Simplified form and the factor-then-cancel routine, the crucial difference between cancelling factors and cancelling terms, multiplying rational expressions including opposite factors and polynomial factors, dividing by multiplying by the reciprocal, and comparing package designs with surface-area-to-volume ratios.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 8 — Rational Functions
Multiply and Divide Rational Expressions
Objectives
Five outcomes. Everything here is fraction arithmetic with polynomials in place of numbers.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-581 — the lesson these objectives are drawn from
Warm-up
You can simplify 15 over 65 and multiply two numerical fractions.
Discussion prompt
Simplify 15 over 65. What exactly did you divide out, and would the same move work on the expression 4 times x plus 3, over x minus 5 times x plus 3?
Hint: Write both as products first.
Answer:
\[ \frac{15}{65} = \frac{3 \cdot 5}{13 \cdot 5} = \frac{3}{13} \]
\[ \frac{4(x+3)}{(x-5)(x+3)} = \frac{4}{x-5} \]
The same move, with a binomial in place of the 5. Everything in this lesson is fraction arithmetic you already know, with polynomials taking the place of numbers — and the only new skill is factoring first so the shared piece becomes visible.
Concept
A rational expression is in simplified form when its numerator and denominator share no factor other than 1 and negative 1. Simplifying always takes two steps: factor both parts, then divide out whatever they have in common.
simplified form of a rational expression — The form in which the numerator and denominator have no common factors other than 1 and negative 1.
\[ \frac{ac}{bc} = \frac{a}{b}, \quad b \neq 0, \; c \neq 0 \]
The property is the ordinary rule for fractions. What makes it feel new is that the common factor is a binomial rather than a number, and binomials hide until you factor.
Figure (svg): A rational expression simplified by factoring and then dividing out a common factor
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-573
Section
Section 1
Concept
To simplify, factor the numerator and the denominator completely, then divide out every factor they share. If they share nothing, the expression is already simplified and no further move is available.
\[ \frac{x^2-2x-15}{x^2-9} = \frac{(x+3)(x-5)}{(x+3)(x-3)} = \frac{x-5}{x-3} \]
Not every expression simplifies. Four over x times x plus 2 is already in simplified form, and trying to force a cancellation there produces a wrong answer rather than a simpler one.
Figure (svg): A rational expression simplified by factoring and then dividing out a common factor
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-573 — Simplifying Rational Expressions
Picture it
Example 1, in three lines.
Figure (svg): A rational expression simplified by factoring and then dividing out a common factor
The shared factor was x plus 3, invisible in the original form and obvious once both parts were factored.
Worked example
Example 1.
\[ \text{Simplify } \frac{x^2-2x-15}{x^2-9}. \]
Factor the numerator
Why: Two numbers multiplying to negative 15 and adding to negative 2.
\[ (x + 3) (x - 5) \]
Factor the denominator
Why: A difference of two squares.
\[ (x + 3) (x - 3) \]
Divide out the common factor
Why: X plus 3 appears in both products.
\[ \text{cancel } x + 3 \]
Write the simplified form
Why: Nothing further is shared.
\[ \frac{x - 5}{x - 3} \]
Figure (svg): A rational expression simplified by factoring and then dividing out a common factor
\[ \frac{x-5}{x-3} \]
Verify: test at a convenient value
Why: At x equal to 1 the original is 1 minus 2 minus 15, over 1 minus 9, which is negative 16 over negative 8, or 2. The simplified form gives negative 4 over negative 2, also 2. They agree, as they must at every value except x equal to negative 3, where the original is undefined.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-573
Matching
Factor both parts first.
Match the pairs
Why: All four numerators and denominators involve the numbers 2 and 7, and only factoring reveals which pieces actually match. The last one has nothing shared at all, which is a legitimate answer rather than a failure.
Worked example
Guided Practice 1 to 6.
\[ \text{Simplify } \frac{2(x+1)}{(x+1)(x+3)}, \; \frac{40x+20}{10x+30}, \; \frac{4}{x(x+2)}, \; \frac{x+4}{x^2-16}, \; \frac{x^2-2x-3}{x^2-x-6}, \; \frac{2x^2+10x}{3x^2+16x+5}. \]
First two
Why: Cancel x plus 1; then factor out 20 and 10 respectively.
\[ \frac{2}{x + 3}; 2(2 x + 1) / (x + 3) \]
Third
Why: Nothing is shared between 4 and the denominator.
Fourth and fifth
Why: The denominator factors as x plus 4 times x minus 4; then cancel x minus 3.
\[ \frac{1}{x - 4}; \frac{x + 1}{x + 2} \]
Sixth
Why: Two x times x plus 5, over 3x plus 1 times x plus 5.
\[ 2 x / (3 x + 1) \]
Figure (svg): The solution to Worked example six to simplify, one that will not shown as a ladder of expressions, one row per algebraic move
\[ \frac{2}{x+3}, \; \frac{2(2x+1)}{x+3}, \; \text{as is}, \; \frac{1}{x-4}, \; \frac{x+1}{x+2}, \; \frac{2x}{3x+1} \]
Verify: check the last denominator's factoring
Why: Three x plus 1 times x plus 5 expands to 3x squared plus 15x plus x plus 5, which is 3x squared plus 16x plus 5 — the original. Verifying a factorisation by expanding takes ten seconds and catches the error that would otherwise cancel the wrong thing.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 574-574
Trap
\[ \frac{x^2+7x}{x^2} \]
Cancel the x squared on both sides
Why: The same symbols appear above and below, so they are struck out.
\[ = 7x \quad \text{(wrong)} \]
At x equal to 1 the original is 8 over 1, which is 8, while 7x gives 7. The x squared on top is a TERM, not a factor of the whole numerator.
\[ \frac{x^2+7x}{x^2} = \frac{x(x+7)}{x \cdot x} \]
Factor first, then cancel one x
Why: Only after factoring is x a factor of the entire numerator.
\[ = \frac{x+7}{x} \]
At x equal to 1 this gives 8, matching. Factoring is not a formality here — it is what turns a term into a factor and makes the cancellation legal.
Fill the middle
Example 1.
Fill in the blanks
x^2-2x-15 = (x+3)(x-5)
Why: Three and negative 5 multiply to negative 15 and add to negative 2, so the numerator is x plus 3 times x minus 5. The x plus 3 is what will cancel against the denominator.
Sorting
Factor before deciding.
Sort into buckets
Sort each expression.
The fourth is the book's printed warning: x minus 5 over x minus 3 does not become negative 5 over negative 3, because those numbers are terms rather than factors.
Prediction
Commit before reasoning.
Predict first
The expression x squared minus 9, over x plus 3, simplifies to x minus 3. Do they agree at every value?
Correct: Everywhere except x equal to -3, where the original is undefined.
\[ \frac{x^2-9}{x+3} = x-3 \text{ for } x \neq -3 \]
Why: At x equal to negative 3 the original has a zero denominator and has no value, while x minus 3 gives negative 6 quite happily. Cancelling removes the restriction from the formula but not from the function, which is why the graph has a hole there — exactly the situation Lesson 8.3 distinguished from a vertical asymptote. Simplified forms are equal to the original on the original's domain, and nowhere else.
Section
Section 2
Concept
Cancelling divides the numerator and the denominator by the same quantity. That is only valid when the quantity multiplies everything above and everything below — in other words, when it is a factor of each, not merely a term inside one.
\[ \frac{x-5}{x-3} \neq \frac{-5}{-3} \]
The test is quick: substitute a number. If the two forms disagree at even one value, the cancellation was illegal.
Figure (svg): A correct cancellation of a factor beside an incorrect cancellation of a term
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-573 — Avoid errors
Picture it
The same-looking move, applied correctly and incorrectly.
Figure (svg): A correct cancellation of a factor beside an incorrect cancellation of a term
On the left the x multiplies the whole numerator once it has been factored out. On the right the 5 and the 3 are added and subtracted, never multiplied, so nothing can go.
Worked example
The book's worked comparison.
\[ \text{Explain why } \frac{x^2+7x}{x^2} \text{ simplifies but } \frac{x-5}{x-3} \text{ does not.} \]
Factor the first numerator
Why: X is a factor of both terms, so it comes out front.
\[ x(x + 7) \]
Cancel one x
Why: It now multiplies the whole numerator and the whole denominator.
\[ \frac{x + 7}{x} \]
Look at the second
Why: X minus 5 is a sum, not a product; there is nothing to factor out.
Test the illegal move
Why: At x equal to 4 the original is negative 1 and negative 5 over negative 3 is about 1.67.
Figure (svg): A correct cancellation of a factor beside an incorrect cancellation of a term
\[ \frac{x(x+7)}{x \cdot x} = \frac{x+7}{x}; \qquad \frac{x-5}{x-3} \text{ as is} \]
Verify: test the legal one too
Why: At x equal to 2 the original is 4 plus 14, over 4, which is 18 over 4, or 4.5. The simplified form gives 9 over 2, also 4.5. Agreement at a test value is not a proof, but disagreement is an immediate disproof — which is why it is worth thirty seconds whenever a cancellation feels uncertain.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-573
Sorting
Ask whether the piece is a factor of the whole.
Sort into buckets
Sort each proposed cancellation.
The fifth needed one extra move first, factoring out negative 1, before its cancellation became visible — but it is legal all the same.
Worked example
The move used in Example 4.
\[ \text{Simplify } \frac{1-x}{x-1} \text{ and } \frac{3-x}{x-3}. \]
Factor out negative 1 from the first numerator
Why: One minus x is negative 1 times x minus 1.
\[ (-1) (x - 1) \]
Cancel
Why: The x minus 1 now appears above and below.
\[ -1 \]
Repeat for the second
Why: Three minus x is negative 1 times x minus 3.
\[ -1 \]
State the general rule
Why: Any binomial over its reverse gives negative 1.
\[ \frac{a - b}{b - a} = -1 \]
Figure (svg): Two binomials that differ only in sign, rewritten so a cancellation becomes possible
\[ \frac{1-x}{x-1} = -1; \qquad \frac{3-x}{x-3} = -1 \]
Verify: test each at one value
Why: At x equal to 5 the first is negative 4 over 4, which is negative 1. At x equal to 7 the second is negative 4 over 4, again negative 1. Two binomials in reversed order are opposites, and a quantity divided by its opposite is always negative 1 — provided it is not zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 575-575
Error analysis
A student simplifies a quotient of two binomials.
Annotate
On: \( \frac{x+4}{x+7} = \frac{4}{7} \)
Cancelling divides both parts by the same quantity, and dividing x plus 4 by x does not leave 4. The expression is already in simplified form.
Fill the middle
The opposite-factor move.
Fill in the blanks
1 - x = (-1)(x - 1)
Why: Factoring out negative 1 reverses both signs inside, turning 1 minus x into negative 1 times x minus 1. That single move makes a cancellation possible where none seemed available.
Two truths and a lie
All three are about cancelling.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is false. In x plus 4 over x plus 7 the x appears in both, but it is added rather than multiplied, so nothing cancels — substituting 3 gives 0.7 against four sevenths. The rule is about factors, and only factoring completely reveals which pieces qualify.
Prediction
Commit before reasoning.
Predict first
You are unsure whether a cancellation was legal. What is the fastest check?
Correct: Substitute one convenient number into both forms and compare.
\[ \text{at } x=4: \; \frac{x-5}{x-3} = -1 \text{ but } \frac{-5}{-3} \approx 1.67 \]
Why: It takes seconds and is decisive in one direction: if the two forms disagree at any value, the cancellation was wrong. Agreement at one value is not a proof — two different expressions can happen to meet — but it is strong evidence, and testing a second value makes it stronger. Choosing a small integer that makes neither denominator zero keeps the arithmetic easy.
Section
Section 3
Concept
Multiply numerators, multiply denominators, and simplify. In practice the order is reversed: factor everything first, so the cancelling can happen before any expanding.
\[ \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd} \]
Multiplying out first and factoring afterwards gives the same answer but far more work. Every quantity that was going to cancel would have to be recovered from a much larger polynomial.
Figure (svg): Two rational expressions multiplied by factoring, combining, and cancelling
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 575-575 — Multiplying Rational Expressions
Picture it
Example 4, including the opposite-factor move.
Figure (svg): Two rational expressions multiplied by factoring, combining, and cancelling
Three separate cancellations happened at once, and what survived was a single factor times negative 1.
Worked example
Example 3, a multiple-choice item.
\[ \text{Multiply } \frac{12x^4y}{2xy^3} \cdot \frac{5x^2y^5}{6y}. \]
Multiply across
Why: Numerators together and denominators together.
\[ 60 x ^{6} y ^{6} / (12 x y ^{4}) \]
Divide the coefficients
Why: Sixty over 12 is 5.
\[ \text{coefficient } 5 \]
Subtract the exponents on x
Why: Six minus 1.
\[ x ^{5} \]
Subtract the exponents on y
Why: Six minus 4.
\[ y ^{2} \]
Figure (svg): The solution to Worked example multiply two monomial fractions shown as a ladder of expressions, one row per algebraic move
\[ 5x^5y^2 \]
Verify: check with a value
Why: At x equal to 1 and y equal to 1 the original is 12 over 2 times 5 over 6, which is 6 times five sixths, or 5. The answer gives 5 as well. With monomials the exponent rules of Lesson 5.1 do all the cancelling, which is why no factoring was needed here.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 575-575
Ranking
The efficient order, not the literal one.
Put in order
Why: Step five comes last, and by then there is usually almost nothing left to multiply. Doing it first turns a two-line problem into a page of expanding followed by factoring a fourth-degree polynomial.
Worked example
Example 4.
\[ \text{Multiply } \frac{3x-3x^2}{x^2+4x-5} \cdot \frac{x^2+x-20}{3x}. \]
Factor every part
Why: Three x times 1 minus x; x minus 1 times x plus 5; x plus 5 times x minus 4.
Rewrite the opposite factor
Why: One minus x becomes negative 1 times x minus 1.
\[ (-1) (x - 1) \]
Write one fraction and cancel
Why: Three x, x minus 1 and x plus 5 all appear above and below.
Simplify what is left
Why: Negative 1 times x minus 4.
\[ -x + 4 \]
Figure (svg): Two rational expressions multiplied by factoring, combining, and cancelling
\[ -x+4 \]
Verify: test at a safe value
Why: At x equal to 2 the first fraction is 6 minus 12, over 4 plus 8 minus 5, which is negative 6 over 7. The second is 4 plus 2 minus 20, over 6, which is negative 14 over 6. Their product is 84 over 42, or 2. And the answer gives negative 2 plus 4, which is 2. They agree.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 575-575
Trap
\[ \frac{3x-3x^2}{x^2+4x-5} \cdot \frac{x^2+x-20}{3x} \]
Expand both products first
Why: Multiplying numerators and denominators is taken as the literal first step.
\[ = \frac{3x^4 - \dots}{3x^3 + \dots} \quad \text{(a mess)} \]
Now both parts are fourth-degree polynomials that have to be factored from scratch, and the factors that were about to cancel are buried.
\[ = \frac{3x(1-x)}{(x-1)(x+5)} \cdot \frac{(x+5)(x-4)}{3x} \]
Factor first, cancel second, and never expand at all
Why: The factored form is where cancelling is possible, so reaching it is the priority.
\[ = (-1)(x-4) = -x+4 \]
The rule says multiply then simplify, but the efficient order is factor, cancel, and only then multiply the few survivors. That is true for numerical fractions too.
Fill the middle
Example 3.
Fill in the blanks
\frac5___ = ___x^5y^2
Why: Sixty over 12 is 5, and the exponents subtract: 6 minus 1 on x and 6 minus 4 on y. With monomials the whole simplification is the quotient rule from Lesson 5.1.
Matching
Factor before multiplying.
Match the pairs
Why: The first two are pure exponent arithmetic; the last two need factoring, including the difference of two cubes in the fourth. In every case cancelling happened before any multiplying out.
Prediction
Commit before reasoning.
Predict first
Example 4's answer is negative x plus 4. Where did the negative come from?
Correct: From rewriting 1 - x as (-1)(x - 1) so it could cancel.
\[ 1-x = (-1)(x-1) \;\Longrightarrow\; \text{the } -1 \text{ survives} \]
Why: The numerator held 1 minus x and the denominator held x minus 1 — opposites rather than equals. Pulling out negative 1 made the cancellation possible and left that factor behind in the answer. Missing this move is the single most common error here: it produces x minus 4 instead of 4 minus x, an answer wrong by a sign at every value.
Section
Section 4
Concept
To divide by a rational expression, multiply by its reciprocal. After the flip everything proceeds exactly as for multiplication: factor, combine, cancel. A polynomial divisor becomes 1 over that polynomial.
\[ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc} \]
The flip has to happen before any cancelling. Cancelling across a division sign, while the second fraction is still the wrong way up, cancels the wrong things.
Figure (svg): A division of rational expressions turned into a multiplication by the reciprocal
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 576-576 — Dividing Rational Expressions
Picture it
Example 6: a division that collapses to a plain number.
Figure (svg): A division of rational expressions turned into a multiplication by the reciprocal
Everything containing a variable cancelled, leaving 7 over 2. That happens more often than you would expect when the same factors appear on both sides.
Worked example
Example 6.
\[ \text{Divide } \frac{7x}{2x-10} \div \frac{x^2-6x}{x^2-11x+30}. \]
Multiply by the reciprocal
Why: Flip the second fraction and change the sign to multiplication.
\[ \cdot \frac{x ^{2} - 11 x + 30}{x ^{2} - 6 x} \]
Factor everything
Why: Two x minus 10 is 2 times x minus 5; the quadratic is x minus 5 times x minus 6.
Write one fraction
Why: All numerators over all denominators.
Cancel and simplify
Why: The x, the x minus 5 and the x minus 6 all go.
\[ \frac{7}{2} \]
Figure (svg): A division of rational expressions turned into a multiplication by the reciprocal
\[ \frac{7}{2} \]
Verify: test at a safe value
Why: At x equal to 2 the first fraction is 14 over negative 6, or negative seven thirds. The second is 4 minus 12, over 4 minus 22 plus 30, which is negative 8 over 12, or negative two thirds. Dividing gives seven thirds times three halves, which is 7 over 2. The answer is genuinely constant, not just simplified.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 576-576
Fill the middle
Example 7.
Fill in the blanks
\div (3x^2+5x) \;\text1\; \cdot \frac___}___
Why: Any polynomial is that polynomial over 1, so its reciprocal is 1 over the polynomial. Writing the invisible denominator explicitly is what makes the flip obvious.
Worked example
Example 7.
\[ \text{Divide } \frac{6x^2+x-15}{4x^2} \div (3x^2+5x). \]
Write the polynomial as a fraction
Why: Any polynomial is itself over 1, so its reciprocal is 1 over it.
\[ \cdot \frac{1}{3 x ^{2} + 5 x} \]
Factor the numerator
Why: Three x plus 5 times 2x minus 3.
\[ (3 x + 5) (2 x - 3) \]
Factor the new denominator
Why: X times 3x plus 5.
\[ x(3 x + 5) \]
Cancel and simplify
Why: The 3x plus 5 goes, and the x joins the 4x squared.
\[ \frac{2 x - 3}{4 x ^{3}} \]
Figure (svg): The solution to Worked example divide by a polynomial shown as a ladder of expressions, one row per algebraic move
\[ \frac{2x-3}{4x^3} \]
Verify: check the factorisation
Why: Three x plus 5 times 2x minus 3 expands to 6x squared minus 9x plus 10x minus 15, which is 6x squared plus x minus 15 — the original numerator. And at x equal to 1 the whole original is 6 plus 1 minus 15, over 4, divided by 8, which is negative 8 over 32, or negative 0.25. The answer gives negative 1 over 4, matching.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 577-577
Error analysis
A student divides two rational expressions.
Annotate
On: \( \frac{7x}{2x-10} \div \frac{x^2-6x}{x^2-11x+30}: \; \text{cancel the } x \text{ terms first} \)
The flip is the only step that distinguishes division from multiplication, and doing anything before it acts on the wrong arrangement.
Comparison
Fill the blanks. One extra move.
Comparison matrix
| Step | Multiplying | Dividing |
|---|---|---|
| First move | factor everything | flip the second fraction |
| Second move | write one fraction | factor everything |
| Cancelling | after combining into one fraction | after the flip and the factoring |
| A polynomial factor | write it over 1 | write 1 over it |
The whole difference is the first row. Once the flip is done, a division problem is indistinguishable from a multiplication problem.
Sorting
Only the divisor gets inverted.
Sort into buckets
Sort each expression by what happens to it in a division.
Flipping the wrong fraction gives an answer that is the reciprocal of the right one, which is easy to spot with a single test value.
Prediction
Commit before reasoning.
Predict first
Example 6 came out as 7 over 2, with no variable at all. Is that suspicious?
Correct: No — every variable factor happened to appear on both sides, so all of them cancelled.
\[ \frac{7x(x-5)(x-6)}{2(x-5)(x)(x-6)} = \frac{7}{2}, \; x \neq 0,5,6 \]
Why: After the flip, the factors x, x minus 5 and x minus 6 each appeared once above and once below. Testing at x equal to 2 gives 7 over 2, and at x equal to 3 it gives 7 over 2 again — the expression really is constant wherever it is defined. It is still undefined at 0, 5 and 6, where an original denominator vanishes, so the constant function has three holes in it.
Section
Section 5
Concept
One measure of packaging efficiency is the ratio of surface area to volume: the smaller the ratio, the less material per unit of contents. Writing both ratios and simplifying them makes the comparison a matter of reading two numerators.
\[ \frac{S}{V} = \frac{2s^2+4sh}{s^2h} = \frac{2s+4h}{sh} \]
The point of simplifying here is not tidiness but comparability. Two unsimplified ratios with different denominators cannot be compared at a glance; two simplified ones with the same denominator can.
Figure (svg): Two popcorn tins with the same square base and different heights, with their efficiency ratios
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 574-574 — Multi-step problem
Picture it
Example 2: the old tin beside the new one.
Figure (svg): Two popcorn tins with the same square base and different heights, with their efficiency ratios
Both simplified ratios have denominator s times h. The old tin's numerator is 2s plus 4h and the new tin's is s plus 4h, so the old one is the larger ratio and the less efficient tin.
Worked example
Example 2.
\[ \text{A tin has square base } s \text{ and height } h. \text{ Compare it with one of height } 2h. \]
Write both surface areas
Why: Two square ends plus four rectangular sides.
\[ 2 s ^{2} + 4 s h\text{ and } 2 s ^{2} + 8 s h \]
Write both volumes
Why: Base area times height.
\[ s ^{2} h\text{ and } 2 s ^{2} h \]
Form and simplify each ratio
Why: Factor out s from the first, 2s from the second.
\[ \frac{2 s + 4 h}{s h}\text{ and } \frac{s + 4 h}{s h} \]
Compare
Why: The denominators match and are positive, so compare numerators.
Figure (svg): Two popcorn tins with the same square base and different heights, with their efficiency ratios
\[ \frac{2s+4h}{sh} > \frac{s+4h}{sh} \]
Verify: test with actual numbers
Why: Take s equal to 10 and h equal to 10. The old ratio is 60 over 100, or 0.6; the new is 50 over 100, or 0.5. The taller tin really does use less material per unit of contents, and the algebra says so for every positive s and h rather than only for these.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 574-574
Fill the middle
Example 2, Step 2.
Fill in the blanks
\fracsh___ = \frac______ = \frac______}
Why: Factoring an s out of both parts and cancelling leaves s times h underneath. Both designs simplify to that same denominator, which is what makes them comparable.
Worked example
Guided Practice 7.
\[ \text{Now keep the height } h \text{ and double the base side to } 2s. \text{ Find the ratio.} \]
Write the surface area
Why: Two ends of side 2s, and four sides of width 2s.
\[ 8 s ^{2} + 8 s h \]
Write the volume
Why: The base area is 4s squared.
\[ 4 s ^{2} h \]
Form the ratio and factor
Why: Eight s times s plus h, over 4s times sh.
\[ 8 s(s + h) / (4 s ^{2} h) \]
Simplify
Why: Cancel 4s from both parts.
\[ 2(s + h) / (s h) \]
Figure (svg): The solution to Worked example a wider tin instead shown as a ladder of expressions, one row per algebraic move
\[ \frac{2(s+h)}{sh} = \frac{2s+2h}{sh} \]
Verify: compare all three designs
Why: Writing each numerator over the same denominator sh gives 2s plus 4h for the old, s plus 4h for the taller, and 2s plus 2h for the wider. With s equal to h equal to 10 those are 0.6, 0.5 and 0.4 — so the wider tin is the most efficient of the three. Putting every ratio over a common denominator is what made a three-way comparison possible.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 574-574
Trap
\[ \frac{2s^2+4sh}{s^2h} \text{ against } \frac{2s^2+8sh}{2s^2h} \]
Compare the numerators as they stand
Why: The second numerator is larger, so the second tin is judged worse.
\[ 2s^2+8sh > 2s^2+4sh \;\Longrightarrow\; \text{new tin worse} \quad \text{(wrong)} \]
The denominators are different too — the second is twice the first — so comparing numerators alone means nothing.
\[ \frac{2s+4h}{sh} \text{ against } \frac{s+4h}{sh} \]
Simplify both first, so the denominators match
Why: Only then does comparing numerators decide the question.
\[ 2s+4h > s+4h \;\Longrightarrow\; \text{old tin worse} \]
Simplifying reversed the conclusion. This is the same reason numerical fractions are given a common denominator before being compared.
Ranking
Most efficient first, taking s equal to h.
Put in order
Why: With s equal to h equal to 10 the ratios are 0.4, 0.5, 0.6, 0.8 and 1.2. Every design that makes the tin larger in some direction improves the ratio, because volume grows faster than surface area — which is the same scaling fact behind Lesson 7.7's bird exponent of 2.5 rather than 3.
Sorting
Bigger in any direction lowers the ratio.
Sort into buckets
Sort each change to the tin.
This is why bulk packaging is more material-efficient than single servings, and why small animals lose heat faster than large ones — the same ratio, in a different setting.
Prediction
Commit before reasoning.
Predict first
Doubling every dimension of the tin multiplies its surface area by 4 and its volume by 8. What happens to the ratio?
Correct: It halves, since 4 over 8 is one half.
\[ \frac{S}{V} \propto \frac{L^2}{L^3} = \frac{1}{L} \]
Why: Surface area scales with the square of the linear size and volume with the cube, so the ratio scales with the reciprocal of the size. Doubling every dimension therefore halves the material used per unit of contents. That single fact explains bulk packaging, why elephants have small ears relative to their bodies and mice have large ones, and why a warehouse is cheaper to heat per cubic metre than a shed.
Comparison
Fill the blanks. All four begin with factoring.
Comparison matrix
| Task | First move | Then |
|---|---|---|
| Simplify | factor both parts | divide out common factors |
| Multiply | factor everything | combine, cancel, then multiply survivors |
| Divide | flip the second fraction | then factor, combine and cancel |
| Divide by a polynomial | write it as 1 over the polynomial | then proceed as for multiplying |
Only the third row's first move is new. Everything else is the same factor-and-cancel routine applied to a slightly different arrangement.
Pattern
One routine, with one optional flip at the start.
Cancel factors, never terms. If the piece you want to remove is added to something rather than multiplying it, it cannot go.
OpenStax Algebra and Trigonometry 2e, §1.6 Rational Expressions §1.6
Check
Simplifying. Factor first.
Check your understanding
Simplify (x^2 - 2x - 15)/(x^2 - 9).
Answer: A
Why: Both parts share the factor x + 3, which cancels.
Check
Multiplying. Watch the opposite factor.
Check your understanding
Simplify (3x - 3x^2)/(x^2 + 4x - 5) times (x^2 + x - 20)/(3x).
Answer: A
Why: Rewriting 1 - x as (-1)(x - 1) leaves a factor of -1 in the answer.
Check
Dividing. Flip first.
Check your understanding
Simplify 7x/(2x - 10) divided by (x^2 - 6x)/(x^2 - 11x + 30).
Answer: A
Why: After flipping, the x, the x - 5 and the x - 6 all cancel.
Real world
A cell absorbs nutrients through its surface and consumes them throughout its volume. For a spherical cell of radius r, the surface area is 4 pi r squared and the volume is four thirds pi r cubed.
Discussion prompt
Simplify the ratio of surface area to volume, and explain why cells are microscopic rather than the size of a grape.
Hint: Cancel the pi and as many factors of r as you can.
Answer:
\[ \frac{S}{V} = \frac{4\pi r^2}{\tfrac{4}{3}\pi r^3} = \frac{4\pi r^2 \cdot 3}{4\pi r^3} = \frac{3}{r} \]
The ratio simplifies to 3 over r — an inverse variation, exactly the shape of Lesson 8.1. A cell twice as wide has half the absorbing surface per unit of the volume it must feed.
A cell of radius 10 micrometres has a ratio of 0.3 per micrometre; one the size of a grape, at 10 millimetres, has 0.0003 — a thousand times worse. Its interior would starve long before nutrients diffused that far. That is why large organisms are built from enormous numbers of tiny cells rather than a few big ones, and why the cells that do grow large, like a bird's egg, are mostly inert stored food rather than active cytoplasm. All of it falls out of one cancellation.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does x plus 4, over x plus 7, simplify to 4 over 7?
Correct: No — the x values are terms in a sum, not factors, so nothing cancels.
\[ \text{at } x = 3: \; \frac{7}{10} = 0.7 \neq \frac{4}{7} \approx 0.57 \]
Why: Cancelling divides the whole numerator and the whole denominator by the same quantity, and dividing x plus 4 by x does not leave 4. Testing settles it: at x equal to 3 the original is 7 over 10, or 0.7, while four sevenths is about 0.57. The expression is already in simplified form. The legal version of this move looks almost identical — x squared plus 7x, over x squared, does simplify — but only because factoring turns the x into a factor of the entire numerator first.
Explain it
They keep crossing out matching symbols wherever they see them.
Discussion prompt
In four sentences or fewer, explain when you may cross something out in a fraction and when you may not.
Hint: Think about what crossing out really does.
Answer:
Crossing out means dividing the top and the bottom by the same thing, and that only works if the thing multiplies everything on each side.
So in 6 times 5, over 7 times 5, the 5 can go, because it multiplies the whole top and the whole bottom. But in 6 plus 5, over 7 plus 5, it cannot, because there it is being added. Factor first, and then whatever you see multiplying both parts is fair game.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For factoring, do ten quadratics with no fractions around them until it is automatic. For opposite factors, circle every binomial and check whether any two are reverses of each other before cancelling. For flipping, rewrite the whole problem as a multiplication on its own line before starting. For word problems, write both quantities, form the ratio, and simplify before comparing anything.
Connect it up
Paper. Fifteen minutes.
Draw it
Build an operations page. Top left: write the simplifying property with a numerical example and an algebraic one side by side, then write out the terms-versus-factors warning with a numerical test that shows the illegal version failing. Top right: work Example 1 in full and one of your own invention. Middle left: work Example 4 in full, circling the opposite factor and writing beside it where the negative one in the answer came from. Middle right: work Example 6 in full, boxing the flip as the first step. Bottom: write the two tins' surface areas and volumes, form both ratios, simplify both, and write one sentence explaining why the comparison was impossible before simplifying. In a margin, write your own test-value check for any one of these.
If any of your cancellations struck out a symbol that was added rather than multiplied, redo that line — and test it at a value to see the disagreement for yourself.
Recap
Five things, all built on factoring.
| If you see | Then |
|---|---|
| A quotient of polynomials | Factor both parts before anything else |
| A factor above and below | Divide it out |
| A term that looks the same above and below | Leave it; only factors cancel |
| 1 - x above and x - 1 below | Rewrite as -1 times x - 1; the quotient is -1 |
| A division sign | Flip the second expression first |
| A polynomial divisor | Write it as 1 over that polynomial |
Lesson 8.5 adds and subtracts rational expressions, where a common denominator becomes necessary — the one thing multiplication and division never needed.
McDougal Littell Algebra 2 (Texas Edition), Ch. 8 Rational Functions — Lesson 8.4 Multiply and Divide Rational Expressions §8.4, pp. 573-581 — everything on these slides traces back here
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