Multiplying and Dividing Rational Expressions

Multiplying and dividing rational expressions uses the same rules as number fractions: multiply across, and to divide, multiply by the reciprocal. The algebra is mostly factoring and cancelling, plus one new trap with restrictions when you divide.

Key ideas

Multiplying

$$\frac{A}{B} \times \frac{C}{D} = \frac{AC}{BD}$$

1. Factor every numerator and denominator.
2. State the restrictions (every denominator factor $\ne 0$).
3. Cancel common factors, across the fractions too, then multiply what’s left.

Cancelling before multiplying keeps the numbers small, like $\dfrac{4}{9} \times \dfrac{3}{8} = \dfrac{1}{3} \times \dfrac{1}{2} = \dfrac{1}{6}$.

Dividing

$$\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \times \frac{D}{C}$$

Keep the first fraction, flip the second, and multiply.

Restrictions when dividing

When you flip $\dfrac{C}{D}$, its numerator $C$ moves into a denominator. So the restrictions come from:

• the denominators $B$ and $D$, and
• the numerator $C$ of the fraction you’re dividing by.

Find all of them before you cancel.

Worked examples

Example 1: Multiplying monomials

Simplify $\dfrac{6x^2}{5y} \times \dfrac{10y^3}{3x}$ and state the restrictions.

Solution.

$$\frac{6x^2}{5y} \times \frac{10y^3}{3x} = \frac{60x^2y^3}{15xy} = 4xy^2, \qquad x \ne 0,\ y \ne 0$$

Example 2: Multiplying with factoring

Simplify $\dfrac{x^2 - 4}{x^2 + 3x} \times \dfrac{x + 3}{x - 2}$ and state the restrictions.

Solution. Factor everything:

$$\frac{(x - 2)(x + 2)}{x(x + 3)} \times \frac{x + 3}{x - 2}$$

Restrictions from the denominators $x$, $x + 3$, and $x - 2$: $x \ne 0, -3, 2$.

Cancel $(x - 2)$ and $(x + 3)$:

$$\frac{x + 2}{x}, \qquad x \ne 0, -3, 2$$

Example 3: Dividing

Simplify $\dfrac{x^2 - x - 6}{x^2 - 9} \div \dfrac{x + 2}{x + 3}$ and state the restrictions.

Solution. Factor and flip the second fraction:

$$\frac{(x - 3)(x + 2)}{(x - 3)(x + 3)} \times \frac{x + 3}{x + 2}$$

Restrictions:

• from the first denominator: $x \ne 3, -3$
• from the second denominator: $x \ne -3$
• from the second numerator (it moved to the bottom): $x \ne -2$

Everything cancels:

$$1, \qquad x \ne 3, -3, -2$$

Example 4: Dividing with opposites

Simplify $\dfrac{2x - 6}{x + 1} \div \dfrac{9 - x^2}{x^2 + x}$ and state the restrictions.

Solution. Factor, writing $9 - x^2 = (3 - x)(3 + x) = -(x - 3)(x + 3)$:

$$\frac{2(x - 3)}{x + 1} \times \frac{x(x + 1)}{-(x - 3)(x + 3)}$$

Restrictions: $x \ne -1$ and $x \ne 0$ (denominators), and $x \ne 3, -3$ (the flipped numerator $9 - x^2$).

Cancel $(x - 3)$ and $(x + 1)$:

$$\frac{-2x}{x + 3}, \qquad x \ne -3, -1, 0, 3$$

Common mistakes

Forgetting to flip. Dividing by $\dfrac{x + 2}{x + 3}$ means multiplying by $\dfrac{x + 3}{x + 2}$.

Flipping the wrong fraction. Only the second fraction (the divisor) is flipped.

Missing the divisor’s numerator restriction. In Example 3, $x = -2$ makes the divisor equal to $0$, and you can’t divide by zero. That restriction is easy to lose once the fraction is flipped.

Cancelling before factoring. In $\dfrac{x^2 - 4}{x^2 + 3x}$, you can’t cancel the $x^2$‘s. Factor first.

Losing a negative from opposites. $9 - x^2$ factors as $-(x - 3)(x + 3)$. The $-1$ ends up in the answer.

Practice

1. (Warm-up) Simplify $\dfrac{3a}{4b} \times \dfrac{8b^2}{9a^2}$ and state the restrictions.

Solution

$$\frac{24ab^2}{36a^2b} = \frac{2b}{3a}, \qquad a \ne 0,\ b \ne 0$$

2. (Warm-up) Simplify $\dfrac{5x}{7} \div \dfrac{10x^2}{21}$ and state the restriction.

Solution

$$\frac{5x}{7} \times \frac{21}{10x^2} = \frac{105x}{70x^2} = \frac{3}{2x}, \qquad x \ne 0$$

3. (Warm-up) State all the restrictions on $\dfrac{x + 1}{x - 2} \div \dfrac{x + 5}{x + 4}$.

Solution

Denominators: $x \ne 2$ and $x \ne -4$. Divisor’s numerator: $x \ne -5$.

So $x \ne 2, -4, -5$.

4. (Core) Simplify $\dfrac{x^2 - 9}{2x + 6} \times \dfrac{4}{x - 3}$ and state the restrictions.

Solution

$$\frac{(x - 3)(x + 3)}{2(x + 3)} \times \frac{4}{x - 3} = \frac{4}{2} = 2, \qquad x \ne -3, 3$$

5. (Core) Simplify $\dfrac{x^2 + 5x + 6}{x^2 - 1} \times \dfrac{x - 1}{x + 2}$ and state the restrictions.

Solution

$$\frac{(x + 2)(x + 3)}{(x - 1)(x + 1)} \times \frac{x - 1}{x + 2} = \frac{x + 3}{x + 1}, \qquad x \ne 1, -1, -2$$

6. (Core) Simplify $\dfrac{x^2 - 16}{x^2 + 2x} \div \dfrac{x - 4}{x}$ and state the restrictions.

Solution

$$\frac{(x - 4)(x + 4)}{x(x + 2)} \times \frac{x}{x - 4} = \frac{x + 4}{x + 2}$$

Restrictions: $x \ne 0, -2$ (denominators) and $x \ne 4$ (divisor’s numerator). So $x \ne 0, -2, 4$.

7. (Core) Simplify $\dfrac{x^2 - 5x + 6}{x^2 - 4x + 4} \div \dfrac{x^2 - 9}{x^2 + x - 6}$ and state the restrictions.

Solution

$$\frac{(x - 2)(x - 3)}{(x - 2)^2} \times \frac{(x + 3)(x - 2)}{(x - 3)(x + 3)} = 1$$

Restrictions: $x \ne 2$ (first denominator), $x \ne -3, 2$ (second denominator), and $x \ne 3, -3$ (divisor’s numerator). So the answer is $1$, with $x \ne 2, 3, -3$.

8. (Challenge) A rectangle has area $\dfrac{x^2 - 1}{x + 3}$ m² and width $\dfrac{x - 1}{x^2 + 3x}$ m. Find a simplified expression for its length.

Solution

Length = area ÷ width:

$$\frac{(x - 1)(x + 1)}{x + 3} \times \frac{x(x + 3)}{x - 1} = x(x + 1)$$

The length is $x(x + 1)$ m, with $x \ne -3, 0, 1$.

9. (Challenge) Simplify $\dfrac{a^2 - b^2}{a^2 + ab} \times \dfrac{a}{a - b}$ and state the restrictions.

Solution

$$\frac{(a - b)(a + b)}{a(a + b)} \times \frac{a}{a - b} = 1$$

Restrictions: $a \ne 0$, $a \ne -b$, and $a \ne b$.