off the original denominator first, then simplify.
Why: The simplified form is a plain polynomial with no denominator at all, so it looks completely safe.
Trap
Simplify first, then read the domain off whatever is left.
\[ \frac{x^{2}-9}{x-3} \]
Cancel, then report that every input is allowed
Why: The simplified form is a plain polynomial with no denominator at all, so it looks completely safe.
\[ \frac{(x-3)(x+3)}{x-3} = x+3 \quad \text{(all real numbers?)} \]
Test the value three in both forms
Why: The simplified form returns six. The original returns zero over zero, which is undefined. The two forms disagree at exactly one point, so the claim about the domain is false.
\[ \text{original: } \frac{0}{0} \quad \text{vs} \quad x+3 = 6 \]
Read the restriction off the original denominator first, then simplify.
\[ \frac{x^{2}-9}{x-3} \]
Restriction first
Why: The denominator is zero when the variable is three, so three is thrown out before any cancelling can hide it.
\[ x \ne 3 \]
Now simplify and carry the restriction along
Why: The two forms agree at every input where both are defined, and the excluded value rides along with the answer. That single excluded point is called a hole.
\[ \frac{x^{2}-9}{x-3} = x+3, \qquad x \ne 3 \]
Notation
Annotate
From Trap: the hole does not heal — read this one piece at a time. What is each part doing?
On: \( \frac{x^{2}-9}{x-3} \)
Pattern
Every single time, in this order:
Do this before you touch anything else. It takes twenty seconds and it is worth points on every rational-expression problem you will ever see.
Picture it
Animation
Shows: The cancelled factor leaves a hole — a rendered Manim animation.
Rendered with Manim.
Takeaway: The expressions are equal everywhere except the point you cancelled away.
Elimination
Eliminate the wrong options
Which values must be excluded from the domain of this expression?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The denominator factors as (x - 4)(x + 3), so it is zero at 4 and at -3. Both values are excluded. The expression does simplify to 1 over (x + 3), but 4 was already illegal in the original and stays illegal.
Check
Factor the denominator on paper before you choose.
\[ \frac{x-4}{x^{2}-x-12} \]
Check your understanding
Which values must be excluded from the domain of this expression?
Answer: A
Why: The denominator factors as (x - 4)(x + 3), so it is zero at 4 and at -3. Both values are excluded. The expression does simplify to 1 over (x + 3), but 4 was already illegal in the original and stays illegal.
Section
Part 2
Concept
To simplify a rational expression means to divide out a factor that the top and the bottom share.
\[ \frac{a \cdot c}{b \cdot c} = \frac{a}{b} \cdot \frac{c}{c} = \frac{a}{b} \cdot 1 = \frac{a}{b} \]
That middle step is the whole justification. A shared factor makes a copy of one, and multiplying by one changes nothing. Nothing is being deleted - it is being separated out and recognized as one.
lowest terms — A rational expression whose numerator and denominator share no common factor other than one. That is the target form for every answer in this deck.
Intuition
Reduce six ninths and you get two thirds. Ask yourself what you actually did there.
\[ \frac{6}{9} = \frac{2 \cdot 3}{3 \cdot 3} = \frac{2}{3} \]
You rewrote both numbers as products and removed a matching three. You did not remove a digit you happened to see in both places.
Nobody reduces sixteen sixty-fourths by crossing out the sixes. That is a joke precisely because it is not a legal move - and it is exactly the move students make with letters.
Concept
factor — Something being multiplied. In a product, each piece is a factor of the whole thing.
term — Something being added or subtracted. In a sum, each piece is a term - and a term is only part of the numerator, never a factor of all of it.
| Expression | Its pieces are | May you cancel one? |
|---|---|---|
| 3(x + 2) | factors 3 and (x + 2) | yes |
| 3x + 2 | terms 3x and 2 | no |
| (x - 1)(x + 4) | factors (x - 1) and (x + 4) | yes |
| x squared plus 9 | terms x squared and 9 | no |
This is why the very first move in every problem in this deck is factor everything completely. Until it is a product, nothing may be cancelled.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of rational expression, domain restriction, lowest terms, factor, term as Rational Expressions uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Picture it
Animation
Shows: A hole versus an asymptote — a rendered Manim animation.
Rendered with Manim.
Takeaway: A surviving zero downstairs is a wall; a cancelled one is just a missing point.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The number on top and the number on the bottom look like a matching pair.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: It feels like the same move that turns six ninths into two thirds - a matching number on top and bottom.
Ask what the numerator really is before touching anything.
Why: It feels like the same move that turns six ninths into two thirds - a matching number on top and bottom.
Trap
The number on top and the number on the bottom look like a matching pair.
\[ \frac{x+5}{5} \]
Cross out the two fives and report the leftover
Why: It feels like the same move that turns six ninths into two thirds - a matching number on top and bottom.
\[ \frac{x+5}{5} \;\longrightarrow\; x \]
Test the answer at five
Why: The original gives ten over five, which is two. The claimed answer gives five. Two and five are different numbers, so the cancellation destroyed the expression.
\[ \frac{5+5}{5} = 2 \qquad \text{but} \qquad x = 5 \]
Ask what the numerator really is before touching anything.
\[ \frac{x+5}{5} \]
The five on top is a term, not a factor of the whole numerator
Why: The numerator is a sum. Nothing multiplies the entire top, so there is no common factor to divide out. This expression is already in lowest terms.
\[ \frac{x+5}{5} = \frac{x}{5} + 1 \]
Check what a real common factor looks like
Why: Here five multiplies the entire numerator, so it is a genuine factor and may be divided out. Test at five: thirty over five is six, and x plus one is six. They agree, which is what a legal cancellation always does.
\[ \frac{5x+5}{5} = \frac{5(x+1)}{5} = x+1 \]
Break the constraint
Discussion prompt
The rule this trap just fixed:
The numerator is a sum. Nothing multiplies the entire top, so there is no common factor to divide out. This expression is already in lowest terms.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Answer:
It feels like the same move that turns six ninths into two thirds - a matching number on top and bottom.
Estimation
Predict first
Simplify completely and state the restrictions.
Commit before you compute: what does Worked example: simplify by factoring both parts come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by substituting a legal test value
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Use zero, which breaks no denominator.
Worked example
Simplify completely and state the restrictions.
\[ \frac{x^{2}-9}{x^{2}+x-6} \]
Factor the numerator as a difference of squares
Why: Both pieces are perfect squares separated by a minus sign, so the difference-of-squares pattern applies directly.
\[ x^{2}-9 = (x-3)(x+3) \]
Factor the denominator
Why: Two numbers multiplying to negative six and adding to one are three and negative two.
\[ x^{2}+x-6 = (x+3)(x-2) \]
State the restrictions from the factored denominator now
Why: Do this before cancelling, while the evidence is still on the page.
\[ x \ne -3, \qquad x \ne 2 \]
Divide out the common factor
Why: The binomial x plus three multiplies the entire top and the entire bottom, so it is a true common factor and makes a copy of one.
\[ \frac{(x-3)(x+3)}{(x+3)(x-2)} = \frac{x-3}{x-2} \]
Verify by substituting a legal test value
Why: Use zero, which breaks no denominator. The original gives negative nine over negative six, which is three halves. The answer gives negative three over negative two, which is also three halves. They agree.
\[ x=0:\quad \frac{-9}{-6} = \frac{3}{2} \qquad \frac{-3}{-2} = \frac{3}{2} \]
Picture it
Animation
Shows: Each line of the worked example "simplify by factoring both parts", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use zero, which breaks no denominator. The original gives negative nine over negative six, which is three halves. The answer gives negative three over negative two, which is also three halves. They agree.
Fill the middle
Fill in the blanks
From Worked example: pull the greatest common factor first — finish the line. Write what belongs on the right of the equals sign before you look.
2x^2x(x+5)+10x = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Both terms on top share a factor of two and a factor of the variable.
Worked example
Simplify completely and state the restrictions.
\[ \frac{2x^{2}+10x}{x^{2}+7x+10} \]
Factor the greatest common factor out of the numerator
Why: Both terms on top share a factor of two and a factor of the variable. Always take the greatest common factor before looking for anything fancier.
\[ 2x^{2}+10x = 2x(x+5) \]
Factor the trinomial on the bottom
Why: Two numbers multiplying to ten and adding to seven are five and two.
\[ x^{2}+7x+10 = (x+5)(x+2) \]
Record the restrictions
Why: The denominator is zero at negative five and at negative two, so both are excluded from the start.
\[ x \ne -5, \qquad x \ne -2 \]
Divide out the shared binomial
Why: The factor x plus five multiplies all of the top and all of the bottom. The two on top does not cancel with anything - there is no two on the bottom.
\[ \frac{2x(x+5)}{(x+5)(x+2)} = \frac{2x}{x+2} \]
Verify at a legal test value
Why: Use one. The original gives twelve over eighteen, which reduces to two thirds. The answer gives two over three. They agree, so the simplification is sound.
\[ x=1:\quad \frac{12}{18} = \frac{2}{3} \qquad \frac{2}{3} \]
Picture it
Animation
Shows: Each line of the worked example "pull the greatest common factor first", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use one. The original gives twelve over eighteen, which reduces to two thirds. The answer gives two over three. They agree, so the simplification is sound.
Concept
Sometimes the top and bottom contain binomials that are not equal but are opposites - each is the other with every sign flipped.
\[ 3-x \quad \text{and} \quad x-3 \]
They are not the same factor, so you may not simply cancel them. But factoring a negative one out of either turns it into the other.
\[ 3-x = -(x-3) \qquad \Longrightarrow \qquad \frac{3-x}{x-3} = -1 \]
So a pair of opposite binomials cancels to negative one. Pull the negative out where you can see it rather than trying to cancel in your head.
Estimation
Predict first
Simplify completely and state the restrictions.
Commit before you compute: what does Worked example: opposites in the numerator come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify at a legal test value
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Use zero. The original gives three over negative nine, which is negative one third.
Worked example
Simplify completely and state the restrictions.
\[ \frac{3-x}{x^{2}-9} \]
Factor the denominator as a difference of squares
Why: Nine is a perfect square, so the pattern applies. Now the restrictions are visible.
\[ x^{2}-9 = (x-3)(x+3), \qquad x \ne 3, \; x \ne -3 \]
Factor negative one out of the numerator
Why: The numerator is the opposite of the factor x minus three. Pulling out negative one rewrites it so the shared factor is literally identical, which is the only situation in which cancelling is legal.
\[ 3-x = -1(x-3) \]
Divide out the matching binomial
Why: Now x minus three appears as a factor of both the top and the bottom, and the negative one stays behind as a factor of the answer.
\[ \frac{-1(x-3)}{(x-3)(x+3)} = \frac{-1}{x+3} \]
Verify at a legal test value
Why: Use zero. The original gives three over negative nine, which is negative one third. The answer gives negative one over three, also negative one third. They agree.
\[ x=0:\quad \frac{3}{-9} = -\frac{1}{3} \qquad \frac{-1}{3} = -\frac{1}{3} \]
Pattern
The recipe never changes:
If either part will not factor, you are done - the expression was already in lowest terms.
Picture it
Animation
Shows: Multiplying: factor, then cancel — a rendered Manim animation.
Rendered with Manim.
Takeaway: Expanding first creates work you then have to undo.
Check
Only one of these four is a legal simplification. Ask of each: is the thing being cancelled a factor of the entire numerator?
Check your understanding
Which of these simplifications is correct?
Answer: A
Why: The numerator 3x + 6 factors as 3(x + 2), so 3 really is a factor of the entire top and divides out with the 3 on the bottom, leaving x + 2. Test at x = 1: (3 + 6) over 3 is 3, and 1 + 2 is 3.
Section
Part 3
Concept
Multiplication is the friendliest operation here. Numerator times numerator, denominator times denominator. No common denominator is needed and none is wanted.
\[ \frac{A}{B} \cdot \frac{C}{D} = \frac{AC}{BD} \]
There is one word of warning attached to that rule: do not actually multiply it out yet.
If you expand first you get a big polynomial that you then have to factor all over again just to reduce it. Factor first, cancel, and multiply only what survives.
Intuition
Watch this happen with plain numbers and the point is obvious.
\[ \frac{2}{3} \cdot \frac{9}{4} = \frac{18}{12} = \frac{3}{2} \]
That is the expand-then-reduce route: you made an eighteen and a twelve and then had to find that they share a six.
\[ \frac{2}{3} \cdot \frac{9}{4} = \frac{2 \cdot 3 \cdot 3}{3 \cdot 2 \cdot 2} = \frac{3}{2} \]
Same answer, half the arithmetic. With polynomials the saving is not half - it is the difference between a two-line problem and a page of expanding.
Step zero
Discussion prompt
Worked example: multiplying two rational expressions — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Factor every numerator and every denominator
Answer:
Worked example
Multiply and simplify. State the restrictions.
\[ \frac{x^{2}-4}{x^{2}+6x+9} \cdot \frac{x+3}{x-2} \]
Factor every numerator and every denominator
Why: Nothing may be cancelled until each piece is written as a product. The first numerator is a difference of squares; the first denominator is a perfect-square trinomial.
\[ x^{2}-4 = (x-2)(x+2), \qquad x^{2}+6x+9 = (x+3)^{2} \]
Record the restrictions from the original denominators
Why: The first denominator is zero when the variable is negative three; the second is zero when it is two. Write them down now, before anything cancels.
\[ x \ne -3, \qquad x \ne 2 \]
Write the product as one fraction, still factored
Why: Multiplying across is allowed, but leave every factor visible so you can see what matches.
\[ \frac{(x-2)(x+2)(x+3)}{(x+3)^{2}(x-2)} \]
Divide out the factors that appear on both sides of the bar
Why: The binomial x minus two appears once on top and once on the bottom. The binomial x plus three appears once on top and twice on the bottom, so one copy survives below.
\[ = \frac{x+2}{x+3}, \qquad x \ne -3, \; x \ne 2 \]
Verify at a legal test value
Why: Use zero, which breaks no denominator. The original is negative four ninths times negative three halves, which is two thirds. The answer is two over three. They agree.
\[ x=0:\quad \frac{-4}{9} \cdot \frac{3}{-2} = \frac{12}{18} = \frac{2}{3} \qquad \frac{0+2}{0+3} = \frac{2}{3} \]
Concept
There is no separate division procedure. You convert division into multiplication and then use the rule you already know.
\[ \frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} \]
Flip the divisor - the fraction after the division sign - and change the sign to multiplication. The first fraction never moves.
Do the flip before you cancel anything. Cancelling across a division sign is not a legal move; cancelling across a multiplication sign is.
Picture it
Animation
Shows: Dividing is multiplying by the reciprocal — a rendered Manim animation.
Rendered with Manim.
Takeaway: Flip the second fraction, then proceed exactly as for multiplication.
Intuition
Ask what dividing by a half means. How many halves fit inside three? Six of them. Dividing by a half doubled the number.
\[ 3 \div \frac{1}{2} = 3 \cdot 2 = 6 \]
Dividing by a small piece gives a big count. That is why the fraction turns upside down: the reciprocal is exactly the number that undoes it.
\[ \frac{C}{D} \cdot \frac{D}{C} = 1 \]
So multiplying by the reciprocal cancels the divisor out. Nothing magical happens - it is the same fact that lets you undo a multiplication by dividing.
Hypothesis
Predict first
Worked example: dividing two rational expressions is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Factor everything in sight, including the divisor
Why: The first numerator is a difference of squares; the first denominator is a trinomial whose numbers multiply to six and add to five.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Divide and simplify. State the restrictions.
\[ \frac{x^{2}-1}{x^{2}+5x+6} \div \frac{x-1}{x+2} \]
Factor everything in sight, including the divisor
Why: The first numerator is a difference of squares; the first denominator is a trinomial whose numbers multiply to six and add to five.
\[ x^{2}-1 = (x-1)(x+1), \qquad x^{2}+5x+6 = (x+2)(x+3) \]
Record the restrictions, including the sneaky one
Why: The two visible denominators are zero at negative three and negative two. But you also may not divide by zero, so the divisor's numerator cannot vanish either - that rules out one.
\[ x \ne -3, \qquad x \ne -2, \qquad x \ne 1 \]
Flip the divisor and change to multiplication
Why: Only the fraction after the division sign turns over. Do it now, before any cancelling.
\[ \frac{(x-1)(x+1)}{(x+2)(x+3)} \cdot \frac{x+2}{x-1} \]
Divide out common factors, then multiply what is left
Why: The binomials x minus one and x plus two each appear once above and once below the bar, so each makes a copy of one.
\[ = \frac{x+1}{x+3}, \qquad x \ne -3, \; x \ne -2, \; x \ne 1 \]
Verify at a legal test value
Why: Use zero. The original is negative one sixth divided by negative one half, which is one third. The answer is one over three. They agree.
\[ x=0:\quad \frac{-1}{6} \div \frac{-1}{2} = \frac{-1}{6} \cdot \frac{2}{-1} = \frac{1}{3} \qquad \frac{0+1}{0+3} = \frac{1}{3} \]
Picture it
Animation
Shows: Each line of the worked example "dividing two rational expressions", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use zero. The original is negative one sixth divided by negative one half, which is one third. The answer is one over three. They agree.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Division means a fraction turns over, so turn over the one you reach first.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Reading left to right, the first fraction is the one your eye lands on, and the rule is remembered as just flip and multiply without saying which one flips.
Only the divisor turns over - the fraction that comes after the division sign.
Why: Reading left to right, the first fraction is the one your eye lands on, and the rule is remembered as just flip and multiply without saying which one flips.
Trap
Division means a fraction turns over, so turn over the one you reach first.
\[ \frac{x+1}{x+2} \div \frac{x+3}{x+4} \]
Invert the leading fraction and multiply
Why: Reading left to right, the first fraction is the one your eye lands on, and the rule is remembered as just flip and multiply without saying which one flips.
\[ \frac{x+2}{x+1} \cdot \frac{x+3}{x+4} \]
Test the result at zero
Why: The original is one half divided by three quarters, which is two thirds. This route gives two times three fourths, which is three halves. Two thirds and three halves are reciprocals of each other, not equal - the whole expression came out upside down.
\[ \text{original: } \frac{2}{3} \qquad \text{this route: } \frac{2}{1} \cdot \frac{3}{4} = \frac{3}{2} \]
Only the divisor turns over - the fraction that comes after the division sign.
\[ \frac{x+1}{x+2} \div \frac{x+3}{x+4} \]
Invert the second fraction and multiply
Why: The dividend is what you are cutting up; the divisor is what you are cutting it into. Only the thing doing the cutting becomes a reciprocal.
\[ \frac{x+1}{x+2} \cdot \frac{x+4}{x+3} \]
Check at zero
Why: This gives one half times four thirds, which is four sixths, or two thirds - exactly the value of the original expression at zero. The order is confirmed correct.
\[ \frac{1}{2} \cdot \frac{4}{3} = \frac{4}{6} = \frac{2}{3} \]
Translation
\( \frac{x+2}{x+1} \cdot \frac{x+3}{x+4} \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Pattern
One recipe covers both operations:
Notice what is not on the list: finding a common denominator. That is only for adding and subtracting.
Edge cases
Discussion prompt
Pattern: multiplying and dividing works on the cases you have just seen. Push it to the edge: what is the most degenerate input it still handles — empty, zero, one item, everything equal — and what is the first case where it stops being true? Name the case, not just "it breaks".
Hint: Try the smallest legal input, then the largest, then the one where two things collide. Methods are specified at their edges; the middle takes care of itself.
Answer:
Notice what is not on the list: finding a common denominator. That is only for adding and subtracting.
Check
Factor both differences of squares first, then look for matching factors.
\[ \frac{x^{2}-4}{x^{2}-9} \cdot \frac{x+3}{x+2} \]
Check your understanding
Which is the simplified product?
Answer: A
Why: Factored, the product is (x - 2)(x + 2)(x + 3) over (x - 3)(x + 3)(x + 2). The (x + 2) and the (x + 3) each divide out, leaving (x - 2) over (x - 3). Test at x = 0: the original is (-4)/(-9) times 3/2, which is 2/3, and (0 - 2)/(0 - 3) is also 2/3.
Section
Part 4
Concept
When the denominators are already identical, adding is easy: combine the numerators and keep the one denominator underneath.
\[ \frac{A}{C} + \frac{B}{C} = \frac{A+B}{C} \]
The denominator is never added. It is the name of the piece you are counting - three sevenths plus two sevenths is five sevenths, not five fourteenths.
After combining, always factor the new numerator. Adding often creates a factor that then cancels, and an unsimplified answer is usually marked wrong.
Worked example
Combine and simplify. State the restrictions.
\[ \frac{x^{2}}{x+2} - \frac{4}{x+2} \]
Take the restriction off the shared denominator
Why: Both denominators are the same binomial, and it is zero when the variable is negative two.
\[ x \ne -2 \]
Subtract the numerators over the single denominator
Why: The denominators already match, so only the tops combine. Nothing happens to the bottom at all.
\[ \frac{x^{2}-4}{x+2} \]
Factor the new numerator
Why: Combining produced a difference of squares that was not visible before. This is why you never stop at the combined form.
\[ \frac{(x-2)(x+2)}{x+2} \]
Divide out the common factor
Why: The binomial x plus two multiplies the whole top and is the whole bottom, so it makes a copy of one. The restriction rides along with the answer.
\[ = x-2, \qquad x \ne -2 \]
Verify at a legal test value
Why: Use zero. The original is zero over two minus four over two, which is negative two. The answer is zero minus two, also negative two. They agree.
\[ x=0:\quad \frac{0}{2} - \frac{4}{2} = -2 \qquad 0-2 = -2 \]
Picture it
Animation
Shows: Each line of the worked example "matching denominators", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use zero. The original is zero over two minus four over two, which is negative two. The answer is zero minus two, also negative two. They agree.
Anomaly
Predict first
A student writes this, and it looks reasonable:
Subtract the second numerator from the first, straight across.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The subtraction sign sits right next to the x, so it feels like it belongs to the x alone.
Put the entire second numerator inside parentheses before you subtract it.
Why: The subtraction sign sits right next to the x, so it feels like it belongs to the x alone. The negative four is copied down exactly as it appears.
Trap
Subtract the second numerator from the first, straight across.
\[ \frac{3x+2}{x-1} - \frac{x-4}{x-1} \]
Carry the minus onto the leading term and copy the rest
Why: The subtraction sign sits right next to the x, so it feels like it belongs to the x alone. The negative four is copied down exactly as it appears.
\[ 3x+2-x-4 = 2x-2 \;\Longrightarrow\; \frac{2(x-1)}{x-1} = 2 \]
Test the result at zero
Why: The original is two over negative one minus negative four over negative one, which is negative two minus four, or negative six. This route claims the answer is always two, no matter the input. That is not the same expression at all.
\[ \text{original at } x=0: \; -2-4 = -6 \qquad \text{this route: } 2 \]
Put the entire second numerator inside parentheses before you subtract it.
\[ \frac{3x+2}{x-1} - \frac{x-4}{x-1} \]
Wrap the whole numerator, then distribute the minus to every term inside
Why: You are subtracting the entire fraction, which means subtracting its entire numerator. Every term inside the parentheses changes sign - including the ones that were already negative.
\[ \frac{(3x+2)-(x-4)}{x-1} = \frac{3x+2-x+4}{x-1} \]
Combine like terms
Why: The two constants are now plus two and plus four, not plus two and minus four. That single sign is the whole difference between the columns.
\[ = \frac{2x+6}{x-1}, \qquad x \ne 1 \]
Check at zero
Why: This gives six over negative one, which is negative six - exactly what the original expression evaluates to at zero. The parentheses saved the problem.
\[ \frac{2(0)+6}{0-1} = -6 \]
Notation
Annotate
From Trap: the minus sign only hits the first term — read this one piece at a time. What is each part doing?
On: \( \frac{2(0)+6}{0-1} = -6 \)
Concept
You cannot combine numerators until the pieces are the same size. Halves and thirds do not stack; sixths and sixths do.
\[ \frac{A}{B} + \frac{C}{D} \ne \frac{A+C}{B+D} \]
Denominators are never added, never multiplied straight across for a sum, and never averaged. They must be made to match, and then only the tops combine.
least common denominator — The smallest expression that every denominator in the problem divides into evenly. Built from the factored denominators by taking each distinct factor to the highest power it appears anywhere.
Intuition
Add one sixth and one fourth. You did not add the sixes and fours together.
\[ 6 = 2 \cdot 3, \qquad 4 = 2 \cdot 2 \]
You broke both denominators into their factors, took each distinct factor as many times as the greediest one needed it, and got twelve.
\[ \text{LCD} = 2 \cdot 2 \cdot 3 = 12, \qquad \frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \]
That is the entire polynomial procedure. The only change is that the factors are now binomials instead of primes.
Concept
Factor every denominator, then walk the list of distinct factors that appear anywhere.
For each one, ask: what is the largest number of copies any single denominator has? Take that many. Multiply them all together.
\[ (x-3)(x+2) \;\text{and}\; (x-3)(x+3) \;\Longrightarrow\; \text{LCD} = (x-3)(x+2)(x+3) \]
\[ (x+1) \;\text{and}\; (x+1)^{2} \;\Longrightarrow\; \text{LCD} = (x+1)^{2} \]
Leave the least common denominator factored. You are about to cancel against it, and an expanded denominator hides every factor you need.
Picture it
Animation
Shows: Adding needs a common denominator — a rendered Manim animation.
Rendered with Manim.
Takeaway: The LCD is the product of the distinct factors, each to its highest power.
Concept
Once you have the target denominator, ask of each fraction: what factors is it missing?
Multiply the numerator and the denominator by exactly those missing factors. Since top and bottom get the same thing, you are multiplying by one and the value is untouched.
\[ \frac{5}{(x-3)(x+2)} \cdot \frac{x+3}{x+3} = \frac{5(x+3)}{(x-3)(x+2)(x+3)} \]
Leave the new numerator in factored form for one beat, then distribute it. Distributing too early is where sign errors are born.
Ranking
Put in order
Put the moves of Worked example: add with unlike denominators into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. You cannot see a common denominator through an unfactored trinomial.
Worked example
Add and simplify. State the restrictions.
\[ \frac{5}{x^{2}-x-6} + \frac{2}{x^{2}-9} \]
Factor both denominators
Why: You cannot see a common denominator through an unfactored trinomial. Two numbers multiplying to negative six and adding to negative one are negative three and two.
\[ x^{2}-x-6 = (x-3)(x+2), \qquad x^{2}-9 = (x-3)(x+3) \]
Record every restriction now
Why: Three distinct factors appear across the two denominators, so three values are excluded.
\[ x \ne 3, \qquad x \ne -2, \qquad x \ne -3 \]
Build the least common denominator from the distinct factors
Why: The factor x minus three is shared, so it is used once, not twice. The other two factors each appear once.
\[ \text{LCD} = (x-3)(x+2)(x+3) \]
Build each fraction up to that denominator
Why: The first fraction is missing x plus three; the second is missing x plus two. Multiply each numerator and denominator by its missing factor.
\[ \frac{5(x+3)}{(x-3)(x+2)(x+3)} + \frac{2(x+2)}{(x-3)(x+2)(x+3)} \]
Distribute and add the numerators
Why: Only the tops combine. Five times x plus three is five x plus fifteen; two times x plus two is two x plus four.
\[ = \frac{5x+15+2x+4}{(x-3)(x+2)(x+3)} = \frac{7x+19}{(x-3)(x+2)(x+3)} \]
Verify at a legal test value
Why: Use zero. The original is negative five sixths plus negative two ninths, which over eighteen is negative fifteen eighteenths plus negative four eighteenths, or negative nineteen eighteenths. The answer gives nineteen over negative eighteen. They agree, and the numerator shares no factor with the denominator, so this is final.
\[ x=0:\quad -\frac{5}{6}-\frac{2}{9} = -\frac{19}{18} \qquad \frac{19}{(-3)(2)(3)} = -\frac{19}{18} \]
Picture it
Animation
Shows: Each line of the worked example "add with unlike denominators", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use zero. The original is negative five sixths plus negative two ninths, which over eighteen is negative fifteen eighteenths plus negative four eighteenths, or negative nineteen eighteenths. The answer gives nineteen over negative eighteen. They agree, and the numerator shares no factor with the denominator, so this is final.
Concept
Multiplying the denominators together is never wrong. Any common denominator gets a correct answer; the least one just gets there with less arithmetic.
In the last example the product would have carried a second copy of the shared factor, and you would have had to cancel it back out at the end.
\[ \text{product} = (x-3)^{2}(x+2)(x+3) \qquad \text{LCD} = (x-3)(x+2)(x+3) \]
So if you are stuck, use the product and simplify hard at the end. What you may not do is use something smaller than the least common denominator - that is not a common denominator at all.
Anomaly
Predict first
A student writes this, and it looks reasonable:
The second denominator already contains the binomial, so one copy should be enough.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: The factored denominators both show an x plus two, so it looks like taking it a single time is enough to cover both.
Take each distinct factor to the highest power it reaches in any one denominator.
Why: The factored denominators both show an x plus two, so it looks like taking it a single time is enough to cover both.
Trap
The second denominator already contains the binomial, so one copy should be enough.
\[ \frac{5}{x^{2}-4} - \frac{3}{x^{2}+4x+4} \]
Factor, then reuse the shared binomial only once
Why: The factored denominators both show an x plus two, so it looks like taking it a single time is enough to cover both.
\[ \frac{5}{(x-2)(x+2)} - \frac{3}{(x+2)^{2}}, \qquad \text{LCD} \overset{?}{=} (x-2)(x+2) \]
Force the second fraction into that denominator
Why: The second fraction needs an x minus two, so it is multiplied on - but its extra x plus two has nowhere to go and quietly gets dropped.
\[ \frac{5-3(x-2)}{(x-2)(x+2)} = \frac{11-3x}{(x-2)(x+2)} \]
Test at zero
Why: The original is five over negative four minus three over four, which is negative one and a quarter minus three quarters, or negative two. This route gives eleven over negative four, which is negative two and three quarters. The chosen denominator was smaller than either original denominator required, so a whole factor was thrown away.
\[ \text{original: } -2 \qquad \text{this route: } \frac{11}{-4} = -2.75 \]
Take each distinct factor to the highest power it reaches in any one denominator.
\[ \frac{5}{(x-2)(x+2)} - \frac{3}{(x+2)^{2}} \]
The binomial x plus two appears squared, so the LCD carries it squared
Why: A common denominator must be divisible by every original denominator. One copy of x plus two is not divisible by two copies, so one copy cannot be the common denominator.
\[ \text{LCD} = (x-2)(x+2)^{2} \]
Build both fractions up and subtract with parentheses
Why: The first is missing one x plus two; the second is missing the x minus two. Then five x plus ten minus three x plus six leaves two x plus sixteen.
\[ \frac{5(x+2)-3(x-2)}{(x-2)(x+2)^{2}} = \frac{2x+16}{(x-2)(x+2)^{2}} = \frac{2(x+8)}{(x-2)(x+2)^{2}} \]
Check at zero
Why: Two times eight is sixteen, over negative two times four, which is negative eight - giving negative two, exactly the original value. Using the plain product instead would also work here, but it would carry an extra x plus two you would have to cancel back out.
\[ \frac{2(0+8)}{(0-2)(0+2)^{2}} = \frac{16}{-8} = -2 \]
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Estimation
Predict first
Subtract and simplify. State the restrictions.
Commit before you compute: what does Worked example: subtract with unlike denominators come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify at a legal test value
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Use zero. The original is three over negative two minus two over one, which is negative one and a half minus two, or negative three and a half.
Worked example
Subtract and simplify. State the restrictions.
\[ \frac{3}{x-2} - \frac{2}{x+1} \]
Record the restrictions
Why: The first denominator is zero at two, the second at negative one. Neither input is allowed.
\[ x \ne 2, \qquad x \ne -1 \]
The denominators share no factor, so the least common denominator is their product
Why: Two different linear binomials have nothing in common, so nothing can be reused and the smallest option is simply both of them multiplied.
\[ \text{LCD} = (x-2)(x+1) \]
Build each fraction up to the LCD
Why: The first is missing x plus one; the second is missing x minus two. Each is multiplied top and bottom by its missing factor, which is multiplying by one.
\[ \frac{3(x+1)}{(x-2)(x+1)} - \frac{2(x-2)}{(x-2)(x+1)} \]
Subtract the numerators with the second one in parentheses
Why: Three x plus three minus the quantity two x minus four gives three x plus three minus two x plus four. Both terms of the second numerator flipped sign.
\[ = \frac{(3x+3)-(2x-4)}{(x-2)(x+1)} = \frac{x+7}{(x-2)(x+1)} \]
Verify at a legal test value
Why: Use zero. The original is three over negative two minus two over one, which is negative one and a half minus two, or negative three and a half. The answer gives seven over negative two, also negative three and a half.
\[ x=0:\quad -\frac{3}{2}-2 = -\frac{7}{2} \qquad \frac{0+7}{(0-2)(0+1)} = -\frac{7}{2} \]
Picture it
Animation
Shows: Each line of the worked example "subtract with unlike denominators", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use zero. The original is three over negative two minus two over one, which is negative one and a half minus two, or negative three and a half. The answer gives seven over negative two, also negative three and a half.
Concept
Sometimes the two denominators look different but are actually opposites - each is the other with every sign flipped.
\[ x-5 \qquad \text{and} \qquad 5-x \]
Do not build a giant common denominator out of both. Factor a negative one out of one of them and the two become the same.
\[ \frac{3}{5-x} = \frac{3}{-(x-5)} = \frac{-3}{x-5} \]
A negative sign can live in the numerator, in the denominator, or out front of the whole fraction. Move it to the numerator, where it is easiest to keep track of.
Step zero
Discussion prompt
Worked example: opposite denominators — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Record the restriction
Answer:
Worked example
Add and simplify. State the restrictions.
\[ \frac{x}{x-5} + \frac{3}{5-x} \]
Record the restriction
Why: Both denominators vanish at the same input, five, because they are opposites of one another. There is only one excluded value.
\[ x \ne 5 \]
Factor a negative one out of the second denominator
Why: Five minus x is the opposite of x minus five, so pulling out a negative one rewrites it as the same binomial the first fraction already uses.
\[ 5-x = -(x-5) \]
Move that negative one up into the numerator
Why: A fraction with a negative denominator equals the same fraction with a negative numerator. Now both denominators are literally identical.
\[ \frac{3}{5-x} = \frac{-3}{x-5} \]
Add the numerators over the shared denominator
Why: No building up was needed at all - the denominators matched once the negative was pulled out. This is far less work than the product of the two.
\[ \frac{x}{x-5} + \frac{-3}{x-5} = \frac{x-3}{x-5}, \qquad x \ne 5 \]
Verify at a legal test value
Why: Use zero. The original is zero over negative five plus three over five, which is three fifths. The answer is negative three over negative five, also three fifths.
\[ x=0:\quad 0 + \frac{3}{5} = \frac{3}{5} \qquad \frac{0-3}{0-5} = \frac{3}{5} \]
Pattern
Six moves, in this order:
The denominator stays factored the whole way through. Expanding it just hides the cancellation waiting at the end.
Picture it
Animation
Shows: Always check against the domain — a rendered Manim animation.
Rendered with Manim.
Takeaway: Multiplying by a variable denominator can create solutions that were forbidden.
Check
Build both fractions up to the common denominator first, and mind the minus sign.
\[ \frac{5}{x-1} - \frac{3}{x+2} \]
Check your understanding
Which is the correct difference?
Answer: A
Why: The LCD is (x - 1)(x + 2). Building up gives 5(x + 2) minus 3(x - 1) on top, which is 5x + 10 - 3x + 3, or 2x + 13. Test at x = 0: the original is -5 minus 1.5, which is -6.5, and 13 over (-1)(2) is also -6.5.
Section
Part 5
Concept
A complex fraction is a fraction whose numerator, denominator, or both contain fractions of their own.
\[ \frac{\;\dfrac{1}{x}+\dfrac{1}{3}\;}{\;\dfrac{1}{x}-\dfrac{1}{3}\;} \]
complex fraction — A quotient in which the numerator or the denominator (or both) is itself a fraction or a sum of fractions. Also called a compound fraction. Simplifying it means rewriting it as a single ordinary rational expression.
The word complex here has nothing to do with complex numbers. It just means layered.
Intuition
Nothing new is going on. The main fraction bar - the long one - means the top divided by the bottom, exactly as it always has.
\[ \frac{\;\dfrac{2}{3}\;}{\;\dfrac{4}{9}\;} = \frac{2}{3} \div \frac{4}{9} = \frac{2}{3} \cdot \frac{9}{4} = \frac{3}{2} \]
So there are exactly two honest strategies, and both end in a division you already know how to do.
Method one: squash the top into one fraction, squash the bottom into one fraction, then divide. Method two: multiply the top and bottom by the least common denominator of all the little fractions, which wipes them out in one stroke.
Both give the same answer every time. Method one is safer when the pieces are already close to combined; method two is usually faster and creates fewer sign errors.
Worked example
Simplify. State the restrictions.
\[ \frac{\;\dfrac{1}{x}+\dfrac{1}{3}\;}{\;\dfrac{1}{x}-\dfrac{1}{3}\;} \]
Hunt down every denominator, including the hidden one
Why: The little denominators rule out zero. The whole bottom is also a denominator, and it vanishes when one over x equals one third - that is, at three.
\[ x \ne 0, \qquad x \ne 3 \]
Combine the numerator into a single fraction
Why: The two little denominators are x and three, so their least common denominator is three x. Build both up and add.
\[ \frac{1}{x}+\frac{1}{3} = \frac{3}{3x}+\frac{x}{3x} = \frac{3+x}{3x} \]
Combine the denominator into a single fraction the same way
Why: Same least common denominator, but now the numerators subtract. Three minus x, not x minus three - the order follows the original.
\[ \frac{1}{x}-\frac{1}{3} = \frac{3}{3x}-\frac{x}{3x} = \frac{3-x}{3x} \]
Divide: flip the bottom fraction and multiply
Why: The long bar was a division sign all along. The three x appears on top and bottom of the product, so it divides straight out.
\[ \frac{3+x}{3x} \cdot \frac{3x}{3-x} = \frac{3+x}{3-x} \]
Verify at a legal test value
Why: Use one. The top is one plus one third, which is four thirds; the bottom is one minus one third, which is two thirds; their quotient is two. The answer gives four over two, also two.
\[ x=1:\quad \frac{\;4/3\;}{\;2/3\;} = 2 \qquad \frac{3+1}{3-1} = 2 \]
Picture it
Animation
Shows: Each line of the worked example "method one, combine then divide", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use one. The top is one plus one third, which is four thirds; the bottom is one minus one third, which is two thirds; their quotient is two. The answer gives four over two, also two.
Concept
Collect every little denominator in the whole picture - top and bottom - and find their least common denominator.
Then multiply the numerator and the denominator of the big fraction by it. Same thing top and bottom means you multiplied by one, so the value is unchanged.
\[ \frac{N}{D} = \frac{N \cdot \text{LCD}}{D \cdot \text{LCD}} \]
Every little fraction gets absorbed at once, because the least common denominator is built to clear all of them. What is left is an ordinary rational expression.
Distribute that multiplier to every term on the top and every term on the bottom. Missing one term is the only way this method goes wrong.
Explain it
Discussion prompt
Explain Method two: multiply by the LCD of everything to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Collect every little denominator in the whole picture - top and bottom - and find their least common denominator.
Fill the middle
Fill in the blanks
From Worked example: method two, clear the little fractions — finish the line. Write what belongs on the right of the equals sign before you look.
\textx^{2} = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The little denominators exclude zero.
Worked example
Simplify. State the restrictions.
\[ \frac{\;1-\dfrac{1}{x}\;}{\;1-\dfrac{1}{x^{2}}\;} \]
List the restrictions
Why: The little denominators exclude zero. The whole bottom is zero when one equals one over x squared, which happens at one and at negative one.
\[ x \ne 0, \qquad x \ne 1, \qquad x \ne -1 \]
Find the least common denominator of every little fraction
Why: The little denominators are x and x squared. The highest power of x anywhere is the second, so that is the multiplier.
\[ \text{LCD} = x^{2} \]
Multiply the top and the bottom of the big fraction by it
Why: Distribute to every term. On top, x squared times one is x squared and x squared times one over x is x. On the bottom, x squared times one over x squared is one.
\[ \frac{x^{2}\left(1-\dfrac{1}{x}\right)}{x^{2}\left(1-\dfrac{1}{x^{2}}\right)} = \frac{x^{2}-x}{x^{2}-1} \]
Factor both parts and divide out the shared factor
Why: The top has a common factor of x; the bottom is a difference of squares. The binomial x minus one is a factor of each.
\[ \frac{x(x-1)}{(x-1)(x+1)} = \frac{x}{x+1} \]
Verify at a legal test value
Why: Use two. The top is one minus one half, which is one half; the bottom is one minus one quarter, which is three quarters; their quotient is two thirds. The answer gives two over three.
\[ x=2:\quad \frac{\;1/2\;}{\;3/4\;} = \frac{2}{3} \qquad \frac{2}{2+1} = \frac{2}{3} \]
Picture it
Animation
Shows: Each line of the worked example "method two, clear the little fractions", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Use two. The top is one minus one half, which is one half; the bottom is one minus one quarter, which is three quarters; their quotient is two thirds. The answer gives two over three.
Fill the middle
Fill in the blanks
From Worked example: a complex fraction that needs opposites — finish the line. Write what belongs on the right of the equals sign before you look.
\frac-\frac{1}{5x}, \qquad x \ne 0, \; x \ne 5___ = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The little denominator x rules out zero, and the main denominator x minus five rules out five.
Worked example
Simplify. State the restrictions.
\[ \frac{\;\dfrac{1}{x}-\dfrac{1}{5}\;}{\;x-5\;} \]
List the restrictions
Why: The little denominator x rules out zero, and the main denominator x minus five rules out five.
\[ x \ne 0, \qquad x \ne 5 \]
The only little denominators are x and five, so use their product
Why: They share no factor, so the least common denominator of the little fractions is five x.
\[ \text{LCD} = 5x \]
Multiply the top and the bottom by five x
Why: On top, five x times one over x is five and five x times one fifth is x. On the bottom, the whole binomial gets multiplied - it does not vanish, it just picks up a factor.
\[ \frac{5x\left(\dfrac{1}{x}-\dfrac{1}{5}\right)}{5x(x-5)} = \frac{5-x}{5x(x-5)} \]
Recognize the opposites and pull out a negative one
Why: Five minus x on top is the opposite of x minus five on the bottom. Rewriting the top makes the shared factor identical so it can be divided out, leaving the negative behind.
\[ \frac{-(x-5)}{5x(x-5)} = -\frac{1}{5x}, \qquad x \ne 0, \; x \ne 5 \]
Verify at a legal test value
Why: Use one. The top is one minus one fifth, which is four fifths; the bottom is one minus five, which is negative four; their quotient is negative one fifth. The answer gives negative one over five.
\[ x=1:\quad \frac{\;4/5\;}{-4} = -\frac{1}{5} \qquad -\frac{1}{5(1)} = -\frac{1}{5} \]
Pattern
Pick one method and commit to it. Method two, in four moves:
Method one, if you prefer it: combine the top into one fraction, combine the bottom into one fraction, then flip the bottom and multiply. Same answer, different bookkeeping.
Whichever you use, do not cancel a piece of the top against a piece of the bottom while the little fractions are still there. Nothing is a factor yet.
Picture it
Animation
Shows: Clearing a complex fraction — a rendered Manim animation.
Rendered with Manim.
Takeaway: Multiply top and bottom by the LCD of the small fractions.
Check
Multiply the top and the bottom by the least common denominator of the little fractions, then read off the result.
\[ \frac{\;\dfrac{1}{x}+\dfrac{1}{2}\;}{\;\dfrac{1}{x}-\dfrac{1}{2}\;} \]
Check your understanding
Which is the simplified form?
Answer: A
Why: The little denominators are x and 2, so the LCD is 2x. Multiplying top and bottom by 2x gives (2 + x) over (2 - x). Test at x = 1: the top is 1.5, the bottom is 0.5, and the quotient is 3, which matches (2 + 1)/(2 - 1) = 3.
Section
Part 6
Concept
Every calculus course opens with the same expression, and it is a rational expression. Simplifying it is a job for exactly the skills in this deck.
\[ \frac{f(x+h)-f(x)}{h}, \qquad h \ne 0 \]
difference quotient — The change in a function's output divided by the change in its input. It measures the average rate of change of the function between two nearby inputs.
The whole difficulty is that the numerator always starts out with a common factor hidden inside it. Your job is to expand, collect, and expose that factor so it cancels the denominator.
Analogy
Discussion prompt
Explain One rational expression you will meet again by analogy to something with no College Algebra in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Every calculus course opens with the same expression, and it is a rational expression. Simplifying it is a job for exactly the skills in this deck.
Picture it
Animation
Shows: Rates add, times do not — a rendered Manim animation.
Rendered with Manim.
Takeaway: The rate is what combines, which is why the reciprocals appear.
Picture it
Figure (svg): A curve with two marked points joined by a straight secant line, the horizontal gap labeled h and the vertical gap labeled as the change in output
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Take two points on a curve and connect them with a straight line. The difference quotient is the slope of that line: rise over run.
Intuition
Take two points on a curve and connect them with a straight line. The difference quotient is the slope of that line: rise over run.
Figure (svg): A curve with two marked points joined by a straight secant line, the horizontal gap labeled h and the vertical gap labeled as the change in output
The run is the gap between the two inputs, which is why the denominator is just that gap. The rise is the difference between the two outputs, which is the numerator.
Calculus then asks what happens as the two points slide together. You cannot answer that until the gap has been cancelled out of the bottom - which is precisely the algebra you are about to do.
Counterexample
Discussion prompt
Take two points on a curve and connect them with a straight line. The difference quotient is the slope of that line: rise over run.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The run is the gap between the two inputs, which is why the denominator is just that gap. The rise is the difference between the two outputs, which is the numerator.
Step zero
Discussion prompt
Worked example: the difference quotient of a square — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Substitute the whole input into every occurrence of the variable
Answer:
Worked example
Simplify the difference quotient for this function.
\[ f(x) = x^{2}, \qquad \frac{f(x+h)-f(x)}{h} \]
Substitute the whole input into every occurrence of the variable
Why: The function squares whatever it is handed. Here it is handed x plus h, so the entire binomial gets squared - and squaring a binomial produces a middle term.
\[ f(x+h) = (x+h)^{2} = x^{2}+2xh+h^{2} \]
Subtract the original output
Why: The two squared terms cancel each other. They always do, and that is exactly what leaves every surviving term carrying an h.
\[ (x^{2}+2xh+h^{2}) - x^{2} = 2xh+h^{2} \]
Factor the gap out of the numerator
Why: Both surviving terms contain an h, so it is a genuine common factor of the entire numerator - which is the only thing that makes cancelling legal.
\[ 2xh+h^{2} = h(2x+h) \]
Divide out the gap
Why: The gap is not zero - it is the distance between two different inputs - so dividing it out is allowed. This is the whole reason the numerator had to be factored first.
\[ \frac{h(2x+h)}{h} = 2x+h, \qquad h \ne 0 \]
Verify with numbers
Why: Take the input three and a gap of one half. Then the function gives twelve and a quarter at three and a half, and nine at three, so the quotient is three and a quarter divided by one half, which is six and a half. The formula gives two times three plus one half, also six and a half.
\[ \frac{12.25-9}{0.5} = 6.5 \qquad 2(3)+0.5 = 6.5 \]
Picture it
Animation
Shows: Each line of the worked example "the difference quotient of a square", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take the input three and a gap of one half. Then the function gives twelve and a quarter at three and a half, and nine at three, so the quotient is three and a quarter divided by one half, which is six and a half. The formula gives two times three plus one half, also six and a half.
Ranking
Put in order
Put the moves of Worked example: the difference quotient of a reciprocal into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The function returns one over whatever it is handed, so the shifted input goes into the denominator - all of it, as a single quantity.
Worked example
Simplify the difference quotient for the reciprocal function. This one needs a common denominator inside the numerator.
\[ f(x) = \frac{1}{x}, \qquad \frac{f(x+h)-f(x)}{h} \]
Write both outputs
Why: The function returns one over whatever it is handed, so the shifted input goes into the denominator - all of it, as a single quantity.
\[ f(x+h) = \frac{1}{x+h}, \qquad f(x) = \frac{1}{x} \]
Combine the numerator over a common denominator
Why: The two little denominators share no factor, so their least common denominator is the product. Build each up and subtract with the second numerator in parentheses.
\[ \frac{1}{x+h}-\frac{1}{x} = \frac{x-(x+h)}{x(x+h)} = \frac{-h}{x(x+h)} \]
Divide that by the gap, which means multiplying by its reciprocal
Why: The whole thing is a complex fraction, so the outer bar is a division sign. Flip the divisor, which is just the gap over one.
\[ \frac{-h}{x(x+h)} \cdot \frac{1}{h} = \frac{-h}{h \cdot x(x+h)} \]
Divide out the gap
Why: The gap is now a factor of the entire numerator and the entire denominator, so it makes a copy of one and disappears. The negative stays behind.
\[ = \frac{-1}{x(x+h)}, \qquad h \ne 0, \; x \ne 0, \; x+h \ne 0 \]
Verify with numbers
Why: Take the input two and a gap of one. The function gives one third at three and one half at two, so the numerator is negative one sixth and the quotient is negative one sixth. The formula gives negative one over two times three, also negative one sixth.
\[ \frac{\frac{1}{3}-\frac{1}{2}}{1} = -\frac{1}{6} \qquad \frac{-1}{2(3)} = -\frac{1}{6} \]
Pattern
The same four moves work for every function you will be handed:
If the gap will not factor out, something went wrong upstream - almost always an unexpanded square or a minus sign that did not reach every term. Go back and check those two things first.
Picture it
Animation
Shows: Simplifying is not solving — a rendered Manim animation.
Rendered with Manim.
Takeaway: Clearing denominators on an expression changes its value.
Elimination
Eliminate the wrong options
Which is the simplified difference quotient?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: The shifted output is x squared + 2xh + h squared + 1. Subtracting x squared + 1 leaves 2xh + h squared, which factors as h(2x + h), and dividing by h gives 2x + h. Test with x = 3 and h = 0.5: f(3.5) = 13.25, f(3) = 10, and 3.25 divided by 0.5 is 6.5, which equals 2(3) + 0.5.
Check
Substitute carefully, subtract the whole original output, and factor before you cancel.
\[ f(x) = x^{2}+1, \qquad \frac{f(x+h)-f(x)}{h} \]
Check your understanding
Which is the simplified difference quotient?
Answer: A
Why: The shifted output is x squared + 2xh + h squared + 1. Subtracting x squared + 1 leaves 2xh + h squared, which factors as h(2x + h), and dividing by h gives 2x + h. Test with x = 3 and h = 0.5: f(3.5) = 13.25, f(3) = 10, and 3.25 divided by 0.5 is 6.5, which equals 2(3) + 0.5.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — What They Are and Where They Break · Simplifying: Factors Only · Multiplying and Dividing · Adding and Subtracting · Complex Fractions · The Difference Quotient. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
Every single problem in this deck started the same way: factor everything. Nothing else is legal until the pieces are products.
| If you see | The first move is |
|---|---|
| A product or a quotient | Factor both parts, then cancel, then multiply |
| A sum or a difference | Factor the denominators, then build the least common denominator |
| Denominators that are opposites | Pull out a negative one from one of them |
| A fraction inside a fraction | Multiply top and bottom by the LCD of the little fractions |
| A difference quotient | Substitute in parentheses, expand, factor out the gap |
And the habit that carries into every later chapter: after you finish, drop one legal number into the original and into your answer. If they disagree, you found your own mistake before the grader did.
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