The Central Idea
A rational expression has the form
\[ \boxed{\frac{P(x)}{Q(x)}} \]where \(P(x)\) and \(Q(x)\) are polynomials and
\[ \boxed{Q(x)\neq0}. \]The condition \(Q(x)\neq0\) is not optional. It determines which values of the variable belong to the domain of the expression.
What Is a Rational Expression?
A rational expression is a quotient of polynomial expressions.
Examples include
\[ \frac{x+3}{x-5}, \] \[ \frac{2x^2-7x+1}{x^2+4}, \]and
\[ \frac{5}{x^3}. \]In each case, both the numerator and denominator are polynomials.
Numerator and Denominator
In
\[ \frac{P(x)}{Q(x)}, \]\(P(x)\) is the numerator and \(Q(x)\) is the denominator.
For example, in
\[ \frac{x^2-4}{x+7}, \]the numerator is
\[ x^2-4, \]and the denominator is
\[ x+7. \]Rational Expressions Extend Polynomial Expressions
Every polynomial can also be viewed as a rational expression because it can be written over \(1\).
For example,
\[ x^2+3x-4 = \frac{x^2+3x-4}{1}. \]Since the denominator \(1\) can never equal zero, this representation introduces no domain restriction.
Why the Denominator Cannot Be Zero
Division asks how many copies of one quantity fit into another or, equivalently, which quantity multiplied by the divisor produces the dividend.
Consider
\[ \frac{6}{2}=3 \]because
\[ 2\cdot3=6. \]But
\[ \frac{6}{0} \]would require a number \(k\) such that
\[ 0\cdot k=6. \]No such number exists because
\[ 0\cdot k=0 \]for every real number \(k\).
Therefore, division by zero is undefined.
Zero in the Numerator Is Allowed
A zero numerator is different from a zero denominator.
For any nonzero number \(a\),
\[ \frac{0}{a}=0. \]For example,
\[ \frac{0}{5}=0. \]The difficulty occurs only when the denominator becomes zero.
Three Important Cases Involving Zero
Compare the following situations.
If the numerator is zero and the denominator is nonzero,
\[ \frac{0}{5}=0. \]If the numerator is nonzero and the denominator is zero,
\[ \frac{5}{0} \]is undefined.
The expression
\[ \frac{0}{0} \]is also undefined. In some later mathematical settings it is called an indeterminate form, because additional information is required to determine the behavior of a related function or limit.
The Domain of an Expression
The domain is the collection of input values for which an expression or function is defined.
For a rational expression,
\[ \frac{P(x)}{Q(x)}, \]the domain consists of all permitted values of \(x\) for which
\[ Q(x)\neq0. \]Unless another number system is specified, this lesson will use real numbers as the default domain.
Excluded Values
A value that makes the denominator zero is called an excluded value or domain restriction.
Consider
\[ \frac{x+2}{x-3}. \]The denominator is zero when
\[ x-3=0. \]Therefore,
\[ x=3. \]The domain restriction is
\[ \boxed{x\neq3}. \]Finding Domain Restrictions
To determine the domain of a rational expression, focus on the denominator.
Set the denominator equal to zero:
\[ Q(x)=0. \]Solve this equation. Every resulting real value must be excluded from the domain.
Example: Linear Denominator
Determine the domain of
\[ \frac{4x-1}{x+6}. \]Set the denominator equal to zero:
\[ x+6=0. \]Therefore,
\[ x=-6. \]The rational expression is defined for every real number except \(-6\):
\[ \boxed{x\neq-6}. \]Domain in Set-Builder Notation
The domain of
\[ \frac{4x-1}{x+6} \]may be written in set-builder notation as
\[ \boxed{\{x\in\mathbb{R}\mid x\neq-6\}}. \]This reads:
“The set of all real numbers \(x\) such that \(x\) is not equal to \(-6\).”
Domain in Interval Notation
The same domain can be written in interval notation as
\[ \boxed{(-\infty,-6)\cup(-6,\infty)}. \]The value \(-6\) divides the real number line into two intervals.
Parentheses are used because \(-6\) is not included.
Example: Two Excluded Values
Determine the domain of
\[ \frac{x+1}{(x-2)(x+5)}. \]The denominator equals zero when either factor equals zero:
\[ x-2=0 \]or
\[ x+5=0. \]Therefore,
\[ x=2 \]or
\[ x=-5. \]The restrictions are
\[ \boxed{x\neq-5,\qquad x\neq2}. \]Interval Notation with Two Excluded Values
If
\[ x\neq-5 \]and
\[ x\neq2, \]then the real domain is
\[ \boxed{ (-\infty,-5)\cup(-5,2)\cup(2,\infty) }. \]Each excluded value breaks the number line into separate intervals.
Factoring Helps Reveal Restrictions
Sometimes the denominator is not already factored.
Consider
\[ \frac{3x+1}{x^2-9}. \]Factor the denominator:
\[ x^2-9=(x-3)(x+3). \]The denominator is therefore zero at
\[ x=3 \]and
\[ x=-3. \]Thus,
\[ \boxed{x\neq-3,\qquad x\neq3}. \]Why Factoring Is So Useful Here
Polynomial factoring turns a denominator into a product.
The zero-product principle then says that
\[ (x-a)(x-b)=0 \]exactly when
\[ x=a \]or
\[ x=b. \]The factoring methods developed in the previous section now become tools for identifying domain restrictions.
Example: Factor a Quadratic Denominator
Determine the domain of
\[ \frac{x-4}{x^2-5x+6}. \]Factor the denominator:
\[ x^2-5x+6=(x-2)(x-3). \]Therefore,
\[ x=2 \]and
\[ x=3 \]make the denominator zero.
The domain restrictions are
\[ \boxed{x\neq2,\qquad x\neq3}. \]A Denominator May Have No Real Zeros
Consider
\[ \frac{x-1}{x^2+4}. \]For every real number \(x\),
\[ x^2\geq0. \]Therefore,
\[ x^2+4\geq4>0. \]The denominator never equals zero for real \(x\).
Thus the real domain is
\[ \boxed{(-\infty,\infty)}. \]The Number System Matters
The expression
\[ \frac{1}{x^2+1} \]has no excluded real values because
\[ x^2+1>0 \]for every real number \(x\).
Over the complex numbers, however,
\[ x^2+1=0 \]has the solutions
\[ x=\pm i. \]Domain statements therefore depend on the number system being used. In elementary real-variable algebra, real numbers are usually assumed unless otherwise stated.
Restrictions Come from the Denominator, Not the Numerator
Consider
\[ \frac{x-4}{x+2}. \]The numerator equals zero when
\[ x=4. \]This is allowed because the denominator at \(x=4\) is
\[ 4+2=6. \]Therefore,
\[ \frac{4-4}{4+2} = \frac{0}{6} = 0. \]The only excluded value is determined by
\[ x+2=0, \]giving
\[ \boxed{x\neq-2}. \]Evaluating a Rational Expression
Consider
\[ R(x)=\frac{x+5}{x-2}. \]At \(x=4\),
\[ R(4) = \frac{4+5}{4-2} = \frac{9}{2}. \]The input \(4\) is permitted because the denominator is not zero.
Attempting to Evaluate at an Excluded Value
For
\[ R(x)=\frac{x+5}{x-2}, \]try \(x=2\):
\[ R(2) = \frac{2+5}{2-2} = \frac{7}{0}. \]Since division by zero is undefined,
\[ \boxed{R(2)\text{ is undefined}}. \]This is why \(2\) is excluded from the domain.
The Natural Domain
The natural domain of an algebraic expression is the largest collection of values for which the expression is defined without imposing additional restrictions.
For
\[ f(x)=\frac{x+1}{x^2-4}, \]factor the denominator:
\[ x^2-4=(x-2)(x+2). \]The natural real domain is therefore
\[ \boxed{ \mathbb{R}\setminus\{-2,2\} }. \]Domain Restrictions Must Be Found Before Canceling
Consider
\[ \frac{x^2-9}{x-3}. \]The original denominator tells us immediately that
\[ x\neq3. \]Factor the numerator:
\[ x^2-9=(x-3)(x+3). \]Then
\[ \frac{(x-3)(x+3)}{x-3}. \]For allowed values with \(x\neq3\), the common factor may be canceled, leaving
\[ x+3. \]But the original restriction remains:
\[ \boxed{x+3,\qquad x\neq3}. \]Why the Restriction Remains After Canceling
The polynomial
\[ x+3 \]by itself is defined at \(x=3\).
Its value there is
\[ 3+3=6. \]But the original rational expression
\[ \frac{x^2-9}{x-3} \]gives
\[ \frac{0}{0} \]when \(x=3\), so it is undefined there.
Thus the expressions agree everywhere in the original domain, but their natural domains are not identical unless the restriction \(x\neq3\) is retained.
Equivalent Formulas Can Carry Different Natural Domains
This example introduces an important algebraic distinction.
\[ \frac{x^2-9}{x-3} = x+3 \]is valid only under the condition
\[ x\neq3. \]A more precise statement is
\[ \boxed{ \frac{x^2-9}{x-3}=x+3 \quad\text{for }x\neq3 }. \]Algebraic simplification preserves values on the original domain. It does not automatically restore values that were excluded from the original expression.
Canceling Does Not Mean Erasing a Restriction
A common factor may disappear from the written formula after simplification, but its original zero remains excluded.
For example,
\[ \frac{(x-5)(x+2)}{x-5} = x+2 \]only when
\[ x\neq5. \]Therefore the simplified form should be interpreted as
\[ \boxed{x+2,\qquad x\neq5}. \]A Preview of Holes
When a common factor cancels but its zero remains excluded from the original domain, the graph of the associated rational function may contain a missing point called a hole or removable discontinuity.
For
\[ f(x)=\frac{x^2-9}{x-3}, \]the simplified formula is
\[ x+3 \]with
\[ x\neq3. \]A later lesson will examine holes and asymptotes systematically.
Restrictions from Repeated Factors
Consider
\[ \frac{x+4}{(x-2)^3(x+1)^2}. \]The denominator equals zero when
\[ x-2=0 \]or
\[ x+1=0. \]Therefore,
\[ \boxed{x\neq2,\qquad x\neq-1}. \]Repetition of a factor affects multiplicity and later graph behavior, but it does not create additional numerical values to exclude.
Restrictions from a Product of Several Factors
Determine the domain of
\[ \frac{2x+1} {x(x-3)(x+4)}. \]The denominator is zero when
\[ x=0, \] \[ x=3, \]or
\[ x=-4. \]Thus,
\[ \boxed{x\neq-4,\qquad x\neq0,\qquad x\neq3}. \]Domain Restrictions from a Perfect Square
Consider
\[ \frac{x+5}{x^2-6x+9}. \]Factor the denominator:
\[ x^2-6x+9=(x-3)^2. \]Therefore the only excluded value is
\[ \boxed{x\neq3}. \]Although the factor \(x-3\) occurs twice, it still produces only one excluded input value.
A Constant Denominator
Consider
\[ \frac{x^2+7x-2}{5}. \]The denominator is the nonzero constant \(5\).
Since
\[ 5\neq0, \]there are no domain restrictions.
The real domain is
\[ \boxed{(-\infty,\infty)}. \]The Zero Polynomial Cannot Serve as a Denominator
A rational expression is defined using a denominator polynomial that is not identically zero.
An expression such as
\[ \frac{x+1}{0} \]is undefined for every value of \(x\).
Likewise,
\[ \frac{x+1}{x-x} \]has denominator
\[ x-x=0 \]for every \(x\), so it has no permissible real inputs.
Expressions with Several Variables
Rational expressions are not limited to one variable.
Consider
\[ \frac{x+y}{x-y}. \]The denominator cannot equal zero:
\[ x-y\neq0. \]Therefore,
\[ \boxed{x\neq y}. \]The domain consists of all ordered pairs \((x,y)\) satisfying that condition.
Multivariable Restrictions Can Form Geometric Sets
For
\[ f(x,y)=\frac{1}{x^2+y^2}, \]we require
\[ x^2+y^2\neq0. \]Over the real numbers,
\[ x^2+y^2=0 \]only when
\[ x=0,\qquad y=0. \]Therefore the domain is all points in the coordinate plane except
\[ \boxed{(0,0)}. \]This is an early example of an algebraic condition describing a geometric restriction.
Another Multivariable Example
Consider
\[ \frac{x+y}{x^2-y^2}. \]Factor the denominator:
\[ x^2-y^2=(x-y)(x+y). \]Therefore,
\[ x-y\neq0 \]and
\[ x+y\neq0. \]Equivalently,
\[ \boxed{x\neq y,\qquad x\neq-y}. \]Rational Expressions and Rational Functions
A rational expression becomes the formula for a rational function when we define a function such as
\[ f(x)=\frac{P(x)}{Q(x)}. \]The natural domain of the function excludes every real zero of \(Q(x)\).
For example,
\[ f(x)=\frac{x+1}{x-4} \]has domain
\[ \boxed{\mathbb{R}\setminus\{4\}}. \]Why Domain Matters Before Graphing
A formula does not automatically define a function at every real number.
For
\[ f(x)=\frac{1}{x-2}, \]there is no graph point with \(x=2\) because the function is not defined there.
Domain restrictions therefore affect the structure of a graph, including possible holes, breaks, and asymptotes.
Why Domain Matters Before Solving Equations
Rational equations frequently require multiplying both sides by denominators.
Before doing so, we record the values that make those denominators zero.
Any proposed solution that violates an original restriction must be rejected.
This principle will become essential when rational equations are studied later in this section.
Restrictions Are Part of the Mathematical Object
Consider the expression
\[ \frac{x^2-4}{x-2}. \]Its algebraic structure includes both the formula and the condition
\[ x\neq2. \]After factoring,
\[ \frac{(x-2)(x+2)}{x-2}, \]we may simplify to
\[ x+2, \]but only while carrying the restriction
\[ \boxed{x\neq2}. \]Good algebra tracks both symbolic transformations and the conditions under which those transformations are valid.
Example: Determine Restrictions Before Simplifying
Consider
\[ \frac{x^2+x-6}{x^2-4}. \]Begin with the denominator:
\[ x^2-4=(x-2)(x+2). \]Therefore,
\[ \boxed{x\neq2,\qquad x\neq-2}. \]Now factor the numerator:
\[ x^2+x-6=(x+3)(x-2). \]The expression becomes
\[ \frac{(x+3)(x-2)}{(x-2)(x+2)}. \]The factor \(x-2\) will cancel in the next lesson, but both original restrictions must still be retained.
Domain First, Simplification Second
A reliable habit is:
- inspect the original denominator;
- factor it when useful;
- find every value that makes it zero;
- record those values as restrictions;
- only then simplify the expression.
This order prevents restrictions from disappearing unnoticed during cancellation.
Common Error: Looking Only at the Numerator
Suppose we are given
\[ \frac{x-7}{x+3}. \]A student sees that the numerator is zero at \(x=7\) and excludes \(7\) from the domain.
That is incorrect.
The denominator determines restrictions:
\[ x+3=0. \]Therefore,
\[ \boxed{x\neq-3}. \]The value \(x=7\) is allowed and makes the entire expression equal zero.
Common Error: Forgetting to Factor the Denominator
Consider
\[ \frac{1}{x^2-x-6}. \]The denominator may not reveal its zeros immediately.
Factor:
\[ x^2-x-6=(x-3)(x+2). \]Therefore,
\[ \boxed{x\neq3,\qquad x\neq-2}. \]Factoring often converts a difficult domain question into a simple zero-product problem.
Common Error: Canceling Terms Instead of Factors
Cancellation applies to multiplicative factors, not individual terms joined by addition or subtraction.
For example,
\[ \frac{x+3}{x} \]cannot be simplified by “canceling the \(x\).”
The numerator \(x+3\) is a sum, not a product containing \(x\) as a factor.
The only immediate domain restriction is
\[ \boxed{x\neq0}. \]Common Error: Losing a Restriction After Cancellation
Suppose
\[ \frac{(x-4)(x+1)}{x-4} \]is simplified to
\[ x+1. \]Writing only \(x+1\) loses information from the original expression.
The correct simplified statement is
\[ \boxed{x+1,\qquad x\neq4}. \]Common Error: Treating Undefined as Zero
Consider
\[ \frac{x-3}{x-3}. \]For \(x\neq3\),
\[ \frac{x-3}{x-3}=1. \]But at \(x=3\),
\[ \frac{0}{0} \]is undefined.
Therefore,
\[ \boxed{ \frac{x-3}{x-3}=1 \quad\text{for }x\neq3 }. \]The expression does not become zero at the excluded value.
Common Error: Assuming Every Denominator Has a Real Zero
Consider
\[ \frac{x}{x^2+9}. \]Solving
\[ x^2+9=0 \]gives
\[ x^2=-9. \]There are no real solutions.
Therefore, in a real-variable setting, the domain is
\[ \boxed{(-\infty,\infty)}. \]A Reliable Procedure for Finding the Domain
- Identify every denominator in the original expression.
- Set each denominator equal to zero.
- Factor when useful.
- Solve for every real value that makes a denominator zero.
- Exclude those values from the domain.
- Record the restrictions before performing any cancellation.
- Express the domain using inequalities, set notation, or interval notation as appropriate.
Example: Full Domain Workflow
Determine the domain of
\[ \frac{x^2+2x-8}{x^2+x-12}. \]Factor the denominator:
\[ x^2+x-12=(x+4)(x-3). \]Set each factor equal to zero:
\[ x+4=0 \]gives
\[ x=-4, \]while
\[ x-3=0 \]gives
\[ x=3. \]Therefore,
\[ \boxed{x\neq-4,\qquad x\neq3}. \]In interval notation,
\[ \boxed{ (-\infty,-4)\cup(-4,3)\cup(3,\infty) }. \]Example: A Restriction That Will Later Cancel
Determine the restrictions for
\[ \frac{x^2-5x+6}{x^2-x-2}. \]Factor the denominator:
\[ x^2-x-2=(x-2)(x+1). \]Therefore,
\[ \boxed{x\neq2,\qquad x\neq-1}. \]The numerator factors as
\[ x^2-5x+6=(x-2)(x-3). \]Although \(x-2\) appears in both numerator and denominator, the restriction \(x\neq2\) came from the original denominator and must remain after any later simplification.
Practice 1: Identify the Rational Expression
Which of the following is a rational expression?
\[ A.\quad \frac{x+2}{x-4} \] \[ B.\quad \sqrt{x}+3 \] \[ C.\quad 2^x \]Assume the standard definition of a rational expression as a quotient of polynomials.
Practice 2: Linear Denominator
Determine the real domain of
\[ \frac{3x+5}{x-7}. \]Practice 3: Two Linear Factors
Determine the real domain of
\[ \frac{x}{(x+2)(x-5)}. \]Practice 4: Factor the Denominator
Determine the real domain of
\[ \frac{x+4}{x^2-16}. \]Practice 5: Quadratic Denominator
Determine the real domain of
\[ \frac{2x-1}{x^2+x-6}. \]Practice 6: No Real Restriction
Determine the real domain of
\[ \frac{x+7}{x^2+16}. \]Practice 7: Preserve the Original Restriction
Determine the domain restriction before simplifying:
\[ \frac{x^2-25}{x-5}. \]Practice 8: Several Restrictions
Determine the real domain of
\[ \frac{x^2+1}{x(x-2)(x+6)}. \]Practice 9: Repeated Factor
Determine the real domain of
\[ \frac{x-1}{(x+3)^2}. \]Practice 10: Identify the Error
A student simplifies
\[ \frac{(x-2)(x+4)}{x-2} \]to
\[ x+4 \]and concludes that the domain is all real numbers.
Explain the error.
Practice Answers
Practice 1
Choice \(A\) is a quotient of two polynomials:
\[ \boxed{\frac{x+2}{x-4}}. \]The expressions involving \(\sqrt{x}\) and \(2^x\) are not rational expressions under this definition.
Practice 2
Set the denominator equal to zero:
\[ x-7=0. \]Therefore,
\[ \boxed{x\neq7}. \]In interval notation,
\[ \boxed{(-\infty,7)\cup(7,\infty)}. \]Practice 3
The denominator is zero when
\[ x+2=0 \]or
\[ x-5=0. \]Therefore,
\[ \boxed{x\neq-2,\qquad x\neq5}. \]Practice 4
Factor the denominator:
\[ x^2-16=(x-4)(x+4). \]Therefore,
\[ \boxed{x\neq-4,\qquad x\neq4}. \]Practice 5
Factor:
\[ x^2+x-6=(x+3)(x-2). \]Therefore,
\[ \boxed{x\neq-3,\qquad x\neq2}. \]Practice 6
Since
\[ x^2+16>0 \]for every real \(x\), the denominator never equals zero.
Thus,
\[ \boxed{(-\infty,\infty)}. \]Practice 7
The original denominator gives
\[ x-5\neq0, \]so
\[ \boxed{x\neq5}. \]Factoring the numerator gives
\[ x^2-25=(x-5)(x+5). \]The expression will simplify to \(x+5\), but only for
\[ \boxed{x\neq5}. \]Practice 8
The denominator is zero when
\[ x=0,\qquad x=2,\qquad x=-6. \]Therefore,
\[ \boxed{x\neq-6,\qquad x\neq0,\qquad x\neq2}. \]Practice 9
The denominator
\[ (x+3)^2 \]is zero only when
\[ x=-3. \]Therefore,
\[ \boxed{x\neq-3}. \]Practice 10
The cancellation is valid only where the canceled denominator factor is nonzero.
The original denominator gives
\[ x-2\neq0, \]so
\[ x\neq2. \]Therefore the correct simplified statement is
\[ \boxed{x+4,\qquad x\neq2}. \]The value \(2\) remains excluded because the original rational expression is undefined there.
What to Remember
A rational expression has the form
\[ \boxed{\frac{P(x)}{Q(x)}} \]where \(P(x)\) and \(Q(x)\) are polynomials and the denominator cannot be zero:
\[ \boxed{Q(x)\neq0}. \]The essential habits are:
- identify the original denominator;
- factor the denominator when useful;
- find every value that makes it zero;
- exclude those values from the domain;
- record restrictions before simplifying;
- retain original restrictions after cancellation;
- distinguish a zero numerator from a zero denominator.
Domain restrictions are not bookkeeping added after the algebra. They are part of the conditions under which the algebraic expression is defined.
Where We Go Next
Rational expressions often contain common factors in their numerators and denominators. Factoring can expose those common factors and allow an expression to be rewritten in a simpler form.
The next lesson, Simplifying and Canceling Rational Expressions, develops the rules for legitimate cancellation, explains why only factors may be canceled, and shows how to simplify while preserving every restriction inherited from the original denominator.