Operate on rational expressions and state restrictions

Ontario Grade 11 Mathematics

Characteristics of Functions

Simplifying, Multiplying, Dividing, Adding, and Subtracting — with Restrictions

You have already worked with fractions in arithmetic — adding, subtracting, multiplying, and dividing them. A rational expression is the algebraic version of a fraction: instead of integers in the numerator and denominator, you have polynomials. Almost every rule you learned for numeric fractions carries over directly. The key new responsibility in algebra is tracking restrictions — values of the variable that would make a denominator equal to zero, which is undefined in mathematics. This lesson builds those skills step by step, starting with what you already know and moving toward more complex operations.

What you will learn

Identify and state the restrictions on the variables in a rational expression.
Simplify rational expressions by factoring and cancelling common factors.
Multiply and divide rational expressions, stating all restrictions.
Add and subtract rational expressions using a common denominator, stating all restrictions.

Prerequisite Bridge: Fractions and Factoring

A rational expression has the form

\frac{P}{Q}

where

P

and

Q

are polynomials and

Q \neq 0

. Examples include

\frac{3x}{x+2}

and

\frac{x^2 - 4}{x^2 - x - 6}

. Because

Q

is a polynomial rather than just a number, it can equal zero for certain values of

x

, and division by zero is never allowed.

To work with rational expressions, you need two Grade 10 skills: factoring polynomials and simplifying numeric fractions. Recall that

\frac{6}{9} = \frac{2 \cdot 3}{3 \cdot 3} = \frac{2}{3}

because the common factor

3

cancels. The exact same logic applies to algebraic factors. Recall also how to factor: common factor, difference of squares

a^2 - b^2 = (a-b)(a+b)

, and trinomial factoring.

Review those factoring patterns briefly before continuing, because every operation in this lesson depends on writing polynomials in fully factored form first.

\frac{P}{Q}, Q ≠ 0

A rational expression is a polynomial divided by a nonzero polynomial.
Division by zero is undefined, so denominators can never equal zero.
Factoring is the essential first step in every operation on rational expressions.

Restrictions on Rational Expressions

A restriction is a value of the variable that makes any denominator equal to zero. You must identify and state these values before or alongside every simplification or operation, because excluding them is part of giving a complete mathematical answer.

To find restrictions, set each distinct denominator expression equal to zero and solve. For example, in

\frac{5}{x - 3}

, set

x - 3 = 0

, giving

x = 3

. The restriction is

x \neq 3

. For

\frac{2x}{x^2 - 9}

, factor the denominator first:

x^2 - 9 = (x-3)(x+3)

, so set each factor to zero:

x = 3

or

x = -3

. The restrictions are

x \neq 3

and

x \neq -3

.

A critical rule: if you cancel a factor during simplification, the restriction from that factor still applies to the simplified expression. The simplified form looks different, but it only equals the original expression when the variable is not at the restricted value. Always state restrictions based on the denominators before cancelling.

Set every denominator equal to zero and solve to find restrictions.
State restrictions using the $\neq$ symbol.
Restrictions from cancelled factors must still be stated in the final answer.
Factor denominators fully before identifying restrictions.

Simplifying Rational Expressions

Simplifying means writing a rational expression in lowest terms by cancelling factors that appear in both the numerator and the denominator. The process mirrors simplifying a numeric fraction: factor both the numerator and denominator completely, identify common factors, and divide them out.

Consider

\frac{x^2 - x - 6}{x^2 - 9}

. Factor the numerator:

x^2 - x - 6 = (x-3)(x+2)

. Factor the denominator:

x^2 - 9 = (x-3)(x+3)

. The common factor is

(x-3)

. Cancel it to get

\frac{x+2}{x+3}

. The restrictions come from the original denominator:

x \neq 3

and

x \neq -3

. Note that

(x-3)

is no longer visible in the denominator after cancelling, but the restriction

x \neq 3

must still be written.

Never cancel terms that are added or subtracted — only factors that are multiplied. For instance,

\frac{x + 4}{x + 7}

cannot be simplified because

x

is a term (added), not a factor of the whole numerator or denominator.

\frac{(x-3)(x+2)}{(x-3)(x+3)} = \frac{x+2}{x+3}, x ≠ 3, x ≠ -3

Factor numerator and denominator fully before cancelling.
Cancel only common multiplicative factors, not common terms.
Keep all restrictions even when the factor that caused them disappears after cancelling.

Multiplying and Dividing Rational Expressions

Multiplying rational expressions follows the same rule as multiplying numeric fractions: multiply the numerators together and multiply the denominators together, then simplify. The best strategy is to factor everything first and cancel before multiplying, so the numbers stay manageable.

For multiplication:

\frac{A}{B} \cdot \frac{C}{D} = \frac{A \cdot C}{B \cdot D}

. Restrictions come from every denominator that appears — both

B

and

D

— before any cancelling takes place.

For division, recall that dividing by a fraction means multiplying by its reciprocal.

\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}

. Once you flip the second fraction, the expression

C

becomes a denominator, so any value that makes

C = 0

is also a restriction. In other words, restrictions in a division problem come from all original denominators and from the numerator of the divisor (the fraction you are dividing by).

\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} · \frac{D}{C}

Factor all polynomials before multiplying or dividing.
To divide, multiply by the reciprocal of the second fraction.
Collect restrictions from every denominator, including the flipped one in division.
Cancel common factors across the entire numerator and denominator after writing one combined fraction.

Adding and Subtracting Rational Expressions

Just like numeric fractions, rational expressions can only be added or subtracted when they share a common denominator. If the denominators are already the same, simply add or subtract the numerators. If they differ, find the lowest common denominator (LCD), rewrite each fraction with that LCD, and then combine the numerators.

The LCD is the smallest expression divisible by every denominator. Build it by taking each unique factor the greatest number of times it appears across all denominators. For example, if one denominator is

(x+2)

and another is

(x+2)(x-1)

, the LCD is

(x+2)(x-1)

.

After finding the LCD, multiply the numerator and denominator of each fraction by whatever factor is missing. Expand and simplify the resulting numerator carefully — especially with subtraction, where the minus sign distributes across the entire second numerator. Factor the final numerator if possible and cancel with the denominator. State all restrictions from every original denominator.

\frac{A}{B} + \frac{C}{D} = \frac{A · D + C · B}{B · D}

A common denominator is required before adding or subtracting.
The LCD contains each unique factor the maximum number of times it appears.
Distribute the minus sign carefully across every term when subtracting.
Restrictions come from every original denominator, not just the LCD.

Summary of Operations on Rational Expressions

Operation	Rule	Key Restriction Source
Simplify	Factor fully; cancel common factors	All original denominators (including cancelled factors)
Multiply	Multiply numerators; multiply denominators; cancel common factors	All denominators before cancelling
Divide	Multiply by reciprocal of second fraction; cancel common factors	All denominators and the numerator of the divisor
Add / Subtract	Find LCD; rewrite with LCD; combine numerators	All original denominators

Worked example

Multiplying and Dividing Rational Expressions

Simplify and state all restrictions:

\frac{2x^2 + 2x}{x^2 - 4} \div \frac{x^2 + x}{x^2 - 4x + 4}

Rewrite the division as multiplication by the reciprocal
Division by a fraction is the same as multiplying by its reciprocal. Flip the second fraction and change the operation to multiplication.
$\frac{2x^2 + 2x}{x^2 - 4} · \frac{x^2 - 4x + 4}{x^2 + x}$
Factor every polynomial completely
Factor each numerator and denominator. Take out common factors first, then apply difference of squares or trinomial factoring where needed. $2x^2 + 2x = 2x(x+1)$ ; $x^2 - 4 = (x-2)(x+2)$ ; $x^2 - 4x + 4 = (x-2)^2$ ; $x^2 + x = x(x+1)$ .
$\frac{2x(x+1)}{(x-2)(x+2)} · \frac{(x-2)^2}{x(x+1)}$
State all restrictions before cancelling
Set every denominator equal to zero: the original denominators are $(x-2)(x+2)$ and $x^2+x = x(x+1)$ , and after flipping, $x^2+x$ becomes a denominator. So restrictions come from $x-2=0$ , $x+2=0$ , $x=0$ , and $x+1=0$ .
$x ≠ 2, x ≠ -2, x ≠ 0, x ≠ -1$
Cancel common factors
Look for factors that appear in both a numerator position and a denominator position across the combined fraction. The factor $x$ appears in $2x$ (numerator) and $x$ (denominator) — cancel once. The factor $(x+1)$ appears in $(x+1)$ (numerator) and $(x+1)$ (denominator) — cancel. The factor $(x-2)$ appears once in $(x-2)^2$ (numerator) and once in $(x-2)$ (denominator) — cancel one copy, leaving one $(x-2)$ in the numerator.
$\frac{2\cancel{x}\cancel{(x+1)}}{\cancel{(x-2)}(x+2)} · \frac{\cancel{(x-2)}^2}{\cancel{x}\cancel{(x+1)}}$
Write the simplified result
After cancelling, the numerator holds $2$ and one remaining $(x-2)$ , while the denominator holds $(x+2)$ . Combine these to write the final simplified expression with its restrictions.
$\frac{2(x-2)}{x+2}, x ≠ 2, x ≠ -2, x ≠ 0, x ≠ -1$

Answer:

\frac{2(x-2)}{x+2}

, where

x \neq 2,\; x \neq -2,\; x \neq 0,\; x \neq -1

Check: Substitute

x = 3

into the original expression: numerator of first fraction

= 2(9)+2(3) = 24

, denominator

= 9-4 = 5

, so first fraction

= \frac{24}{5}

. Second fraction: numerator

= 9+3 = 12

, denominator

= 9-12+4 = 1

, so second fraction

= 12

. Division gives

\frac{24}{5} \div 12 = \frac{24}{60} = \frac{2}{5}

. Now substitute

x = 3

into the simplified answer:

\frac{2(3-2)}{3+2} = \frac{2}{5}

. The values match, confirming the simplification is correct.

Worked example

Adding Rational Expressions with Different Denominators

Simplify and state all restrictions:

\frac{3}{x^2 - x - 2} + \frac{2}{x^2 + x}

Factor each denominator
Factor both denominators completely before looking for the LCD. $x^2 - x - 2 = (x-2)(x+1)$ . $x^2 + x = x(x+1)$ .
$\frac{3}{(x-2)(x+1)} + \frac{2}{x(x+1)}$
State all restrictions
Set each factor in every denominator equal to zero: $x-2=0$ gives $x=2$ ; $x+1=0$ gives $x=-1$ ; $x=0$ gives $x=0$ .
$x ≠ 2, x ≠ -1, x ≠ 0$
Find the lowest common denominator
List the unique factors across both denominators: $(x-2)$ , $(x+1)$ , and $x$ . Each appears at most once across the two denominators, so the LCD is their product.
$\text{LCD} = x(x-2)(x+1)$
Rewrite each fraction with the LCD
The first fraction $\frac{3}{(x-2)(x+1)}$ is missing the factor $x$ from the LCD, so multiply its numerator and denominator by $x$ . The second fraction $\frac{2}{x(x+1)}$ is missing $(x-2)$ , so multiply its numerator and denominator by $(x-2)$ .
$\frac{3x}{x(x-2)(x+1)} + \frac{2(x-2)}{x(x-2)(x+1)}$
Add the numerators and simplify
Since the denominators are now the same, add the numerators: $3x + 2(x-2) = 3x + 2x - 4 = 5x - 4$ . Check whether the numerator $5x - 4$ shares any factor with the denominator $x(x-2)(x+1)$ . Setting $5x - 4 = 0$ gives $x = \frac{4}{5}$ , which is not a root of any denominator factor, so the expression is already fully simplified.
$\frac{5x - 4}{x(x-2)(x+1)}, x ≠ 2, x ≠ -1, x ≠ 0$

Answer:

\frac{5x-4}{x(x-2)(x+1)}

, where

x \neq 2,\; x \neq -1,\; x \neq 0

Check: Substitute

x = 1

into the original:

\frac{3}{1-1-2} + \frac{2}{1+1} = \frac{3}{-2} + \frac{2}{2} = -1.5 + 1 = -0.5

. Substitute

x = 1

into the answer:

\frac{5(1)-4}{1(1-2)(1+1)} = \frac{1}{(1)(-1)(2)} = \frac{1}{-2} = -0.5

. Both give

-0.5

, confirming the answer is correct.

Common mistakes and how to avoid them

Cancelling terms instead of factors, for example writing

\frac{x + 5}{x + 3}

and cancelling the

x

from numerator and denominator.

Correction:

x

is a term here (it is added, not multiplied). Only cancel when the same expression is a factor of the entire numerator and the entire denominator.

\frac{x(x+5)}{x(x+3)}

would allow cancelling

x

, but

\frac{x+5}{x+3}

does not.

Forgetting to state restrictions that come from cancelled factors, believing the restriction disappears once the factor is gone.

Correction: Restrictions are determined by the original denominators. A cancelled factor still contributed a denominator in the original expression, so its zero value is still restricted.

When dividing, only collecting restrictions from the first fraction's denominator and ignoring the numerator of the second fraction.

Correction: After flipping the second fraction, its original numerator becomes a denominator. Set it equal to zero and include those values as restrictions too.

Forgetting to distribute the minus sign across all terms of the second numerator when subtracting rational expressions.

Correction: Write the subtraction as

\frac{A - (B)}{\text{LCD}}

with brackets around the entire second numerator, then distribute the negative sign to every term inside the brackets before combining.

Building an LCD by simply multiplying all denominators together, even when common factors exist, leading to an unnecessarily complicated expression.

Correction: Factor each denominator first. The LCD uses each unique factor the greatest number of times it appears — not the product of all denominators. This keeps the work as simple as possible.

Lesson summary

A rational expression is a fraction whose numerator and denominator are polynomials; the denominator can never equal zero.
Restrictions are the values of the variable that make any denominator zero; state them using $\neq$ and include those from cancelled factors.
To simplify, factor both numerator and denominator fully and cancel common factors — never common terms.
To multiply, factor everything, state restrictions from all denominators, then cancel and multiply. To divide, first rewrite as multiplication by the reciprocal, adding the numerator of the divisor as a source of restrictions.
To add or subtract, factor the denominators, find the LCD using each unique factor the maximum number of times it appears, rewrite each fraction, then combine and simplify the numerator.
Always state the full set of restrictions alongside the simplified expression as part of the complete answer.

Check your understanding

Question 1

What are the restrictions on

\frac{x+3}{x^2 - 5x + 6}

?

$x \neq 3$ only
$x \neq 2$ and $x \neq 3$
$x \neq -2$ and $x \neq -3$
$x \neq 0$

Show answer and explanation

x \neq 2

and

x \neq 3

Factor the denominator:

x^2 - 5x + 6 = (x-2)(x-3)

. Set each factor equal to zero:

x = 2

and

x = 3

. Both values are restricted. The numerator does not affect restrictions.

Question 2

Simplify

\frac{x^2 - 9}{x^2 + 5x + 6}

. Which answer is correct?

$\frac{x-3}{x+2},\; x \neq -2,\; x \neq -3$
$\frac{x+3}{x+2},\; x \neq -2$
$\frac{x-3}{x+2},\; x \neq -2$ only
$\frac{x-3}{x+3},\; x \neq -2,\; x \neq -3$

Show answer and explanation

\frac{x-3}{x+2},\; x \neq -2,\; x \neq -3

Factor:

x^2 - 9 = (x-3)(x+3)

and

x^2 + 5x + 6 = (x+2)(x+3)

. Cancel the common factor

(x+3)

to get

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

. Restrictions from the original denominator:

x \neq -2

and

x \neq -3

(the restriction from the cancelled factor must still be stated).

Question 3

Which step is correct when dividing

\frac{4x}{x-1} \div \frac{2x^2}{x+3}

?

Multiply $\frac{4x}{x-1}$ by $\frac{2x^2}{x+3}$
Multiply $\frac{4x}{x-1}$ by $\frac{x+3}{2x^2}$ and add the restriction $x \neq 0$
Flip the first fraction and multiply by the second
Subtract the numerators and keep the denominator

Show answer and explanation

Multiply

\frac{4x}{x-1}

by

\frac{x+3}{2x^2}

and add the restriction

x \neq 0

To divide, multiply by the reciprocal of the second fraction:

\frac{4x}{x-1} \cdot \frac{x+3}{2x^2}

. The numerator of the divisor,

2x^2

, becomes a denominator after flipping, so

x = 0

is an additional restriction alongside

x \neq 1

and

x \neq -3

.

Question 4

When adding

\frac{5}{x+4} + \frac{3}{x}

, what is the correct LCD?

$x$
$x + 4$
$x(x+4)$
$x^2 + 4$

Show answer and explanation

x(x+4)

The two denominators are

x+4

and

x

. They share no common factors, so the LCD is their product:

x(x+4)

. The expression

x^2 + 4

is incorrect because

(x+4)

and

x

multiply to

x^2 + 4x

, not

x^2 + 4

.

Key terms

Rational expression: An expression of the form P divided by Q, where P and Q are polynomials and Q is not equal to zero.
Restriction: A value of the variable that must be excluded because it makes a denominator equal to zero, making the expression undefined.
Lowest common denominator (LCD): The simplest expression that is divisible by every denominator involved in an addition or subtraction of rational expressions.
Factor: An expression that is multiplied by another expression. In a rational expression, only common factors (not common terms) can be cancelled.
Simplify (lowest terms): To rewrite a rational expression so that the numerator and denominator share no common factors other than 1.
Reciprocal: The result of swapping the numerator and denominator of a fraction. Used when converting division into multiplication.
Undefined: A mathematical expression has no value when it requires division by zero; rational expressions are undefined at their restricted values.

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