Rational Functions

2. Simplifying Rational Expressions

Learning outcomes
  • I can factor numerators and denominators.
  • I can simplify rational expressions correctly.
  • I can identify restrictions before simplifying.
  • I can recognize equivalent rational expressions.
  • I can verify simplified results.

Introduction

Fractions are not limited to numbers. In algebra, fractions can contain variables, expressions, and polynomials. These are called rational expressions, and simplifying them is an essential skill in mathematics.

Consider the numerical fraction:

12/18 = 2/3

We simplify this fraction by dividing the numerator and denominator by their common factor of 6.

The same principle applies to algebraic fractions. For example:

(x² − 9)/(x − 3)

By factoring the numerator, we can identify a common factor and simplify the expression.

However, simplifying rational expressions requires special care. Unlike ordinary numerical fractions, rational expressions may contain values that make the denominator equal to zero. These values must be identified and excluded, even if the factors responsible are cancelled during simplification.

In this lesson, we will learn how to factor numerators and denominators, simplify rational expressions correctly, identify restrictions, recognize equivalent expressions, and verify our results using algebraic and numerical methods.

What Is a Rational Expression?

A rational expression is a fraction whose numerator and denominator are polynomials.

Its general form is:

\[ \frac{P(x)}{Q(x)} \]

where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial.

Examples include:

  • (x + 3)/(x − 5)

  • (x² − 16)/(x + 4)

  • (2x² + 6x)/(4x)

  • (x² + 5x + 6)/(x² − 9)

The numerator is the expression above the fraction bar, while the denominator is the expression below it.

The fundamental rule is that the denominator cannot equal zero.

Factoring Numerators and Denominators

Factoring is the process of rewriting an expression as a product of simpler expressions.

For example:

x² + 5x + 6 = (x + 2)(x + 3)

Factoring is especially useful when simplifying rational expressions because it reveals common factors that can be cancelled.

Common Factoring Techniques

Technique

Example

Factored form

Greatest common factor

6x² + 9x

3x(2x + 3)

Difference of squares

x² − 25

(x − 5)(x + 5)

Simple quadratic trinomial

x² + 7x + 12

(x + 3)(x + 4)

Quadratic with negative constant

x² − x − 12

(x − 4)(x + 3)

Perfect square trinomial

x² + 6x + 9

(x + 3)²

Grouping

x³ + 2x² + 3x + 6

(x + 2)(x² + 3)

Factoring Out the Greatest Common Factor

Consider:

6x² + 12x

Both terms contain the common factor 6x.

Therefore:

6x² + 12x = 6x(x + 2)

This technique should generally be checked first when factoring a polynomial.

Factoring a Difference of Squares

The identity is:

\[ a^2-b^2=(a-b)(a+b) \]

For example:

x² − 49 = (x − 7)(x + 7)

And:

4x² − 25 = (2x − 5)(2x + 5)

Factoring Quadratic Trinomials

Consider:

x² + 8x + 15

We need two numbers that:

  • Multiply to give 15.

  • Add to give 8.

The numbers are 3 and 5.

Therefore:

x² + 8x + 15 = (x + 3)(x + 5)

Factoring Quadratics with Leading Coefficients

Consider:

2x² + 7x + 3

We seek two binomial factors whose product gives the original quadratic.

2x² + 7x + 3 = (2x + 1)(x + 3)

Check by expanding:

(2x + 1)(x + 3)

= 2x² + 6x + x + 3

= 2x² + 7x + 3

This confirms the factorization.

The Fundamental Rule of Simplifying Rational Expressions

To simplify a rational expression, divide the numerator and denominator by their common nonzero factors.

For example:

\[ \frac{3(x+2)}{5(x+2)} \]

When x ≠ −2, the common factor (x + 2) is nonzero and can be cancelled.

The simplified expression is:

3/5, where x ≠ −2.

The restriction is important because the original expression is undefined at x = −2.

Only factors can be cancelled, not individual terms.

For example:

(x + 3)/(x + 5)

The x terms cannot be cancelled because they are part of separate sums.

By contrast:

x(x + 3)/[x(x + 5)]

contains a common factor of x that can be cancelled, provided x ≠ 0.

A Reliable Four-Step Simplification Method

  1. Find restrictions: Set the original denominator equal to zero and solve.

  2. Factor completely: Factor both the numerator and denominator wherever possible.

  3. Cancel common factors: Divide out factors appearing in both the numerator and denominator.

  4. Verify: Check the simplified result by algebraic reasoning or substituting permitted values.

This order prevents the common mistake of losing excluded values during cancellation.

Worked Example 1: Simplifying with a Common Monomial Factor

Simplify:

\[ \frac{12x^2}{18x} \]

Step 1: Identify restrictions.

18x = 0

x = 0

Therefore, x ≠ 0.

Step 2: Factor numerator and denominator.

12x² = 6x(2x)

18x = 6x(3)

Rewrite:

\[ \frac{6x(2x)}{6x(3)} \]

Step 3: Cancel 6x.

\[ \frac{2x}{3} \]

Step 4: Verify with x = 2.

Original:

12(2²)/[18(2)] = 48/36 = 4/3

Simplified:

2(2)/3 = 4/3

Both expressions give the same result.

Final answer: 2x/3, where x ≠ 0.

Worked Example 2: Simplifying Using the Difference of Squares

Simplify:

\[ \frac{x^2-16}{x-4} \]

Step 1: Find restrictions.

x − 4 = 0

x = 4

Therefore, x ≠ 4.

Step 2: Factor the numerator.

x² − 16 = (x − 4)(x + 4)

Step 3: Rewrite and cancel.

\[ \frac{(x-4)(x+4)}{x-4}=x+4 \]

Step 4: Verify with x = 6.

Original:

(36 − 16)/(6 − 4) = 20/2 = 10

Simplified:

6 + 4 = 10

The results agree.

Final answer: x + 4, where x ≠ 4.

Worked Example 3: Factoring Both Numerator and Denominator

Simplify:

\[ \frac{x^2+5x+6}{x^2+x-6} \]

Step 1: Find restrictions from the original denominator.

x² + x − 6 = (x + 3)(x − 2)

Therefore:

x ≠ −3 and x ≠ 2.

Step 2: Factor the numerator.

x² + 5x + 6 = (x + 2)(x + 3)

Step 3: Rewrite the expression.

\[ \frac{(x+2)(x+3)}{(x+3)(x-2)} \]

Step 4: Cancel the common factor (x + 3).

\[ \frac{x+2}{x-2} \]

Step 5: Verify with x = 0.

Original:

(0 + 0 + 6)/(0 + 0 − 6) = −1

Simplified:

(0 + 2)/(0 − 2) = −1

Both results agree.

Final answer: (x + 2)/(x − 2), where x ≠ −3 and x ≠ 2.

Worked Example 4: Simplifying with Numerical Coefficients

Simplify:

\[ \frac{6x^2-18x}{9x} \]

Step 1: Find restrictions.

9x = 0

x = 0

Therefore, x ≠ 0.

Step 2: Factor the numerator.

6x² − 18x = 6x(x − 3)

Step 3: Rewrite.

\[ \frac{6x(x-3)}{9x} \]

Step 4: Cancel x and simplify 6/9.

\[ \frac{2(x-3)}{3} \]

Final answer: 2(x − 3)/3, where x ≠ 0.

Worked Example 5: More Than One Common Factor

Simplify:

\[ \frac{x^2-5x+6}{x^2-4} \]

Step 1: Find restrictions.

x² − 4 = (x − 2)(x + 2)

Therefore:

x ≠ 2 and x ≠ −2.

Step 2: Factor the numerator.

x² − 5x + 6 = (x − 2)(x − 3)

Step 3: Rewrite.

\[ \frac{(x-2)(x-3)}{(x-2)(x+2)} \]

Step 4: Cancel (x − 2).

\[ \frac{x-3}{x+2} \]

Final answer: (x − 3)/(x + 2), where x ≠ 2 and x ≠ −2.

Notice that although only one factor was cancelled, both original restrictions remain.

Worked Example 6: Recognizing When an Expression Cannot Be Simplified

Consider:

\[ \frac{x^2-16}{x^2-4} \]

Factor both parts:

\[ \frac{(x-4)(x+4)}{(x-2)(x+2)} \]

There are no common factors.

Therefore, the expression cannot be simplified by cancellation.

The original denominator is zero at x = 2 and x = −2.

Final answer: (x² − 16)/(x² − 4), where x ≠ −2 and x ≠ 2.

An expression does not always become simpler just because its numerator and denominator can be factored.

Recognizing Equivalent Rational Expressions

Two rational expressions are equivalent on a specified domain if they produce the same output for every input in that domain.

For example:

\[ \frac{x^2-9}{x-3}=x+3 \]

is true whenever x ≠ 3.

The expression on the left is undefined at x = 3, while x + 3 is defined there.

Therefore, the expressions agree for all x ≠ 3, but they are not identical as functions with their natural domains.

Worked Example 7: Determining Equivalence

Determine whether the following expressions are equivalent:

A(x) = (x² + 4x + 3)/(x + 1)

B(x) = x + 3

Factor A(x):

A(x) = (x + 1)(x + 3)/(x + 1)

Cancel the common factor:

A(x) = x + 3, where x ≠ −1.

Therefore, A(x) and B(x) give the same output whenever x ≠ −1.

However, B(x) is also defined at x = −1.

Conclusion: They are equivalent on the restricted domain x ≠ −1, but they do not have the same natural domain.

Using Cross-Multiplication to Check Equivalence

Suppose we want to compare:

A(x) = (x + 2)/(x − 1)

B(x) = (x² + x − 2)/(x² − 2x + 1)

Factor B(x):

Numerator:

x² + x − 2 = (x + 2)(x − 1)

Denominator:

x² − 2x + 1 = (x − 1)²

Therefore:

\[ B(x)=\frac{(x+2)(x-1)}{(x-1)^2} \]

Cancel one factor of (x − 1):

B(x) = (x + 2)/(x − 1)

Both expressions have the restriction x ≠ 1.

Therefore, A(x) and B(x) are equivalent rational functions with the same domain.

Cross-multiplication offers another method:

For A/B and C/D, check whether:

AD = BC

This establishes equality wherever both original denominators are nonzero.

Verifying Simplified Results

Simplifying an expression is not complete until we are confident that the factorization and cancellation were performed correctly.

There are three useful verification methods.

Method 1: Expanding the Factors

Suppose:

\[ \frac{(x+2)(x+3)}{x+3}=x+2 \]

Expanding the numerator gives:

(x + 2)(x + 3) = x² + 5x + 6

Therefore, the original expression is:

(x² + 5x + 6)/(x + 3)

The factorization is correct, and cancellation is valid for x ≠ −3.

Method 2: Substituting Numerical Values

Consider:

(x² − 9)/(x − 3) = x + 3, where x ≠ 3.

Choose x = 5.

Original:

(25 − 9)/(5 − 3) = 16/2 = 8

Simplified:

5 + 3 = 8

The results agree.

A numerical check can detect errors, but testing one or two values does not prove that two expressions are equivalent for every permitted input.

Method 3: Checking Restrictions

After simplification, return to the original denominator.

For example:

\[ \frac{x^2-4}{x-2}=x+2 \]

The original denominator requires x ≠ 2.

Therefore, the correct statement is:

(x² − 4)/(x − 2) = x + 2, where x ≠ 2.

This confirms that the simplified form has preserved the original restriction.

Graphical Interpretation of Simplification

Simplification can reveal important information about the graph of a rational function.

Consider:

\[ f(x)=\frac{x^2-1}{x-1} \]

Factor:

f(x) = (x − 1)(x + 1)/(x − 1)

Simplify:

f(x) = x + 1, where x ≠ 1.

The graph is the straight line y = x + 1, except for the missing point (1, 2).

This missing point is called a hole or removable discontinuity.

The simplified equation describes the same line everywhere except at the excluded value.

A graphing calculator may display the complete line without making the hole obvious, so algebraic restrictions must always be checked.

Real-World Applications

Application 1: Simplifying Area Ratios

Suppose the area of a rectangular garden is:

A = x² + 5x + 6

and its width is:

W = x + 2

The length is:

L = A/W

Therefore:

\[ L=\frac{x^2+5x+6}{x+2} \]

Factor the numerator:

L = (x + 2)(x + 3)/(x + 2)

Simplify:

L = x + 3, where x ≠ −2.

For a physical garden, the dimensions must also be positive, so x must be greater than −2.

Application 2: Average Production Cost

A factory has a total production cost:

C(x) = 5x² + 20x

where x is the number of items produced.

The average cost per item is:

\[ A(x)=\frac{5x^2+20x}{x} \]

Factor the numerator:

A(x) = 5x(x + 4)/x

Simplify:

A(x) = 5(x + 4), where x ≠ 0.

For 10 items:

A(10) = 5(14) = $70 per item.

The simplified form makes calculations easier.

In the real-world context, x must be a positive whole number.

Application 3: Simplifying Physics Formulae

Suppose an object moves a distance:

d = 3t² + 6t

over a time interval t, with t greater than zero.

Its average speed over the interval is:

v = d/t

Therefore:

\[ v=\frac{3t^2+6t}{t} \]

Factor:

v = 3t(t + 2)/t

Simplify:

v = 3(t + 2), where t ≠ 0.

For t = 4 seconds:

v = 3(6) = 18 m/s.

This illustrates how simplifying rational expressions can make scientific formulae easier to use.

Common Mistakes

Mistake 1: Cancelling Terms Instead of Factors

Incorrect:

(x + 4)/(x + 6) = 4/6

The x terms cannot be cancelled because they are not common factors.

Correct:

The expression (x + 4)/(x + 6) cannot be simplified by cancellation.

Mistake 2: Forgetting to Factor Completely

Consider:

(x² + 5x + 6)/(x² + x − 6)

Without factoring, common factors may not be obvious.

Correct factorization:

(x + 2)(x + 3)/[(x + 3)(x − 2)]

Simplified:

(x + 2)/(x − 2), where x ≠ −3 and x ≠ 2.

Mistake 3: Losing Excluded Values

Incorrect:

(x² − 9)/(x − 3) = x + 3, with no restriction.

Correct:

x + 3, where x ≠ 3.

Mistake 4: Incorrect Factorization

Incorrect:

x² + 7x + 12 = (x + 2)(x + 6)

Expanding gives:

x² + 8x + 12

Correct:

x² + 7x + 12 = (x + 3)(x + 4)

Mistake 5: Assuming Every Expression Can Be Simplified

Not every rational expression contains common factors.

For example:

(x + 2)/(x + 5)

cannot be simplified by cancellation.

Mistake 6: Relying Only on Numerical Verification

Two different expressions might give the same answer for a particular input.

Numerical substitution is useful for checking, but algebraic verification is needed to establish equivalence for all permitted values.

Key Terms

  • Rational Expression: A quotient of two polynomials.

  • Numerator: The polynomial above the fraction bar.

  • Denominator: The polynomial below the fraction bar.

  • Factor: An expression that is multiplied by another expression.

  • Common Factor: A factor shared by two or more expressions.

  • Greatest Common Factor: The largest factor shared by all terms.

  • Factoring: Rewriting an expression as a product of factors.

  • Difference of Squares: A factoring pattern of the form a² − b² = (a − b)(a + b).

  • Simplification: Rewriting an expression in a less complicated equivalent form.

  • Excluded Value: A value that makes the original denominator zero.

  • Domain Restriction: A condition excluding inputs that make an expression undefined.

  • Equivalent Rational Expressions: Expressions that give the same values over a specified common domain.

  • Verification: Checking whether a mathematical result is correct.

  • Removable Discontinuity: A missing point in a graph caused by a cancelled denominator factor.

  • Hole: The missing point associated with a removable discontinuity.

Key Takeaways

  1. Rational expressions are simplified by factoring numerators and denominators and cancelling common factors.

  2. Always identify excluded values from the original denominator before simplifying.

  3. Only common factors can be cancelled; individual terms cannot.

  4. The greatest common factor should be checked before using other factoring techniques.

  5. Difference-of-squares and quadratic factoring are especially useful for simplifying rational expressions.

  6. Cancelled denominator factors still produce restrictions in the original expression.

  7. Equivalent rational expressions produce the same outputs over their common permitted domain.

  8. Algebraic expansion, substitution, and checking restrictions can help verify simplified results.

  9. Simplification can reveal holes and other important features of rational-function graphs.

  10. Rational-expression simplification is useful in geometry, economics, physics, and other mathematical applications.

Check Your Understanding

Question 1: Common Factors

Simplify and state restrictions:

\[ \frac{15x^2}{20x} \]

Question 2: Difference of Squares

Simplify:

\[ \frac{x^2-49}{x-7} \]

State any excluded values.

Question 3: Factoring Quadratics

Simplify:

\[ \frac{x^2+6x+8}{x^2+5x+6} \]

State all restrictions.

Question 4: Numerical and Algebraic Factors

Simplify:

\[ \frac{8x^2+16x}{12x} \]

State any restrictions.

Question 5: Recognizing Equivalence

Determine whether the following expressions are equivalent over their common domain:

A(x) = (x² − 25)/(x − 5)

B(x) = x + 5

Explain your reasoning and identify any domain differences.

Question 6: Verification

A student simplifies:

\[ \frac{x^2+3x+2}{x+1}=x+2 \]

a) Verify the factorization.

b) State the excluded value.

c) Check the result using x = 3.

Question 7: Multi-Step Simplification

Simplify completely:

\[ \frac{2x^2-8x}{x^2-16} \]

State all restrictions.

Question 8: Real-World Application

A rectangular field has area:

A = x² + 9x + 20

Its width is:

W = x + 4

a) Write an expression for its length.

b) Simplify the expression.

c) State the mathematical restriction.

d) Calculate the length when x = 6.

 

Fully Worked Answers

Answer 1

Given:

15x²/(20x)

Step 1: Find restrictions.

20x = 0 gives x = 0.

Step 2: Simplify.

15x²/(20x) = 3x/4

Final answer: 3x/4, where x ≠ 0.

Answer 2

Given:

(x² − 49)/(x − 7)

Factor:

(x − 7)(x + 7)/(x − 7)

Cancel:

x + 7

The original denominator is zero at x = 7.

Final answer: x + 7, where x ≠ 7.

Answer 3

Given:

(x² + 6x + 8)/(x² + 5x + 6)

Factor the numerator:

x² + 6x + 8 = (x + 2)(x + 4)

Factor the denominator:

x² + 5x + 6 = (x + 2)(x + 3)

Rewrite:

\[ \frac{(x+2)(x+4)}{(x+2)(x+3)} \]

Cancel (x + 2):

\[ \frac{x+4}{x+3} \]

Restrictions from the original denominator:

x ≠ −2 and x ≠ −3.

Final answer: (x + 4)/(x + 3), where x ≠ −2 and x ≠ −3.

Answer 4

Given:

(8x² + 16x)/(12x)

Factor:

8x(x + 2)/(12x)

Cancel x and simplify 8/12:

2(x + 2)/3

The original denominator requires x ≠ 0.

Final answer: 2(x + 2)/3, where x ≠ 0.

Answer 5

A(x) = (x² − 25)/(x − 5)

Factor:

A(x) = (x − 5)(x + 5)/(x − 5)

Simplify:

A(x) = x + 5, where x ≠ 5.

B(x) = x + 5 is defined for all real numbers.

Therefore, A(x) and B(x) are equivalent for x ≠ 5, but their natural domains differ.

Answer 6

Given:

(x² + 3x + 2)/(x + 1)

a) Factor the numerator.

x² + 3x + 2 = (x + 1)(x + 2)

Therefore:

(x + 1)(x + 2)/(x + 1) = x + 2

b) Restriction.

x + 1 = 0

x = −1

Therefore, x ≠ −1.

c) Verify with x = 3.

Original:

(9 + 9 + 2)/(3 + 1) = 20/4 = 5

Simplified:

3 + 2 = 5

Both results agree.

Answer 7

Given:

(2x² − 8x)/(x² − 16)

Step 1: Find restrictions.

x² − 16 = (x − 4)(x + 4)

Therefore:

x ≠ 4 and x ≠ −4.

Step 2: Factor the numerator.

2x² − 8x = 2x(x − 4)

Step 3: Rewrite.

\[ \frac{2x(x-4)}{(x-4)(x+4)} \]

Step 4: Cancel (x − 4).

\[ \frac{2x}{x+4} \]

Final answer: 2x/(x + 4), where x ≠ −4 and x ≠ 4.

Answer 8

a) Write the length expression.

Length = Area / Width

L = (x² + 9x + 20)/(x + 4)

b) Simplify.

Factor the numerator:

x² + 9x + 20 = (x + 4)(x + 5)

Therefore:

L = (x + 4)(x + 5)/(x + 4)

L = x + 5

c) Restriction.

x + 4 ≠ 0

x ≠ −4

For the physical field, both dimensions must be positive, so x greater than −4 is required.

d) Substitute x = 6.

L = 6 + 5

Length = 11 units.

 

This completes Simplifying Rational Expressions, with detailed coverage of factoring, cancellation, restrictions, equivalent expressions, verification, and practical applications.