The Central Idea
Suppose a rational expression contains the same nonzero factor in both its numerator and denominator:
\[ \frac{ac}{bc}. \]Because multiplication is associative and \(c\neq0\),
\[ \frac{ac}{bc} = \frac{a}{b}\cdot\frac{c}{c}. \]Since
\[ \frac{c}{c}=1, \]we obtain
\[ \boxed{ \frac{ac}{bc}=\frac{a}{b} \quad\text{for }c\neq0 }. \]This is the algebra behind cancellation.
Cancellation Is Division by a Common Factor
Canceling is not a special operation that makes symbols disappear.
It is ordinary division of both numerator and denominator by the same nonzero factor.
For example,
\[ \frac{6x}{9x} \]has a common factor of \(3x\). Divide numerator and denominator by \(3x\):
\[ \frac{6x\div3x}{9x\div3x} = \frac{2}{3}, \]provided
\[ x\neq0. \]Therefore,
\[ \boxed{ \frac{6x}{9x}=\frac23 \quad\text{for }x\neq0 }. \]Why the Common Factor Must Be Nonzero
The cancellation
\[ \frac{ac}{bc} \to \frac{a}{b} \]relies on the step
\[ \frac{c}{c}=1. \]This is true only when
\[ c\neq0. \]If \(c=0\), then
\[ \frac{c}{c}=\frac00, \]which is undefined.
Domain restrictions therefore remain essential even when a factor later cancels.
Find Restrictions Before Simplifying
Consider
\[ \frac{x^2-9}{x-3}. \]Before doing anything else, inspect the original denominator:
\[ x-3\neq0. \]Therefore,
\[ \boxed{x\neq3}. \]Only after recording that restriction should we simplify.
Factor Before You Cancel
The numerator
\[ x^2-9 \]is a difference of squares:
\[ x^2-9=(x-3)(x+3). \]Therefore,
\[ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3}. \]Now the common factor \(x-3\) is visible.
Cancel the Common Factor
For \(x\neq3\),
\[ \frac{(x-3)(x+3)}{x-3} = (x+3)\frac{x-3}{x-3}. \]Since
\[ \frac{x-3}{x-3}=1, \]we get
\[ \boxed{x+3,\qquad x\neq3}. \]The simplified formula is \(x+3\), but the original domain restriction remains.
Factors Cancel; Terms Do Not
This distinction is the most important rule in simplifying rational expressions.
In
\[ \frac{x(x+4)}{x}, \]\(x\) is a factor of the entire numerator and denominator, so it may be canceled:
\[ \boxed{x+4,\qquad x\neq0}. \]But in
\[ \frac{x+4}{x}, \]the \(x\) in the numerator is only one term of a sum. It is not a factor of the entire numerator.
Nothing cancels.
Why Terms Cannot Be Canceled Across Addition
Consider the incorrect step
\[ \frac{x+4}{x} \not= 1+4. \]The numerator is
\[ x+4, \]not
\[ x\cdot4. \]Cancellation is based on multiplicative factors. Addition does not create a common factor automatically.
If desired, the expression can be separated correctly:
\[ \frac{x+4}{x} = \frac{x}{x}+\frac4x = 1+\frac4x, \]with
\[ x\neq0. \]Example: Simplifying Numerical and Variable Factors
Simplify
\[ \frac{18x^3}{24x^2}. \]The greatest common numerical factor is \(6\), and both numerator and denominator contain \(x^2\):
\[ \frac{18x^3}{24x^2} = \frac{6x^2(3x)}{6x^2(4)}. \]Cancel \(6x^2\):
\[ \boxed{ \frac{18x^3}{24x^2} = \frac{3x}{4}, \qquad x\neq0 }. \]Using Exponent Laws to Simplify
Monomial rational expressions can also be simplified using the quotient rule for exponents:
\[ \frac{x^m}{x^n}=x^{m-n}, \qquad x\neq0. \]For example,
\[ \frac{x^7}{x^3} = x^{7-3} = x^4, \]with the original restriction
\[ x\neq0. \]Example: Several Variables
Simplify
\[ \frac{15x^4y^3}{25x^2y}. \]Reduce the numerical coefficients:
\[ \frac{15}{25}=\frac35. \]Subtract exponents for the common variable factors:
\[ \frac{x^4}{x^2}=x^2, \] \[ \frac{y^3}{y}=y^2. \]Therefore,
\[ \boxed{ \frac{15x^4y^3}{25x^2y} = \frac{3x^2y^2}{5}, \qquad x\neq0,\ y\neq0 }. \]Canceling a Common Binomial Factor
A common factor does not have to be a single number or variable.
Consider
\[ \frac{(x+5)(x-2)}{(x+5)(x+7)}. \]The common factor is the entire binomial
\[ x+5. \]Canceling it gives
\[ \boxed{ \frac{x-2}{x+7}, \qquad x\neq-5,\ -7 }. \]Both restrictions come from the original denominator.
Factorization Reveals Hidden Common Factors
Consider
\[ \frac{x^2+5x+6}{x^2+4x+3}. \]No common factor is immediately visible in expanded form.
Factor both polynomials:
\[ x^2+5x+6=(x+2)(x+3), \] \[ x^2+4x+3=(x+1)(x+3). \]Therefore,
\[ \frac{x^2+5x+6}{x^2+4x+3} = \frac{(x+2)(x+3)}{(x+1)(x+3)}. \]Cancel \(x+3\):
\[ \boxed{ \frac{x+2}{x+1}, \qquad x\neq-3,\ -1 }. \]Why We Factor the Numerator and Denominator Completely
Partial factoring can hide a common factor.
Suppose we want to simplify
\[ \frac{2x^2+8x}{2x^2+12x+16}. \]Factor the numerator completely:
\[ 2x^2+8x=2x(x+4). \]Factor the denominator completely:
\[ 2x^2+12x+16 = 2(x^2+6x+8) = 2(x+2)(x+4). \]Now the full structure is visible.
Example: Complete Factorization Before Cancellation
Continuing with
\[ \frac{2x^2+8x}{2x^2+12x+16}, \]the original denominator factors as
\[ 2(x+2)(x+4), \]so
\[ x\neq-2,\qquad x\neq-4. \]Then
\[ \frac{2x(x+4)}{2(x+2)(x+4)}. \]Cancel the common factors \(2\) and \(x+4\):
\[ \boxed{ \frac{x}{x+2}, \qquad x\neq-2,\ -4 }. \]Difference of Squares in Rational Expressions
Special factoring patterns are especially useful in rational expressions.
Consider
\[ \frac{x^2-16}{x^2-8x+16}. \]Factor the numerator:
\[ x^2-16=(x-4)(x+4). \]Factor the denominator:
\[ x^2-8x+16=(x-4)^2. \]The original restriction is
\[ x\neq4. \]Cancel one factor of \(x-4\):
\[ \boxed{ \frac{x+4}{x-4}, \qquad x\neq4 }. \]Canceling Powers of a Common Factor
Consider
\[ \frac{(x-2)^5}{(x-2)^3}. \]Three copies of \(x-2\) cancel from numerator and denominator:
\[ \frac{(x-2)^5}{(x-2)^3} = (x-2)^{5-3}. \]Therefore,
\[ \boxed{(x-2)^2,\qquad x\neq2}. \]Canceling Only Part of a Power
Simplify
\[ \frac{(x+1)^2(x-5)} {(x+1)^4(x+3)}. \]The factor \(x+1\) appears twice in the numerator and four times in the denominator.
Cancel two copies:
\[ \frac{x-5} {(x+1)^2(x+3)}. \]The original restrictions are
\[ x\neq-1,\qquad x\neq-3. \]Thus,
\[ \boxed{ \frac{x-5}{(x+1)^2(x+3)}, \qquad x\neq-1,\ -3 }. \]Opposite Factors
Sometimes two factors look different only because their signs are reversed.
Since
\[ a-b=-(b-a), \]the factors \(x-3\) and \(3-x\) differ by a factor of \(-1\).
This relationship must be handled before cancellation.
Example: Canceling Opposite Binomials
Simplify
\[ \frac{x-3}{3-x}. \]Rewrite the denominator:
\[ 3-x=-(x-3). \]Then
\[ \frac{x-3}{3-x} = \frac{x-3}{-(x-3)} = -1, \]provided
\[ x\neq3. \]Therefore,
\[ \boxed{-1,\qquad x\neq3}. \]Example: Opposite Factors Inside a Larger Expression
Simplify
\[ \frac{(x+2)(x-5)} {(5-x)(x+7)}. \]Since
\[ 5-x=-(x-5), \]rewrite the denominator:
\[ -(x-5)(x+7). \]Cancel \(x-5\):
\[ \boxed{ -\frac{x+2}{x+7}, \qquad x\neq5,\ -7 }. \]Negative Factors Can Be Moved
A negative sign may be placed in the numerator, denominator, or in front of an entire fraction:
\[ \frac{-a}{b} = -\frac{a}{b} = \frac{a}{-b}. \]For example,
\[ \frac{x-4}{4-x} = -1 \]for
\[ x\neq4. \]Recognizing equivalent sign placements makes many rational expressions easier to simplify.
A Common Factor May Be a Polynomial
Consider
\[ \frac{(x^2+1)(x-4)} {(x^2+1)(x+2)}. \]The entire polynomial
\[ x^2+1 \]is a common factor.
Over the real numbers, \(x^2+1\neq0\) automatically, so it may be canceled for every real \(x\).
The remaining denominator gives the restriction \(x\neq-2\):
\[ \boxed{ \frac{x-4}{x+2}, \qquad x\neq-2 }. \]Example: Common Quadratic Factor
Simplify
\[ \frac{(x^2+3x+2)(x-1)} {(x^2+3x+2)(x+5)}. \]Before canceling, factor the original denominator factor:
\[ x^2+3x+2=(x+1)(x+2). \]Therefore,
\[ x\neq-1,\qquad x\neq-2,\qquad x\neq-5. \]Cancel the common quadratic factor:
\[ \boxed{ \frac{x-1}{x+5}, \qquad x\neq-1,\ -2,\ -5 }. \]Why the Original Domain Must Be Preserved
Consider
\[ f(x)=\frac{(x-2)(x+4)}{x-2}. \]The formula simplifies to
\[ x+4. \]But the original expression is undefined at \(x=2\).
Therefore,
\[ f(x)=x+4 \]only on the domain
\[ x\neq2. \]At \(x=2\), the original expression has a hole rather than a defined function value.
Simplified Form and Original Form Agree on the Original Domain
If
\[ \frac{(x-a)P(x)}{(x-a)Q(x)} \]simplifies to
\[ \frac{P(x)}{Q(x)}, \]then the two formulas have the same value wherever the original expression is defined and the cancellation is valid.
The simplified formula may have a larger natural domain, but that does not retroactively enlarge the domain of the original expression.
Equivalent Expressions with Conditions
A precise algebraic statement includes the condition under which cancellation is valid.
For example,
\[ \boxed{ \frac{x^2-4}{x-2}=x+2 \quad\text{for }x\neq2 }. \]Without the condition, the right side is defined at \(x=2\) while the left side is not.
Algebraic equivalence should therefore be interpreted together with domain information.
Canceling All Variable Factors
Consider
\[ \frac{12x^3y^2}{18x^3y^2}. \]Cancel the common variable factors and reduce the numerical fraction:
\[ \frac{12}{18} = \frac23. \]Thus,
\[ \boxed{ \frac23, \qquad x\neq0,\ y\neq0 }. \]Even though no variable remains in the simplified formula, the original restrictions remain.
When Nothing Cancels
Not every rational expression simplifies.
Consider
\[ \frac{x+2}{x+5}. \]The numerator and denominator are different binomial factors.
There is no common factor other than \(1\), so the expression is already simplified.
Its restriction is
\[ \boxed{x\neq-5}. \]Different-Looking Terms May Still Contain a Common Factor
Consider
\[ \frac{2x+6}{4x+12}. \]At first glance, the numerator and denominator are different sums.
Factor both:
\[ 2x+6=2(x+3), \] \[ 4x+12=4(x+3). \]Then
\[ \frac{2(x+3)}{4(x+3)} = \frac12, \]with
\[ x\neq-3. \]Therefore,
\[ \boxed{\frac12,\qquad x\neq-3}. \]Example: GCF Factoring Before Cancellation
Simplify
\[ \frac{6x^2+12x}{9x^2+18x}. \]Factor the numerator:
\[ 6x^2+12x=6x(x+2). \]Factor the denominator:
\[ 9x^2+18x=9x(x+2). \]The original denominator is zero at
\[ x=0 \]and
\[ x=-2. \]Cancel \(x(x+2)\) and reduce \(6/9\):
\[ \boxed{ \frac23, \qquad x\neq0,\ -2 }. \]Example: Trinomial Factoring Before Cancellation
Simplify
\[ \frac{x^2+7x+12}{x^2+5x+4}. \]Factor:
\[ x^2+7x+12=(x+3)(x+4), \] \[ x^2+5x+4=(x+1)(x+4). \]The original restrictions are
\[ x\neq-1,\qquad x\neq-4. \]Cancel \(x+4\):
\[ \boxed{ \frac{x+3}{x+1}, \qquad x\neq-1,\ -4 }. \]Example: Several Factors
Simplify
\[ \frac{3(x-1)^2(x+5)} {6(x-1)(x+2)(x+5)}. \]Reduce the numerical factors:
\[ \frac36=\frac12. \]Cancel one factor \(x-1\) and one factor \(x+5\):
\[ \frac{x-1}{2(x+2)}. \]The original denominator gives
\[ x\neq1,\qquad x\neq-2,\qquad x\neq-5. \]Thus,
\[ \boxed{ \frac{x-1}{2(x+2)}, \qquad x\neq1,\ -2,\ -5 }. \]Example: Multivariable Polynomial Factors
Simplify
\[ \frac{x^2y-xy^2}{xy}. \]Factor the numerator:
\[ x^2y-xy^2=xy(x-y). \]Therefore,
\[ \frac{xy(x-y)}{xy}. \]Cancel \(xy\):
\[ \boxed{x-y,\qquad x\neq0,\ y\neq0}. \]Example: Factor by Grouping Before Cancellation
Simplify
\[ \frac{x^3+3x^2+2x+6}{x^2+2}. \]Factor the numerator by grouping:
\[ (x^3+3x^2)+(2x+6) \] \[ = x^2(x+3)+2(x+3) \] \[ = (x+3)(x^2+2). \]Therefore,
\[ \frac{(x+3)(x^2+2)}{x^2+2}. \]Since \(x^2+2\neq0\) for real \(x\), it cancels with no real domain restriction:
\[ \boxed{x+3}. \]Simplifying Does Not Mean Expanding
In rational expressions, factored form is often more useful than expanded form.
For example,
\[ \frac{(x+2)(x-3)}{(x+5)(x-3)} \]should not be expanded first.
The common factor is already visible:
\[ \boxed{ \frac{x+2}{x+5}, \qquad x\neq3,\ -5 }. \]Expanding would hide the very structure needed for cancellation.
Simplest Form
A rational expression is generally considered to be in simplest form when:
- the numerator and denominator are factored enough to identify common factors;
- all common nonconstant factors have been canceled;
- numerical coefficients are reduced;
- negative signs are written clearly;
- the original domain restrictions are preserved.
The exact preferred appearance may vary by context, but these principles remain the same.
Common Error: Canceling Terms Across Addition
A student sees
\[ \frac{x+5}{x} \]and tries to cancel the \(x\).
This is invalid because the numerator is a sum.
The \(x\) is not a factor of the entire numerator.
The expression does not simplify by cancellation.
Common Error: Canceling Terms Across Subtraction
The same mistake can occur in
\[ \frac{x-4}{x}. \]The numerator is not the product \(x(-4)\).
Therefore the \(x\) cannot be canceled.
A correct alternate form is
\[ 1-\frac4x, \]with
\[ x\neq0. \]Common Error: Canceling Before Factoring
Consider
\[ \frac{x^2-9}{x-3}. \]It would be incorrect to cancel an \(x\) or a \(3\) from the expanded numerator.
First factor:
\[ x^2-9=(x-3)(x+3). \]Only then is the common factor \(x-3\) available for cancellation.
Common Error: Forgetting Restrictions
Suppose
\[ \frac{x^2-4}{x-2} \]is simplified to
\[ x+2. \]The simplification is incomplete if the original restriction is omitted.
The correct result is
\[ \boxed{x+2,\qquad x\neq2}. \]Common Error: Canceling a Factor That Is Not Common
Consider
\[ \frac{(x+2)(x+3)} {(x+2)(x+5)}. \]Only \(x+2\) is common to the entire numerator and denominator.
After canceling it, we obtain
\[ \frac{x+3}{x+5}. \]The remaining factors \(x+3\) and \(x+5\) are different and cannot be canceled.
Common Error: Treating Opposite Factors as Identical
The factors
\[ x-4 \]and
\[ 4-x \]are not identical.
Instead,
\[ 4-x=-(x-4). \]Therefore,
\[ \frac{x-4}{4-x} = -1, \]not \(1\), with \(x\neq4\).
Common Error: Losing a Negative Sign
Consider
\[ \frac{3-x}{x-3}. \]Since
\[ 3-x=-(x-3), \]the expression becomes
\[ \frac{-(x-3)}{x-3}. \]Therefore,
\[ \boxed{-1,\qquad x\neq3}. \]Common Error: Reducing Exponents Incorrectly
For
\[ \frac{x^5}{x^2}, \]the correct exponent rule is
\[ x^{5-2}=x^3. \]It is not
\[ x^{5/2} \]and not
\[ x^7. \]Division of like bases subtracts exponents.
Common Error: Canceling an Entire Denominator with One Numerator Term
Consider
\[ \frac{x(x+2)+3}{x+2}. \]The denominator \(x+2\) is a factor of the first numerator term \(x(x+2)\), but it is not a factor of the entire numerator
\[ x(x+2)+3. \]Therefore the denominator cannot be canceled from the whole fraction.
A Reliable Simplification Procedure
- Identify the original denominator.
- Find and record all domain restrictions.
- Factor the numerator completely when possible.
- Factor the denominator completely when possible.
- Rewrite opposite factors using a factor of \(-1\) when necessary.
- Identify factors common to the entire numerator and denominator.
- Cancel only those common nonzero factors.
- Reduce numerical coefficients.
- Inspect the result for additional simplification.
- Carry the original domain restrictions into the final answer.
Example: Full Simplification Workflow
Simplify
\[ \frac{x^2-5x+6}{x^2-x-2}. \]First factor the denominator:
\[ x^2-x-2=(x-2)(x+1). \]Therefore,
\[ x\neq2,\qquad x\neq-1. \]Factor the numerator:
\[ x^2-5x+6=(x-2)(x-3). \]Now
\[ \frac{(x-2)(x-3)} {(x-2)(x+1)}. \]Cancel \(x-2\):
\[ \boxed{ \frac{x-3}{x+1}, \qquad x\neq2,\ -1 }. \]Example: Full Workflow with Numerical Factors
Simplify
\[ \frac{6x^2-24} {9x^2-36x+36}. \]Factor the numerator:
\[ 6x^2-24 = 6(x^2-4) = 6(x-2)(x+2). \]Factor the denominator:
\[ 9x^2-36x+36 = 9(x^2-4x+4) = 9(x-2)^2. \]The original restriction is
\[ x\neq2. \]Cancel one factor \(x-2\) and reduce \(6/9\):
\[ \boxed{ \frac{2(x+2)}{3(x-2)}, \qquad x\neq2 }. \]Example: Full Workflow with Opposite Factors
Simplify
\[ \frac{x^2-9} {3-x}. \]Factor the numerator:
\[ x^2-9=(x-3)(x+3). \]Rewrite the denominator:
\[ 3-x=-(x-3). \]Therefore,
\[ \frac{(x-3)(x+3)} {-(x-3)}. \]The original restriction is
\[ x\neq3. \]Cancel \(x-3\):
\[ \boxed{-(x+3),\qquad x\neq3}. \]Checking by Substitution
A quick way to check a simplification is to substitute an allowed value into both forms.
Suppose
\[ \frac{x^2-9}{x-3} \]simplifies to
\[ x+3, \qquad x\neq3. \]At \(x=5\), the original expression gives
\[ \frac{25-9}{5-3} = \frac{16}{2} = 8. \]The simplified expression gives
\[ 5+3=8. \]The values agree.
Checking by Refactoring
A stronger algebraic check is to verify that the canceled factors can be restored.
Suppose
\[ \frac{(x+2)(x-4)} {(x+5)(x-4)} \]simplifies to
\[ \frac{x+2}{x+5}. \]Multiplying numerator and denominator of the simplified expression by the same nonzero factor \(x-4\) recovers the original form.
This confirms the simplification for \(x\neq4\), together with the original restriction \(x\neq-5\).
Practice 1: Monomial Factors
Simplify and state the restrictions:
\[ \frac{20x^4}{30x^2}. \]Practice 2: Shared Binomial
Simplify and state the restrictions:
\[ \frac{(x+3)(x-5)} {(x+3)(x+2)}. \]Practice 3: Factor First
Simplify and state the restrictions:
\[ \frac{x^2-25}{x-5}. \]Practice 4: Factor Both Polynomials
Simplify and state the restrictions:
\[ \frac{x^2+6x+8} {x^2+5x+6}. \]Practice 5: Numerical and Polynomial Factors
Simplify and state the restrictions:
\[ \frac{8x^2+16x} {12x^2+24x}. \]Practice 6: Opposite Factors
Simplify and state the restrictions:
\[ \frac{x-7}{7-x}. \]Practice 7: Several Factors
Simplify and state the restrictions:
\[ \frac{4(x-1)^3(x+2)} {10(x-1)(x+2)^2}. \]Practice 8: Difference of Squares
Simplify and state the restrictions:
\[ \frac{x^2-16} {x^2+x-20}. \]Practice 9: Multivariable Expression
Simplify and state the restrictions:
\[ \frac{18x^3y^2} {24xy}. \]Practice 10: Identify the Error
A student writes
\[ \frac{x+6}{x} = 6. \]Explain why this cancellation is invalid and give a correct equivalent form.
Practice Answers
Practice 1
Reduce \(20/30\) to \(2/3\) and cancel \(x^2\):
\[ \boxed{ \frac{20x^4}{30x^2} = \frac{2x^2}{3}, \qquad x\neq0 }. \]Practice 2
The common factor is \(x+3\):
\[ \boxed{ \frac{x-5}{x+2}, \qquad x\neq-3,\ -2 }. \]Practice 3
Factor the numerator:
\[ x^2-25=(x-5)(x+5). \]Cancel \(x-5\):
\[ \boxed{x+5,\qquad x\neq5}. \]Practice 4
Factor:
\[ x^2+6x+8=(x+2)(x+4), \] \[ x^2+5x+6=(x+2)(x+3). \]Therefore,
\[ \boxed{ \frac{x+4}{x+3}, \qquad x\neq-2,\ -3 }. \]Practice 5
Factor:
\[ 8x^2+16x=8x(x+2), \] \[ 12x^2+24x=12x(x+2). \]Cancel \(x(x+2)\) and reduce \(8/12\):
\[ \boxed{ \frac23, \qquad x\neq0,\ -2 }. \]Practice 6
Since
\[ 7-x=-(x-7), \]we have
\[ \boxed{-1,\qquad x\neq7}. \]Practice 7
Reduce \(4/10\) to \(2/5\), cancel one factor \(x-1\), and cancel one factor \(x+2\):
\[ \boxed{ \frac{2(x-1)^2}{5(x+2)}, \qquad x\neq1,\ -2 }. \]Practice 8
Factor:
\[ x^2-16=(x-4)(x+4), \] \[ x^2+x-20=(x+5)(x-4). \]Therefore,
\[ \boxed{ \frac{x+4}{x+5}, \qquad x\neq4,\ -5 }. \]Practice 9
Reduce the numerical coefficients and subtract exponents:
\[ \frac{18}{24}=\frac34, \] \[ \frac{x^3}{x}=x^2, \] \[ \frac{y^2}{y}=y. \]Thus,
\[ \boxed{ \frac{3x^2y}{4}, \qquad x\neq0,\ y\neq0 }. \]Practice 10
The numerator \(x+6\) is a sum. The \(x\) is not a factor of the entire numerator, so it cannot be canceled.
Split the fraction term by term:
\[ \frac{x+6}{x} = \frac{x}{x}+\frac6x. \]Therefore,
\[ \boxed{ 1+\frac6x, \qquad x\neq0 }. \]What to Remember
Cancellation is based on the identity
\[ \boxed{ \frac{ac}{bc}=\frac{a}{b} \quad\text{when }c\neq0 }. \]The essential habits are:
- find domain restrictions from the original denominator first;
- factor before looking for cancellation;
- cancel only factors common to the entire numerator and denominator;
- never cancel individual terms across addition or subtraction;
- handle opposite factors by extracting \(-1\);
- reduce numerical coefficients and repeated powers carefully;
- preserve every restriction inherited from the original denominator.
The single most important rule is: factors cancel; terms do not.
Where We Go Next
Simplifying a single rational expression prepares us to combine several rational expressions in one calculation. Addition, subtraction, multiplication, and division each require careful attention to factors, denominators, and domain restrictions.
The next lesson, Operations with Rational Expressions, develops systematic methods for multiplying, dividing, adding, and subtracting rational expressions while maintaining valid domains and simplifying the results.