Simplifying rational expressions. Includes the definition of a rational expression and simplest form, the rule for dividing out a common factor, the distinction between factors and terms, factoring the numerator and denominator before cancelling, recognising opposite factors that cancel to negative one, and dividing a polynomial by a binomial.
Subject: Algebra 1 · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Algebra 1 · Chapter 11 — Rational Expressions and Equations
Simplifying 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 11.3 Simplifying Rational Expressions §11.3, pp. 646-651 — the lesson these objectives are drawn from
Warm-up
You have simplified numerical fractions since primary school. The same rule applies here, with polynomials in place of integers.
Discussion prompt
Simplify twelve eighteenths, and say exactly what you divided out. Then try to simplify the fraction whose numerator is x plus four and whose denominator is four.
Hint: What is shared, and is it a factor or a term?
Answer:
\[ \tfrac{12}{18} = \tfrac{6 \cdot 2}{6 \cdot 3} = \tfrac{2}{3} \]
A common factor of six was divided out of both. In x plus four over four, the four in the numerator is a term rather than a factor — it is added, not multiplied — so nothing can be divided out and the expression is already as simple as it gets.
Concept
A rational expression is a fraction whose numerator and denominator are non-zero polynomials. To simplify it, factor both and divide out any factors they have in common.
simplest form of a rational expression — A rational expression whose numerator and denominator have no factors in common other than one.
Only factors may be divided out, never terms.
Figure (svg): The definition of a rational expression with examples
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 646-646
Section
Section 1
Concept
If a, b and c are non-zero polynomials, then a c over b c equals a over b. Dividing both parts by the same non-zero quantity leaves the value unchanged.
\[ \dfrac{ac}{bc} = \dfrac{a}{b} \]
The variables stand for real numbers, so fraction rules apply.
Figure (svg): The definition of a rational expression with examples
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 646-646 — the definition of a rational expression, the Simplifying Rational Expressions rule, and Example 1
Picture it
Polynomials in place of integers.
Figure (svg): The definition of a rational expression with examples
Nothing new is being asserted about fractions. What is new is that finding the shared factor now requires the factoring of Chapter 10.
Worked example
This is Example 1 from the textbook.
\[ \text{Simplify } \dfrac{14x}{7} \text{ and } \dfrac{6x^2}{9x}. \]
Factor the first
Why: Fourteen is two times seven.
\[ \tfrac{2 \cdot 7 \cdot x}{7} \]
Divide out the seven
Why: It appears in both.
\[ 2 x \]
Factor the second
Why: Six is two times three; nine is three times three.
\[ \tfrac{2 \cdot 3 \cdot x \cdot x}{3 \cdot 3 \cdot x} \]
Divide out three and x
Why: Both are shared.
\[ \tfrac{2x}{3} \]
Figure (svg): The definition of a rational expression with examples
\[ 2x, \qquad \dfrac{2x}{3} \]
Verify: test with a number
Why: At x equal to three the first original is forty-two over seven, which is six, and two x is also six. Substituting a value checks a simplification in one line, and it works for every expression in this lesson.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 646-646
Faded example
Factor both parts first.
Fill in the blanks
\dfrac23 = \dfrac______ = \dfrac___x}___}
Why: One three and one x are shared and are divided out, leaving two x over three. Writing both parts as products makes the shared factors visible rather than guessed at.
Worked example
This is Example 2 from the textbook.
\[ \text{Simplify } \dfrac{2(x + 5)}{2}, \; \dfrac{x(x^2 + 6)}{x^2} \text{ and } \dfrac{x + 4}{x}. \]
Take the first
Why: The two is a factor of both.
\[ x + 5 \]
Take the second
Why: One x cancels from each.
\[ \tfrac{x^2 + 6}{x} \]
Look at the third
Why: The x is a term upstairs.
Report it
Why: Already in simplest form.
\[ \tfrac{x + 4}{x} \]
Figure (svg): The test for whether a rational expression is in simplest form
\[ x + 5, \; \dfrac{x^2 + 6}{x}, \; \dfrac{x + 4}{x} \]
Verify: check the second at a value
Why: At x equal to two the original is two times ten over four, which is five, and the answer is ten over two, also five. The third cannot be simplified because its numerator is a sum, not a product, which the next section makes precise.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 647-647
Trap
\[ \dfrac{2(x + 5)}{2} = x + 5 \;\Longrightarrow\; \text{but } \dfrac{2}{2} = 0 \]
Cross the twos out and leave a blank
Why: The cancelled factor seemed to vanish.
A factor divided by itself gives one, not nought. When the whole numerator is cancelled, what remains is one rather than nothing — so two over two is one, not empty.
\[ \dfrac{2}{2} = 1 \]
Replace a fully cancelled part with 1
Why: Division leaves a quotient, and that quotient is one.
This matters most when a numerator cancels completely, which happens often in the next few lessons.
Matching
Divide out the common factors.
Match the pairs
Why: In each case something multiplied both parts and was divided out. The brackets in the last two matter: they mark what is a single factor rather than a collection of terms.
Elimination
No shared factor other than one.
Eliminate the wrong options
Which expression cannot be simplified further?
Survives elimination: A
Why: The x in the numerator of the first is added to four rather than multiplying anything, so it is a term. Nothing multiplies the whole numerator except one, which is exactly what simplest form means.
Socratic
Polynomials are not integers.
Discussion prompt
Explain why the fraction rules carry over from numbers to rational expressions. Then say what extra condition the rule needs here.
Hint: What do the variables stand for?
Answer:
The variables represent real numbers, so for any particular value the expression is simply a fraction of two numbers and the ordinary rule applies. Since it applies at every value, it applies to the expression as a whole — nothing about polynomials makes them behave differently from the numbers they produce.
The extra condition is that the cancelled factor must be non-zero, which is why the rule is stated for non-zero polynomials. With numbers you can see whether a factor is nought; with a polynomial it depends on the value of the variable, and that is precisely what makes excluded values an issue in this chapter.
Section
Section 2
Concept
A factor multiplies the whole numerator or denominator; a term is one piece of a sum. Only factors may be divided out, which is why four x over four simplifies and x plus four over four does not.
This is the single commonest error in the whole chapter.
Figure (svg): Two columns separating factors that may be divided out from terms that may not
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 647-647 — the Study Tip on dividing out factors rather than terms
Picture it
Multiplied against added.
Figure (svg): Two columns separating factors that may be divided out from terms that may not
The two columns differ only in whether a sign of multiplication or addition joins the parts. That small visual difference carries the whole distinction.
Worked example
The Study Tip's own comparison.
\[ \text{Simplify } \dfrac{4x}{4} \text{ and } \dfrac{x + 4}{4}. \]
Look at the first
Why: Four multiplies x.
Divide it out
Why: Four over four is one.
Look at the second
Why: Four is added to x.
Leave it alone
Why: Nothing multiplies both parts.
\[ \tfrac{x + 4}{4} \]
Figure (svg): Two columns separating factors that may be divided out from terms that may not
\[ \dfrac{4x}{4} = x, \qquad \dfrac{x + 4}{4} \text{ is in simplest form} \]
Verify: test both at a value
Why: At x equal to eight the first is thirty-two over four, which is eight, matching x. The second is twelve over four, which is three — and cancelling the fours would have given eight, which is plainly wrong.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 647-647
Sorting
Is it multiplied or added?
Sort into buckets
Sort each highlighted quantity by what it is in its expression.
Every factor here sits outside a bracket or directly against a variable, and every term sits beside a plus sign. Looking for the operation rather than the number is what makes the sort reliable.
Worked example
Sometimes a sum can be factored first.
\[ \text{Can } \dfrac{2x + 4}{2} \text{ be simplified?} \]
Look at the numerator
Why: A sum of two terms.
\[ 2 x + 4 \]
Factor it
Why: Both terms are even.
\[ 2(x + 2) \]
Now the two is a factor
Why: It multiplies the whole bracket.
\[ \tfrac{2(x+2)}{2} \]
Divide it out
Why: Two over two is one.
\[ x + 2 \]
Figure (svg): A rational expression factored before anything is divided out
\[ \dfrac{2x + 4}{2} = x + 2 \]
Verify: test at a value
Why: At x equal to five the original is fourteen over two, which is seven, and x plus two is also seven. Factoring turned a sum into a product, which is what made the cancellation legal — and cancelling before factoring would have given x plus four.
Error analysis
The student simplified a rational expression by cancelling.
Annotate
On: \( \begin{aligned} \frac{x + 4}{4} &= x \\ \text{check at } x = 8: \quad \frac{12}{4} &= 3 \text{, but } x = 8 \end{aligned} \)
Cancelling a term is the most persistent error in this chapter and it is worth having a permanent defence against. Substituting a single value takes ten seconds and catches every instance of it, which is why the habit is worth forming now.
Faded example
A sum can sometimes become a product.
Fill in the blanks
\dfrac22 = \dfrac___})}___ = x + ___
Why: Factoring the numerator turned the two from a term into a factor, and only then could it be divided out. Cancelling before factoring would have left x plus four, which fails a numerical check.
Hypothesis
You suspect a term was divided out.
Predict first
What single check settles it?
Correct: Substitute one convenient value into both the original and the answer.
\[ \tfrac{x+4}{4} \text{ at } x = 8: \; 3 \quad \text{against the claimed } 8 \]
Why: A wrong cancellation changes the value of the expression, so any single value at which both are defined will expose it — usually a small whole number like two or eight. Re-reading tends to reproduce the same misreading, refactoring is slower, and looking simpler is exactly what a wrong answer does. The substitution is the only check here that is independent of how the simplification was performed.
Socratic
It looks like the same operation.
Discussion prompt
Explain why dividing out a factor preserves the value while removing a term does not. Then say what removing a term is really doing.
Hint: What is being divided in each case?
Answer:
Dividing out a factor divides the whole numerator and the whole denominator by the same quantity, which is a legitimate operation on a fraction. Removing a term divides only part of the numerator, leaving the rest untouched, which is not an operation on the fraction at all.
Effectively it replaces the numerator with a different polynomial and then keeps the same denominator, so the result is a different expression that happens to look related. In x plus four over four, dividing everything by four would give x over four plus one — which is a legitimate rewriting and is nothing like x.
Section
Section 3
Concept
A numerator or denominator that is a sum has no factors visible. Factoring it turns the sum into a product, and only then can common factors be identified.
All of Chapter 10's methods are available here.
Figure (svg): A rational expression factored before anything is divided out
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 647-647 — Example 3, Factor Numerator and Denominator
Picture it
Factor, then divide out.
Figure (svg): A rational expression factored before anything is divided out
The middle line does all the work. Once both parts are written as products, the shared factors can simply be read off.
Worked example
This is Example 3 from the textbook.
\[ \text{Simplify } \dfrac{6x^2}{2x^2 + 6x}. \]
Factor the numerator
Why: Six is two times three.
\[ 2 \cdot 3 \cdot x \cdot x \]
Factor the denominator
Why: Common factor two x.
\[ 2 x(x + 3) \]
Divide out the shared factors
Why: A two and an x.
\[ \tfrac{3x}{x + 3} \]
Check for more
Why: Three x and x plus three share nothing.
Figure (svg): A rational expression factored before anything is divided out
\[ \dfrac{6x^2}{2x^2 + 6x} = \dfrac{3x}{x + 3} \]
Verify: test at a value
Why: At x equal to one the original is six over eight, which is three quarters, and the answer is three over four. The x in the denominator cannot be cancelled with the one on top, because it is inside a sum.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 647-647
Faded example
A sum becomes a product.
Fill in the blanks
2x^2 + 6x = 2x(x + 3) \;\Longrightarrow\; \dfrac3___ = \dfrac___x}___
Why: The common factor two x comes out of the denominator, making the shared parts visible. Without that step nothing at all could be cancelled.
Worked example
The same method with Chapter 10 factoring.
\[ \text{Simplify } \dfrac{x^2 - 4}{x^2 + 5x + 6}. \]
Factor the numerator
Why: A difference of two squares.
\[ (x + 2) (x - 2) \]
Factor the denominator
Why: Two and three.
\[ (x + 2) (x + 3) \]
Divide out the shared factor
Why: x plus two appears in both.
\[ \tfrac{x - 2}{x + 3} \]
Note the restriction
Why: The original had two forbidden values.
\[ x \ne -2, -3 \]
Figure (svg): A rational expression factored before anything is divided out
\[ \dfrac{x - 2}{x + 3}, \quad x \ne -2, -3 \]
Verify: test at a value
Why: At x equal to one the original is negative three over twelve, which is negative a quarter, and the answer is negative one over four. Both restrictions come from the original denominator, and the one at negative two survives even though its factor was cancelled.
Trap
\[ \dfrac{6x^2}{2x^2 + 6x} = \dfrac{6x^2}{2x^2} + \dfrac{6x^2}{6x} \]
Split the fraction over the denominator's two terms
Why: The numerator was shared out between them.
A fraction may be split over a sum in its numerator, not in its denominator. This gives three plus x, which at x equal to one is four rather than the correct three quarters.
\[ \dfrac{6x^2}{2x(x + 3)} = \dfrac{3x}{x + 3} \]
Factor the denominator into a product, then cancel
Why: A product is what the rule needs.
Splitting over a denominator is never legal, whatever it looks like.
Translation
Factor both parts first.
Match the pairs
Why: Three of these needed a Chapter 10 factoring — a common factor, a difference of squares and a trinomial. The factoring is the hard part and the cancelling is immediate once it is done.
Elimination
For a rational expression with a sum underneath.
Eliminate the wrong options
What is the correct first step?
Survives elimination: A
Why: Factoring is what makes the shared parts visible and legal to cancel. Option B is the error the whole lesson is arranged around, and it is tempting precisely because the matching symbols are right there on the page.
Socratic
The expression has not changed.
Discussion prompt
Explain what factoring changes about an expression, given that its value is the same. Then say why that change is what the cancelling rule needs.
Hint: The rule is stated for a product.
Answer:
Factoring does not change the value at all; it changes the form, from a sum into a product. The rule for dividing out is stated for a c over b c — a product on both top and bottom — so it can only be applied once the expression is written that way.
Before factoring, a sum has no factors to identify, so the question of what is shared cannot even be asked. Factoring makes the multiplicative structure visible, and it is that structure the rule depends on. This is the same reason the zero-product property in Lesson 10.4 needed a product rather than a sum.
Section
Section 4
Concept
Two factors that are opposites of each other differ only by a factor of negative one. Factoring that out lets them cancel, leaving a negative sign behind.
\[ \dfrac{a - b}{b - a} = -1 \]
It comes from noticing that b minus a is the negative of a minus b.
Figure (svg): Two factors that are opposites of each other
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 648-648 — Example 4, Recognize Opposite Factors, and its Study Tip
Picture it
Factor out the minus first.
Figure (svg): Two factors that are opposites of each other
The two brackets look different enough to seem uncancellable, which is what makes this worth learning as a pattern rather than rediscovering each time.
Worked example
This is Example 4 from the textbook.
\[ \text{Simplify } \dfrac{4 - x^2}{x^2 - x - 2}. \]
Factor both parts
Why: A difference of squares and a trinomial.
\[ \tfrac{(2-x)(2+x)}{(x-2)(x+1)} \]
Spot the opposites
Why: Two minus x and x minus two.
\[ 2 - x = -(x - 2) \]
Factor out the minus
Why: Rewrite the numerator.
\[ \tfrac{-(x-2)(x+2)}{(x-2)(x+1)} \]
Divide out
Why: The shared bracket cancels.
\[ -\tfrac{x+2}{x+1} \]
Figure (svg): Two factors that are opposites of each other
\[ \dfrac{4 - x^2}{x^2 - x - 2} = -\dfrac{x + 2}{x + 1} \]
Verify: test at a value
Why: At x equal to nought the original is four over negative two, which is negative two, and the answer is negative two over one, also negative two. The minus sign is genuinely part of the answer rather than a slip.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 648-648
Faded example
Then the brackets match.
Fill in the blanks
2 - x = -1(x - 2) \;\Longrightarrow\; \dfrac-1___ = ___
Why: Reversing the order of a subtraction is the same as multiplying by negative one. Once that is factored out, the two brackets are identical and cancel in the ordinary way.
Worked example
Guided Practice items 9 and 14.
\[ \text{Simplify } \dfrac{3(4 - m)}{3(m - 4)} \text{ and } \dfrac{10 - 2x}{x - 5}. \]
Cancel the threes in the first
Why: A shared numerical factor.
\[ \tfrac{4 - m}{m - 4} \]
Recognise the opposites
Why: The pattern applies directly.
\[ -1 \]
Factor the second's numerator
Why: Two is common.
\[ \tfrac{2(5 - x)}{x - 5} \]
Recognise the opposites
Why: Five minus x over x minus five.
\[ -2 \]
Figure (svg): Two factors that are opposites of each other
\[ -1, \qquad -2 \]
Verify: test the second at a value
Why: At x equal to three the original is four over negative two, which is negative two. Both answers are plain numbers, which happens whenever the two parts differ only by a constant multiple and a sign.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 648-648
Trap
\[ \dfrac{4 - m}{m - 4} = 1 \]
Cancel the two brackets, since they contain the same numbers
Why: Both mention four and m, so they looked identical.
They are opposites rather than equals. At m equal to six the expression is negative two over two, which is negative one — the sign is not optional.
\[ \dfrac{4 - m}{m - 4} = -1 \]
Factor out negative one before cancelling
Why: Then the brackets really do match.
Writing the minus explicitly, rather than cancelling by eye, is what keeps the sign.
Sorting
Compare the two brackets.
Sort into buckets
Sort each pair of factors by their relationship.
Addition can be reordered freely, so x plus three and three plus x are simply equal. Subtraction cannot, which is exactly why reversing it introduces a factor of negative one.
Prediction
But reversing an addition does not.
Predict first
What is the difference?
Correct: Addition can be reordered freely; subtraction cannot.
\[ x = 5: \; 3 - x = -2, \; x - 3 = 2 \]
Why: Three plus x and x plus three are the same number for every x, because addition may be done in either order. Three minus x and x minus three are not the same: at x equal to five they are negative two and two, which are opposites rather than equals. Written as additions of signed terms the reason is visible — three plus negative x against x plus negative three — and those genuinely differ.
Socratic
It could be worked out each time.
Discussion prompt
Say why opposite factors are worth recognising on sight. Then say what happens to an expression if they are not spotted.
Hint: Would the expression look finished?
Answer:
They are worth recognising because the two brackets look unrelated, so an expression containing them can appear to be in simplest form when it is not. Nothing in the appearance of four minus m over m minus four suggests it is a plain number.
An unspotted pair leaves the expression looking finished but unfactored, which matters most in the next lessons where these expressions get multiplied and added. Missing a cancellation there means carrying a more complicated expression through several more steps, and the opportunity to simplify rarely returns.
Section
Section 5
Concept
To divide a polynomial by a binomial, write the division as a rational expression, factor, and divide out. Any value making an original denominator nought remains excluded.
The simplified form may be defined where the original is not.
Figure (svg): A polynomial division carried out by factoring
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 648-648 — Example 5, Divide a Polynomial by a Binomial
Picture it
The divisor was a factor.
Figure (svg): A polynomial division carried out by factoring
This works only because the divisor divides the numerator exactly. When it does not, the expression stays a fraction and cannot be reduced to a polynomial.
Worked example
This is Example 5 from the textbook.
\[ \text{Divide } x^2 - 2x - 3 \text{ by } x - 3. \]
Write it as a fraction
Why: Numerator over divisor.
\[ \tfrac{x^2 - 2x - 3}{x - 3} \]
Factor the numerator
Why: Three and negative one.
\[ (x - 3) (x + 1) \]
Divide out the divisor
Why: It is a factor of the numerator.
\[ x + 1 \]
State the restriction
Why: The original denominator.
\[ x \ne 3 \]
Figure (svg): A polynomial division carried out by factoring
\[ x + 1, \quad x \ne 3 \]
Verify: multiply back
Why: x minus three times x plus one gives x squared minus two x minus three, which is the numerator we started with. Multiplying back is the natural check for a division, exactly as it is for numbers.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 648-648
Faded example
The divisor is a factor.
Fill in the blanks
\dfrac11 = \dfrac___})}___ = x + ___
Why: Factoring the numerator revealed the divisor as one of its factors, so the division came out exactly. Multiplying the answer by the divisor recovers the numerator, which is the check.
Worked example
The simplified form is defined where the original is not.
\[ \text{Compare } \dfrac{x^2 - 2x - 3}{x - 3} \text{ and } x + 1 \text{ at } x = 3. \]
Substitute into the original
Why: Numerator and denominator both nought.
\[ \tfrac{0}{0} \]
Interpret
Why: Division by nought.
Substitute into the answer
Why: Three plus one.
\[ 4 \]
Draw the conclusion
Why: They differ at exactly one point.
\[ state x \ne 3 \]
Figure (svg): The restriction that survives a simplification
\[ \dfrac{0}{0} \text{ undefined}, \quad x + 1 = 4 \]
Verify: check they agree everywhere else
Why: At x equal to five the original is twelve over two, which is six, and x plus one is also six. The two expressions agree at every value except three, which is why the restriction has to be written alongside the simplified form.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 648-648
Trap
\[ \dfrac{x^2 - 2x - 3}{x - 3} = x + 1 \]
Report the simplified polynomial and stop
Why: The fraction is gone, so the restriction seemed to go with it.
The original expression has no value at three and x plus one has the value four, so the two are not the same function. Writing them as equal without the restriction claims something false about one point.
\[ \dfrac{x^2 - 2x - 3}{x - 3} = x + 1, \quad x \ne 3 \]
Carry the excluded value into the answer
Why: It comes from the original denominator.
The exclusions are read off the original expression, not the simplified one.
Faded example
Read it off the original denominator.
Fill in the blanks
The expression with denominator x minus 3 is undefined when x equals 3, so the simplified answer must carry the restriction x is not 3.
Why: The exclusion comes from the expression as originally written, not from the simplified version. Cancelling a factor removes it from view without removing the restriction it carried.
Elimination
Dividing x squared - 2x - 3 by x - 3.
Eliminate the wrong options
Which is the complete answer?
Survives elimination: A
Why: Both the quotient and its restriction are needed. Option D is worth noting: the exclusion is read off the original denominator, and nothing about the answer itself generates one.
Socratic
It concerns a single point.
Discussion prompt
Say why carrying an excluded value matters even though it affects only one input. Then say where in this chapter it will matter most.
Hint: Think about what happens if the expression is used again.
Answer:
An expression and its simplified form must be interchangeable, and they are not if one is defined at a point where the other is not. Substituting three into the simplified version would give an answer to a question the original expression does not answer, which is a genuine error rather than a technicality.
It matters most when solving equations, in Lesson 11.7, where a candidate solution may be exactly the excluded value. That is the same situation as Lesson 11.1's cross multiplying: correct algebra can produce a value the original expression forbids, and only the recorded restriction catches it.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Situation | May it cancel? | Result |
|---|---|---|
| a shared factor | yes | it becomes 1 |
| a shared term inside a sum | no | the expression is unchanged |
| opposite factors | yes, after factoring out -1 | they become -1 |
The middle row is the one that costs marks and the bottom row is the one that gets missed. Both are settled by looking at whether the parts are multiplied or added.
Pattern
To simplify any rational expression, these five moves cover it.
Step one is recorded before any cancelling, because a factor that gets divided out takes its restriction out of sight but not out of force.
OpenStax Elementary Algebra 2e, §8.1 Simplify Rational Expressions §8.1
Check
Factors, not terms.
Check your understanding
Simplify (x + 4) over 4.
Answer: A
Why: The four in the numerator is added to x rather than multiplying it, so it is a term and cannot be divided out.
Check
Factor first.
Check your understanding
Simplify 6x squared over (2x squared + 6x).
Answer: A
Why: The denominator factors as two x times x plus three, so a two and an x divide out, leaving three x over x plus three.
Check
Opposites cancel to negative one.
Check your understanding
Simplify (4 - m) over (m - 4).
Answer: A
Why: Four minus m is negative one times m minus four, so after factoring out the minus the brackets cancel and negative one remains.
Real world
This is the air pressure question from the lesson opener. Models relating air pressure to altitude are often written as ratios of polynomials, which simplify considerably before being used.
Discussion prompt
Suppose a pressure model contains the expression whose numerator is x squared minus four and whose denominator is x squared plus five x plus six. Simplify it, state its excluded values, and say why both matter before the model is used.
Hint: Factor both parts and watch the denominator.
Answer:
\[ \dfrac{x^2 - 4}{x^2 + 5x + 6} = \dfrac{(x+2)(x-2)}{(x+2)(x+3)} = \dfrac{x - 2}{x + 3} \]
The shared factor of x plus two divides out, leaving a much shorter expression that is far easier to evaluate at many altitudes.
The excluded values are negative two and negative three, both read off the original denominator, and the one at negative two survives even though its factor was cancelled. In a physical model the negative values may fall outside the range of altitudes anyway, but the restrictions must still be recorded — a model reused later, or rearranged, can easily bring a forbidden value back into play.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Simplify (2x + 4) over 2.
Correct: x + 2.
\[ \dfrac{2x + 4}{2} = \dfrac{2(x + 2)}{2} = x + 2 \]
Why: The numerator is a sum, so nothing may be cancelled until it is factored — and it factors as two times x plus two, at which point the two is a genuine factor of both parts and divides out. The first option cancels the two from only one term of the numerator, which is the classic error: at x equal to five the original is fourteen over two, which is seven, while x plus four gives nine. The last option is wrong because the numerator does factor, so a shared factor does exist; it is simply hidden until the factoring is done. Substituting a single value distinguishes all four answers in about ten seconds.
Explain it
They cancelled the fours in x plus four over four and got x.
Discussion prompt
In no more than four sentences, explain the difference between a factor and a term. Then give them a check that takes ten seconds.
Hint: Is the four multiplied or added?
Answer:
A usable answer: you can only divide out something that multiplies the whole numerator and the whole denominator. In x plus four the four is added to x, so it is a term and is only part of the numerator — dividing it out would divide part of the top and all of the bottom.
The check is to put a number in. At x equal to eight the expression is twelve over four, which is three, and their answer gives eight. Any single value exposes a wrong cancellation immediately.
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: Factors and terms are fixed by asking whether the quantity is multiplied or added. Factoring first is fixed by refusing to cancel anything until both parts are products. Opposite factors are fixed by checking whether two brackets contain the same numbers in the reverse order. Exclusions are fixed by writing them down from the original denominator before any cancelling. 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 the simplifying rule and beneath it two expressions that look almost identical, one with a factor and one with a term, working both and marking clearly which cancels and which does not. Underneath, take an expression whose numerator is a sum, factor it, and show that the cancellation only becomes legal after the factoring — then check both the original and the answer at one value of the variable. In the middle, simplify an expression needing a difference of squares in one part and a trinomial in the other, writing the factored form as its own line and ringing the shared factor. Beneath that, work an expression containing opposite factors, showing the step where negative one is factored out, and beside it write the general pattern for a minus b over b minus a. In the lower half, divide a polynomial by a binomial by factoring, state the excluded value, and evaluate both the original and the simplified form at that value to show they differ. Finally, in the margin, write the ten-second check for a suspicious cancellation.
Every simplification on your page should survive a numerical check at some convenient value. If one does not, the cancelled quantity was a term rather than a factor, which is the error to look for first.
Recap
Five things, and the second is the one that costs the most marks.
| If the question says | Your first move is |
|---|---|
| Simplify a rational expression | Factor both parts completely |
| The numerator is a sum | Factor it; do not cancel yet |
| Two brackets are reversed | Factor out negative one |
| Divide by a binomial | Write it as a fraction and factor |
| Give the simplified form | State the excluded values too |
Lesson 11.4 multiplies and divides rational expressions. The work is the same factoring and cancelling done here, applied to two expressions at once — which is why simplifying had to come first.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 11 Rational Expressions and Equations — Lesson 11.3 Simplifying Rational Expressions §11.3, pp. 646-651 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Algebra 1 — $55/session, free consultation.