Multiplying and dividing rational expressions. Includes both rules, factoring before multiplying so that factors cancel across the multiplication sign, treating a polynomial as a fraction with denominator one, dividing by multiplying by the reciprocal of the divisor, dividing by a polynomial, and collecting the excluded values a product or quotient carries.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Multiplying and Dividing Rational Expressions
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 652-657 — the lesson these objectives are drawn from
Warm-up
Lesson 11.3 simplified a single rational expression. This lesson combines two of them, using the fraction rules you already know.
Discussion prompt
Work out two thirds times nine tenths, cancelling before you multiply. Then say how you would divide two thirds by nine tenths.
Hint: Look for factors shared across the multiplication sign.
Answer:
\[ \tfrac{2}{3} \cdot \tfrac{9}{10} = \tfrac{3}{5}, \qquad \tfrac{2}{3} \div \tfrac{9}{10} = \tfrac{2}{3} \cdot \tfrac{10}{9} = \tfrac{20}{27} \]
The three cancelled with the nine and the two with the ten, both across the multiplication sign. Division became multiplication by the flipped second fraction, and both moves carry over to rational expressions unchanged.
Concept
To multiply rational expressions, multiply the numerators and multiply the denominators. To divide, multiply by the reciprocal of the divisor.
divisor — In a division, the expression being divided by. It is the one whose reciprocal is taken when the division is rewritten as a multiplication.
All the parts must be non-zero polynomials.
Figure (svg): The rules for multiplying and dividing rational expressions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 652-652
Section
Section 1
Concept
The product of two rational expressions has the product of the numerators on top and the product of the denominators underneath. The result is then simplified.
\[ \dfrac{a}{b} \cdot \dfrac{c}{d} = \dfrac{ac}{bd} \]
The simplifying is Lesson 11.3's work.
Figure (svg): The rules for multiplying and dividing rational expressions
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 652-652 — the Multiplying and Dividing Rational Expressions rules and Example 1
Picture it
Two rules, both familiar.
Figure (svg): The rules for multiplying and dividing rational expressions
The rules are stated for non-zero polynomials, which is where the excluded values of this chapter come from. Nothing else about them is new.
Worked example
The pattern of Example 1 from the textbook.
\[ \text{Simplify } \dfrac{4x^3}{3x} \cdot \dfrac{6x}{20x^4}. \]
Multiply the numerators
Why: Four x cubed times six x.
\[ 24 x ^{4} \]
Multiply the denominators
Why: Three x times twenty x to the fourth.
\[ 60 x ^{5} \]
Factor both
Why: Look for shared factors.
\[ \tfrac{24x^4}{60x^5} \]
Divide out
Why: Twelve and x to the fourth.
\[ \tfrac{2}{5x} \]
Figure (svg): The rules for multiplying and dividing rational expressions
\[ \dfrac{4x^3}{3x} \cdot \dfrac{6x}{20x^4} = \dfrac{2}{5x} \]
Verify: test at a value
Why: At x equal to one the two original fractions are four thirds and three tenths, whose product is two fifths — and the answer at x equal to one is two fifths. A single substitution checks the whole chain.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 652-652
Faded example
Tops together, bottoms together.
Fill in the blanks
\dfrac2460 \cdot \dfrac______ = \dfrac___x^4}___x^5}
Why: The numerators multiply and the denominators multiply, with the powers of x adding by Lesson 8.1's rule. Simplifying then reduces the result to two over five x.
Worked example
The same product with the work reordered.
\[ \text{Simplify } \dfrac{4x^3}{3x} \cdot \dfrac{6x}{20x^4} \text{ by cancelling first.} \]
Cancel across the sign
Why: Four and twenty share four.
\[ \tfrac{x^3}{3x} \cdot \tfrac{6x}{5x^4} \]
Cancel again
Why: Three and six share three.
\[ \tfrac{x^3}{x} \cdot \tfrac{2x}{5x^4} \]
Collect the powers
Why: Four x's on top, five underneath.
\[ \tfrac{2x^4}{5x^5} \]
Finish
Why: One x survives below.
\[ \tfrac{2}{5x} \]
Figure (svg): A product simplified by factoring before multiplying
\[ \dfrac{2}{5x} \]
Verify: compare the two routes
Why: Both give two over five x, but the second never produced a number larger than six. Cancelling across the multiplication sign before multiplying is always legal and almost always less work.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 652-652
Trap
\[ \dfrac{2}{x} \cdot \dfrac{3}{x} = \dfrac{5}{x} \]
Combine the numerators and keep the denominator
Why: The denominators matched, so they were left alone.
That is the rule for adding fractions, not multiplying them. Multiplying gives six over x squared, and at x equal to one the true product is six while this gives five.
\[ \dfrac{2}{x} \cdot \dfrac{3}{x} = \dfrac{6}{x^2} \]
Multiply both the numerators and the denominators
Why: A common denominator is needed for adding, not for multiplying.
Multiplication is in fact the easier of the two operations, which is why it comes first in this chapter.
Elimination
Adding and multiplying differ.
Eliminate the wrong options
How do you multiply two rational expressions?
Survives elimination: A
Why: Multiplication needs no common denominator at all, which makes it simpler than addition. Option D is the division rule, and confusing the two is easy when both are learnt in the same lesson.
Hypothesis
Both give the same answer.
Predict first
What is the advantage of cancelling before multiplying?
Correct: The numbers stay small, so the arithmetic and factoring are easier.
\[ \tfrac{24x^4}{60x^5} \quad \text{against} \quad \tfrac{2x^4}{5x^5} \]
Why: Multiplying first produced twenty-four over sixty, which then had to be factored to find the common factor of twelve. Cancelling first never produced a number above six, and the same is true with polynomials — multiplying two trinomials before cancelling gives a degree-four expression that then has to be factored again, which is far more work than cancelling the factors that were already visible.
Socratic
Addition does.
Discussion prompt
Explain why multiplying fractions is simpler than adding them. Then say what a common denominator is actually for.
Hint: What does a denominator represent?
Answer:
Multiplying two fractions means taking a part of a part, and that works out directly as the product of the parts over the product of the wholes — no matching is required because the two denominators describe different divisions that simply compound.
A common denominator is needed for adding because two fractions can only be combined into one when they count the same size of piece. Thirds and quarters cannot be added directly because a third and a quarter are different amounts, so both must first be rewritten in twelfths. Multiplication never needs the pieces to match, which is why it comes before addition in this chapter.
Section
Section 2
Concept
Factoring both expressions before multiplying reveals factors that cancel between one numerator and the other denominator. That is far less work than multiplying out and factoring again.
A factor may cancel across the multiplication sign.
Figure (svg): A product simplified by factoring before multiplying
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 653-653 — Example 2 and its Study Tip on simplifying before multiplying
Picture it
Not only within each.
Figure (svg): A product simplified by factoring before multiplying
Because the two numerators are being multiplied together, either of them may cancel with either denominator. That freedom is what the factoring makes available.
Worked example
The method of Example 2 from the textbook.
\[ \text{Simplify } \dfrac{x^2 - 9}{3x} \cdot \dfrac{x}{x + 3}. \]
Factor the first numerator
Why: A difference of squares.
\[ (x + 3) (x - 3) \]
Look across for matches
Why: x plus three appears below on the right.
Cancel the x
Why: It is above right and below left.
\[ \tfrac{x - 3}{3} \]
Multiply what remains
Why: Nothing left to combine.
\[ \tfrac{x - 3}{3} \]
Figure (svg): A product simplified by factoring before multiplying
\[ \dfrac{x^2 - 9}{3x} \cdot \dfrac{x}{x + 3} = \dfrac{x - 3}{3} \]
Verify: test at a value
Why: At x equal to one the first fraction is negative eight thirds and the second is one quarter, so the product is negative two thirds — and the answer gives negative two over three. The two cancellations both crossed the multiplication sign, which the substitution confirms in one line.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 653-653
Faded example
A numerator can meet the other denominator.
Fill in the blanks
\dfrac33 \cdot \dfrac______ = \dfrac___}}___}
Why: The x plus three cancels between the first numerator and the second denominator, and the x cancels between the second numerator and the first denominator. Both cancellations cross the multiplication sign.
Worked example
Working the same product without cancelling, to compare.
\[ \text{Multiply } \dfrac{x^2 - 9}{3x} \cdot \dfrac{x}{x + 3} \text{ out first, then simplify.} \]
Multiply the numerators
Why: x squared minus nine, times x.
\[ x ^{3} - 9 x \]
Multiply the denominators
Why: Three x times x plus three.
\[ 3 x ^{2} + 9 x \]
Factor both
Why: Common factors and a difference of squares.
\[ \tfrac{x(x+3)(x-3)}{3x(x + 3)} \]
Cancel
Why: x and x plus three.
\[ \tfrac{x - 3}{3} \]
Figure (svg): A product simplified by factoring before multiplying
\[ \dfrac{x - 3}{3} \]
Verify: compare the two routes
Why: Both give x minus three over three, but this route required factoring a cubic and a quadratic that had just been created by multiplying. Cancelling first avoided building them in the first place, which is the Study Tip's point.
Error analysis
The student cancelled while multiplying two rational expressions.
Annotate
On: \( \begin{aligned} \frac{x^2 - 9}{3x} \cdot \frac{x}{x + 3} &= \frac{x^2 - 9}{3} \cdot \frac{1}{x + 3} \\ &= \frac{x - 3}{3} \cdot \frac{x + 3}{x + 3} \end{aligned} \)
The error is not a wrong value but a stalled method: the student factored partially and then stopped short of the cancellation the factoring had made available. Factoring completely before looking for matches is what avoids that.
Sorting
In a product of two rational expressions.
Sort into buckets
Sort each pairing by whether the two may cancel with each other.
Any numerator may cancel with any denominator, but two numerators never cancel with each other. Checking which side of the bar each sits on is the whole test.
Elimination
Multiplying two rational expressions.
Eliminate the wrong options
What is the most efficient first step?
Survives elimination: A
Why: Factoring exposes every available cancellation before the expressions grow. Option B does work but multiplies the labour, which the Study Tip warns about explicitly.
Socratic
It seems to belong to one fraction.
Discussion prompt
Explain why a factor in one numerator can cancel with a factor in the other denominator. Then say why two numerators can never cancel with each other.
Hint: Write the product as a single fraction first.
Answer:
Once the product is written as a single fraction, all the numerators are multiplied together on top and all the denominators together underneath. At that point every factor above is available to cancel with every factor below, regardless of which original fraction it came from — the multiplication sign has stopped separating them.
Two numerators both end up above the bar, so there is no division between them and nothing to cancel. Cancelling requires one factor above and one below, and that is the only condition; which fraction each came from is irrelevant once the product is combined.
Section
Section 3
Concept
A polynomial is a rational expression whose denominator is one. Writing it that way lets the multiplication rule apply without any special case.
\[ x + 4 = \dfrac{x + 4}{1} \]
Its reciprocal is one over it.
Figure (svg): A polynomial written as a fraction so the rule applies
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 653-653 — Example 3, Multiply by a Polynomial
Picture it
With denominator one.
Figure (svg): A polynomial written as a fraction so the rule applies
This one rewriting removes every special case for polynomials in this lesson, both for multiplying and for dividing.
Worked example
This is Example 3 from the textbook.
\[ \text{Simplify } \dfrac{7x}{x^2 + 5x + 4} \cdot (x + 4). \]
Write the polynomial over one
Why: Now both are fractions.
\[ \tfrac{x + 4}{1} \]
Factor the denominator
Why: One and four.
\[ (x + 1) (x + 4) \]
Multiply
Why: Numerators and denominators.
\[ \tfrac{7x(x + 4)}{(x+1)(x+4)} \]
Cancel
Why: x plus four appears above and below.
\[ \tfrac{7x}{x + 1} \]
Figure (svg): A polynomial written as a fraction so the rule applies
\[ \dfrac{7x}{x^2 + 5x + 4} \cdot (x + 4) = \dfrac{7x}{x + 1} \]
Verify: test at a value
Why: At x equal to one the fraction is seven over ten and the binomial is five, so the product is seven halves — and the answer gives seven over two. The values agree, which confirms both the factoring and the cancellation.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 653-653
Faded example
Then the rule applies unchanged.
Fill in the blanks
x + 4 = \dfrac11} \;\Longrightarrow\; \dfrac______ \cdot \dfrac______ = \dfrac______}}
Why: Writing the polynomial over one puts it in the numerator, where multiplying belongs. The x plus four then cancels with the matching factor in the denominator.
Worked example
Guided Practice 4, where everything cancels.
\[ \text{Simplify } \dfrac{3}{x + 1} \cdot (2x + 2). \]
Write it over one
Why: Two x plus two.
\[ \tfrac{2x + 2}{1} \]
Factor it
Why: Common factor two.
\[ 2(x + 1) \]
Multiply
Why: Numerators and denominators.
\[ \tfrac{3 \cdot 2(x + 1)}{x + 1} \]
Cancel
Why: x plus one divides out.
\[ 6 \]
Figure (svg): A polynomial written as a fraction so the rule applies
\[ \dfrac{3}{x + 1} \cdot (2x + 2) = 6 \]
Verify: test at two values
Why: At x equal to nought the fraction is three and the binomial is two, giving six; at x equal to three the fraction is three quarters and the binomial is eight, again giving six. The product really is constant, which the cancellation predicted.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 653-653
Trap
\[ \dfrac{7x}{x^2 + 5x + 4} \cdot (x + 4) = \dfrac{7x}{(x^2 + 5x + 4)(x + 4)} \]
Multiply the whole fraction by putting the polynomial underneath
Why: It was outside the fraction, so it went to the bottom.
Multiplying by a quantity makes a fraction larger, so the quantity belongs in the numerator. Putting it underneath divides instead, which is the opposite operation.
\[ \dfrac{7x}{x^2 + 5x + 4} \cdot \dfrac{x + 4}{1} = \dfrac{7x(x+4)}{x^2 + 5x + 4} \]
Write the polynomial as itself over one, then apply the rule
Why: Its numerator is the polynomial and its denominator is one.
The rewriting makes the placement automatic rather than something to decide.
Matching
Factor, then cancel.
Match the pairs
Why: One of these reduces to a plain number, which happens whenever every variable factor cancels. The others keep a variable, and in each case the surviving factor came from a difference of squares or a trinomial.
Prediction
Some of these reduce to a number.
Predict first
What has to happen for a product of rational expressions to be constant?
Correct: Every variable factor above must cancel with one below.
\[ \tfrac{3}{x+1} \cdot 2(x+1) = 6 \]
Why: A product is constant exactly when no variable survives the cancelling, which requires each variable factor in the combined numerator to have a partner in the combined denominator. In three over x plus one times two x plus two, the x plus one on the bottom met the x plus one hidden inside two x plus two, and nothing containing x was left. That is why factoring is what reveals such cases; before factoring, the two x plus two looks unrelated to the denominator.
Socratic
The placement could just be remembered.
Discussion prompt
Say what the rewriting achieves beyond saving you a decision. Then say where it matters even more.
Hint: Think about the reciprocal.
Answer:
It turns a special case into an ordinary one, so a single rule covers every situation rather than a rule plus an exception. Exceptions are what get misremembered under pressure, and this one has a fifty-fifty failure mode: the polynomial goes either on top or underneath.
It matters more when dividing, because the reciprocal of a polynomial is not obvious until it is written as a fraction. Once x plus four is seen as x plus four over one, its reciprocal is plainly one over x plus four — which is exactly what the next section needs.
Section
Section 4
Concept
To divide one rational expression by another, multiply the first by the reciprocal of the second. Only the divisor turns over.
\[ \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c} \]
The divisor is the expression being divided by.
Figure (svg): Division rewritten as multiplication by the reciprocal
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 654-654 — the division rule and Example 4, Divide Rational Expressions
Picture it
Name the divisor first.
Figure (svg): Two columns showing which expression turns over in a division
Deciding which expression is the divisor before touching anything removes the error entirely. It is always the one written after the division sign.
Worked example
This is Example 4 from the textbook.
\[ \text{Simplify } \dfrac{4n}{n - 5} \div \dfrac{n - 9}{n - 5}. \]
Identify the divisor
Why: The second expression.
\[ \tfrac{n - 9}{n - 5} \]
Multiply by its reciprocal
Why: Flip only that one.
\[ \tfrac{4n}{n-5} \cdot \tfrac{n-5}{n-9} \]
Multiply
Why: Numerators and denominators.
\[ \tfrac{4n(n-5)}{(n-5)(n-9)} \]
Cancel
Why: n minus five appears above and below.
\[ \tfrac{4n}{n - 9} \]
Figure (svg): Division rewritten as multiplication by the reciprocal
\[ \dfrac{4n}{n - 5} \div \dfrac{n - 9}{n - 5} = \dfrac{4n}{n - 9} \]
Verify: test at a value
Why: At n equal to one the first expression is four over negative four, which is negative one, and the divisor is negative eight over negative four, which is two — so the quotient is negative a half. The answer gives four over negative eight, also negative a half.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 654-654
Faded example
The first expression is untouched.
Fill in the blanks
\dfrac59 \div \dfrac______ = \dfrac______ \cdot \dfrac___}}___}}
Why: The divisor's numerator and denominator swap places while the first expression stays exactly as written. Flipping both would give the reciprocal of the correct answer.
Worked example
The step that prevents the commonest error.
\[ \text{In } \dfrac{4n}{n-5} \div \dfrac{n-9}{n-5}, \text{ which expression turns over?} \]
Read the order
Why: The division sign separates them.
\[ first \div second \]
Name the divisor
Why: The one after the sign.
\[ \tfrac{n-9}{n-5} \]
Flip only that
Why: Its numerator goes underneath.
\[ \tfrac{n-5}{n-9} \]
Leave the first alone
Why: It is not the divisor.
\[ \tfrac{4n}{n-5} \]
Figure (svg): Two columns showing which expression turns over in a division
\[ \dfrac{4n}{n-5} \cdot \dfrac{n-5}{n-9} \]
Verify: see what flipping both would give
Why: Flipping both would give n minus five over four n times n minus five over n minus nine, which is the reciprocal of the correct answer times an extra factor. Checking at a value would expose it immediately, but naming the divisor first prevents it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 654-654
Trap
\[ \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{b}{a} \cdot \dfrac{d}{c} \]
Turn both fractions over before multiplying
Why: The rule mentions reciprocals, so both were reciprocated.
Only the divisor is inverted. Flipping both gives the reciprocal of the correct quotient, which a single numerical check exposes at once.
\[ \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c} \]
Leave the first expression untouched
Why: It is the one being divided, not the one dividing.
Testing the rule on two thirds divided by nine tenths settles which version is right in ten seconds.
Elimination
In a division of two rational expressions.
Eliminate the wrong options
Which one gets inverted?
Survives elimination: A
Why: The divisor is the expression being divided by, which is always the second one. Naming it explicitly before flipping anything is the reliable defence against option C.
Translation
Flip the second, then multiply.
Match the pairs
Why: In every row the first expression is unchanged and only the second turns over. The last one is worth noting: the reciprocal of one over y is y, which follows from writing y as y over one.
Socratic
It is stated as a rule.
Discussion prompt
Explain why dividing by a fraction is the same as multiplying by its reciprocal. Then say what the reciprocal of a polynomial is.
Hint: What does dividing by a half do?
Answer:
Dividing by a quantity asks how many of that quantity fit into the first. Dividing by a half asks how many halves fit, and the answer is twice as many as the number itself — so dividing by a half is multiplying by two, which is its reciprocal. The same reasoning works for any fraction: dividing by c over d asks how many pieces of that size fit, and each whole contains d over c of them.
The reciprocal of a polynomial is one over that polynomial, which follows immediately once the polynomial is written as itself over one. So dividing by x plus four is multiplying by one over x plus four, and the divisor ends up in the denominator — which is exactly what the next section does.
Section
Section 5
Concept
Dividing by a polynomial means multiplying by one over that polynomial, so the polynomial ends up in the denominator. Every denominator that appears contributes an excluded value.
A flipped divisor's numerator becomes a denominator.
Figure (svg): Dividing a rational expression by a polynomial
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 654-654 — Example 5, Divide by a Polynomial
Picture it
Flip, multiply, factor, cancel.
Figure (svg): Dividing a rational expression by a polynomial
The polynomial that was outside the fraction ends up inside its denominator, which is the whole visible effect of the division.
Worked example
The method of Example 5 from the textbook.
\[ \text{Simplify } \dfrac{x^2 - 4}{x + 1} \div (x + 2). \]
Write the divisor over one
Why: Then flip it.
\[ \tfrac{1}{x + 2} \]
Multiply
Why: The divisor is now a denominator.
\[ \tfrac{x^2 - 4}{(x+1)(x+2)} \]
Factor the numerator
Why: A difference of squares.
\[ (x + 2) (x - 2) \]
Cancel
Why: x plus two appears above and below.
\[ \tfrac{x - 2}{x + 1} \]
Figure (svg): Dividing a rational expression by a polynomial
\[ \dfrac{x^2 - 4}{x + 1} \div (x + 2) = \dfrac{x - 2}{x + 1} \]
Verify: test at a value
Why: At x equal to nought the first expression is negative four and the divisor is two, so the quotient is negative two — and the answer gives negative two over one, also negative two. The substitution confirms the flip as well as the cancelling.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 654-654
Faded example
Multiplying by one over it.
Fill in the blanks
\dfrac12 \div (x + 2) = \dfrac______ \cdot \dfrac___}___ = \dfrac___}}___
Why: The reciprocal of x plus four over one is one over x plus four, so the divisor moves into the denominator. Factoring the numerator then reveals the cancellation.
Worked example
Every denominator that ever appears.
\[ \text{What values are excluded from } \dfrac{x^2 - 4}{x + 1} \div (x + 2)? \]
Take the first denominator
Why: It appears from the start.
\[ x \ne -1 \]
Take the divisor
Why: Dividing by nought is undefined.
\[ x \ne -2 \]
Note where the divisor went
Why: It became a denominator.
Collect them
Why: Both restrictions stand.
\[ x \ne -1, -2 \]
Figure (svg): Where the restrictions on a product come from
\[ x \ne -1 \text{ and } x \ne -2 \]
Verify: check the cancelled restriction
Why: The x plus two cancelled, so the simplified answer is defined at negative two — but the original division was not, because it would have divided by nought. The restriction survives the cancellation exactly as in Lesson 11.3.
Trap
\[ \dfrac{x^2-4}{x+1} \div (x+2) = \dfrac{x-2}{x+1}, \quad x \ne -1 \]
Read the restrictions off the answer's denominator
Why: The answer has only one denominator, so only one value was excluded.
Dividing by x plus two is undefined when that is nought, so negative two is forbidden too — even though its factor cancelled and left no trace in the answer.
\[ x \ne -1 \text{ and } x \ne -2 \]
Collect restrictions from every denominator and every divisor
Why: Before any cancelling.
A divisor is a denominator in waiting, which is why it carries a restriction from the start.
Sorting
Every denominator and every divisor does.
Sort into buckets
Sort each part by whether it forbids a value.
Three of the six forbid values and all three are denominators or divisors. A numerator being nought is perfectly ordinary; it just makes the whole expression nought.
Hypothesis
It is not written as a denominator.
Predict first
Why can the divisor not be nought?
Correct: Dividing by nought is undefined, and the divisor becomes a denominator.
\[ \div (x + 2) \;\Longrightarrow\; \cdot \tfrac{1}{x+2} \;\Longrightarrow\; x \ne -2 \]
Why: A division by a quantity is undefined whenever that quantity is nought, whether or not it is written as a fraction. Rewriting the division as multiplication by the reciprocal makes this visible, since the divisor moves into a denominator where the restriction is obvious. The restriction was there from the start, though — the rewriting reveals it rather than creating it.
Socratic
Each model already describes a population.
Discussion prompt
Say what dividing one population model by another tells you. Then say why simplifying the quotient is worth doing.
Hint: What does a ratio of two quantities measure?
Answer:
The quotient is the ratio of the two populations at any given time, which says how many times larger one colony is than the other. That comparison is often what matters — whether one is pulling ahead, and by how much — and neither model on its own answers it.
Simplifying is worth doing because the raw quotient is a fraction of fractions that hides the answer. After cancelling, the ratio frequently turns out to be something short, sometimes even a constant, which would mean the two colonies grow in fixed proportion. That conclusion is invisible before the cancelling and obvious after it.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Multiplying | Dividing | |
|---|---|---|
| The rule | tops together, bottoms together | flip the divisor, then multiply |
| What changes | nothing; both stay as written | only the second expression |
| Extra restriction | from each denominator | also from the divisor itself |
Division is multiplication with one extra step and one extra restriction. Everything else about the two operations is identical.
Pattern
To multiply or divide any two rational expressions, these five moves cover it.
Step one comes first because a divisor's restriction disappears from view the moment the division is rewritten, and it is the one most often lost.
OpenStax Elementary Algebra 2e, §8.2 Multiply and Divide Rational Expressions §8.2
Check
Straight across.
Check your understanding
What is 2/x times 3/x?
Answer: A
Why: The numerators multiply to six and the denominators multiply to x squared. No common denominator is needed for multiplication.
Check
Only the divisor flips.
Check your understanding
Simplify (4n)/(n - 5) divided by (n - 9)/(n - 5).
Answer: A
Why: Flipping only the divisor gives four n over n minus five times n minus five over n minus nine, and the n minus five cancels.
Check
Collect every restriction.
Check your understanding
Which values are excluded from (x squared - 4)/(x + 1) divided by (x + 2)?
Answer: A
Why: The first denominator forbids negative one and the divisor forbids negative two, since dividing by nought is undefined.
Real world
This is the prairie dog question from the lesson opener. Two colonies are modelled by rational expressions in the time since the study began, and the question is how their sizes compare.
Discussion prompt
Suppose colony A is modelled by 7t over (t squared + 5t + 4) and colony B by 1 over (t + 4). Find the ratio of A to B, simplify it, and say what it tells you.
Hint: Dividing means multiplying by the reciprocal.
Answer:
\[ \dfrac{7t}{t^2 + 5t + 4} \div \dfrac{1}{t + 4} = \dfrac{7t}{(t+1)(t+4)} \cdot \dfrac{t+4}{1} = \dfrac{7t}{t + 1} \]
The ratio simplifies to seven t over t plus one, which grows towards seven as time passes — so colony A ends up roughly seven times the size of colony B.
That conclusion is invisible in the two models as written and obvious after the cancelling, which is the practical point of this lesson. The excluded values are negative one and negative four from the first denominator, and the divisor forbids nothing extra here since its numerator is one; in the situation itself only non-negative times occur, so none of the restrictions bites.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Simplify (3/(x + 1)) times (2x + 2).
Correct: 6.
\[ \dfrac{3}{x+1} \cdot \dfrac{2(x+1)}{1} = 6 \]
Why: Writing two x plus two over one and factoring it as two times x plus one reveals a factor matching the denominator, so the x plus one cancels completely and six remains. The first option multiplies the polynomial into the denominator instead of the numerator, which divides where it should multiply. The third and fourth stop before factoring the binomial, so the available cancellation is never spotted — and this is the usual reason a product looks unsimplifiable when it is not. Testing at two values confirms the answer: at x equal to nought the product is three times two, and at x equal to three it is three quarters times eight, both giving six.
Explain it
They flipped both expressions when dividing.
Discussion prompt
In no more than four sentences, explain which one turns over and why. Then give them a check they can run on numbers.
Hint: Which is the divisor?
Answer:
A usable answer: only the divisor turns over, and the divisor is the expression written after the division sign. The first expression is the one being divided, so it stays exactly as it is.
The check is to try it on numbers you know. Two thirds divided by nine tenths is two thirds times ten ninths, which is twenty twenty-sevenths — and flipping both would give three halves times ten ninths, which is a completely different number.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The flip is fixed by naming the divisor out loud before touching anything. Factoring first is fixed by refusing to multiply until both expressions are products. Polynomials are fixed by writing them over one. Exclusions are fixed by listing them before any cancelling and including the divisor. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page write both rules, and beneath each write one numerical example with small fractions so you can check them mentally. Underneath, take a product of two rational expressions and work it twice: once by multiplying out first and then factoring, and once by factoring and cancelling first, writing both columns side by side and marking the largest number that appeared in each. In the middle, multiply a rational expression by a polynomial, showing the step where the polynomial is written over one, and ring the factor that cancels. Beneath that, divide two rational expressions: circle the divisor before you flip anything, write the flipped version as its own line, and then simplify. In the lower half, divide a rational expression by a polynomial and list every excluded value with a note beside each saying which denominator or divisor it came from, including one whose factor cancels. Finally, in the margin, write which pairs of positions may cancel with each other and which may not.
Your two columns for the product should end identically. If they do not, the cancelling column most likely cancelled two numerators with each other, which the position rule in your margin forbids.
Recap
Five things, and the last is the one that outlives this lesson.
| If the question says | Your first move is |
|---|---|
| Multiply two rational expressions | Factor both, then cancel across |
| One factor is a polynomial | Write it over one |
| Divide two rational expressions | Name the divisor, then flip it |
| Divide by a polynomial | Multiply by one over it |
| State the answer | Include every excluded value |
Lesson 11.5 turns to addition, which needs something multiplication never did: a common denominator. It begins with the easy case where the denominators already match.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.4 Multiplying and Dividing Rational Expressions §11.4, pp. 652-657 — everything on these slides traces back here
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