Simplifying Rational Expressions

A rational expression is a fraction with polynomials on the top and bottom, like $\dfrac{x^2 - 9}{x^2 + 5x + 6}$. Simplifying one works just like reducing a number fraction such as $\dfrac{6}{8} = \dfrac{3}{4}$, with one extra job: keeping track of the values of $x$ that would make the denominator zero.

Key ideas

Restrictions

You can’t divide by zero, so any value that makes the denominator equal to $0$ is not allowed. These are the restrictions.

To find them, set each factor of the denominator equal to $0$. Always find restrictions from the original expression, before cancelling anything.

Simplifying

1. Factor the numerator and the denominator completely.
2. State the restrictions from the factored denominator.
3. Cancel factors that appear in both the numerator and the denominator.

You can only cancel factors (things multiplied), never terms (things added):

$$\frac{(x - 3)(x + 3)}{(x + 2)(x + 3)} = \frac{x - 3}{x + 2} \qquad \text{but} \qquad \frac{x + 3}{x + 5} \ne \frac{3}{5}$$

Factoring tools you’ll need

Type	Example
common factor	$2x^2 - 8x = 2x(x - 4)$
simple trinomial	$x^2 + 5x + 6 = (x + 2)(x + 3)$
complex trinomial	$2x^2 + 5x - 3 = (2x - 1)(x + 3)$
difference of squares	$x^2 - 9 = (x - 3)(x + 3)$
perfect square	$x^2 - 6x + 9 = (x - 3)^2$

Opposites

$a - b$ and $b - a$ are opposites: $b - a = -(a - b)$. So $\dfrac{5 - x}{x - 5} = -1$ (for $x \ne 5$). Watch for this when a factor looks almost the same as another.

Equivalent, with restrictions

The simplified expression equals the original for every allowed value of $x$. At a restricted value, the original is undefined, even if the simplified form isn’t. That’s why the restrictions stay attached to the answer.

Worked examples

Example 1: Monomials

Simplify $\dfrac{12x^3y}{18xy^2}$ and state the restrictions.

Solution. Divide the coefficients by their common factor $6$, and subtract exponents:

$$\frac{12x^3y}{18xy^2} = \frac{2x^2}{3y}, \qquad x \ne 0,\ y \ne 0$$

Both restrictions come from the original denominator $18xy^2$.

Example 2: Trinomials

Simplify $\dfrac{x^2 - 9}{x^2 + 5x + 6}$ and state the restrictions.

Solution.

$$\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{(x - 3)(x + 3)}{(x + 2)(x + 3)}$$

Restrictions: $x + 2 \ne 0$ and $x + 3 \ne 0$, so $x \ne -2$ and $x \ne -3$.

Cancel the common factor $(x + 3)$:

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

Even though $(x + 3)$ cancelled, $x = -3$ is still not allowed.

Example 3: A common factor and a difference of squares

Simplify $\dfrac{2x^2 - 8x}{x^2 - 16}$ and state the restrictions.

Solution.

$$\frac{2x^2 - 8x}{x^2 - 16} = \frac{2x(x - 4)}{(x - 4)(x + 4)} = \frac{2x}{x + 4}, \qquad x \ne 4, -4$$

Example 4: Opposites

Simplify $\dfrac{5 - x}{x^2 - 25}$ and state the restrictions.

Solution. Factor the denominator, and write $5 - x$ as $-(x - 5)$:

$$\frac{5 - x}{x^2 - 25} = \frac{-(x - 5)}{(x - 5)(x + 5)} = \frac{-1}{x + 5}, \qquad x \ne 5, -5$$

Check with $x = 0$: the original is $\dfrac{5}{-25} = -\dfrac{1}{5}$, and the answer is $\dfrac{-1}{5}$. ✓

Common mistakes

Cancelling terms instead of factors. In $\dfrac{x + 3}{x + 5}$, the $x$‘s are terms, not factors. Nothing cancels. Factor first, then cancel only whole factors.

Finding restrictions after cancelling. In Example 2, the simplified form $\dfrac{x - 3}{x + 2}$ only shows $x \ne -2$. You’d lose $x \ne -3$. Find restrictions from the original denominator.

Not factoring completely. $\dfrac{x^3 - 4x}{x^2 - 2x}$ looks stuck until you factor $x^3 - 4x = x(x - 2)(x + 2)$.

Missing opposites. $\dfrac{3 - x}{x - 3}$ simplifies to $-1$, not $1$. Factor out $-1$ to see the match.

Thinking a cancelled expression is “1” in the wrong place. $\dfrac{x + 3}{x}$ is not $3$, and $\dfrac{x}{x + 3}$ is not $\dfrac{1}{3}$.

Practice

1. (Warm-up) Simplify $\dfrac{15a^2b}{25ab^3}$ and state the restrictions.

Solution

$$\frac{15a^2b}{25ab^3} = \frac{3a}{5b^2}, \qquad a \ne 0,\ b \ne 0$$

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

Solution

$x^2 - 4x = x(x - 4)$, so $x \ne 0$ and $x \ne 4$.

3. (Warm-up) Simplify $\dfrac{4x + 12}{8x}$ and state the restriction.

Solution

$$\frac{4(x + 3)}{8x} = \frac{x + 3}{2x}, \qquad x \ne 0$$

(The $x$ in $x + 3$ can’t cancel with the $x$ in the denominator: it’s part of a sum.)

4. (Core) Simplify $\dfrac{x^2 + 7x + 12}{x^2 - 16}$ and state the restrictions.

Solution

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

5. (Core) Simplify $\dfrac{x^2 - 2x - 15}{x^2 + x - 6}$ and state the restrictions.

Solution

$$\frac{(x - 5)(x + 3)}{(x + 3)(x - 2)} = \frac{x - 5}{x - 2}, \qquad x \ne -3, 2$$

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

Solution

$$\frac{-2(x - 3)}{(x - 3)^2} = \frac{-2}{x - 3}, \qquad x \ne 3$$

7. (Core) Simplify $\dfrac{2x^2 + 5x - 3}{4x^2 - 1}$ and state the restrictions.

Solution

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

8. (Core) Are $\dfrac{x^2 - 1}{x - 1}$ and $x + 1$ equivalent? Explain.

Solution

$\dfrac{x^2 - 1}{x - 1} = \dfrac{(x - 1)(x + 1)}{x - 1} = x + 1$, but only for $x \ne 1$.

At $x = 1$, the first expression is undefined ($\tfrac{0}{0}$), while $x + 1 = 2$. So they’re equivalent for every $x$ except $1$: the correct statement is $\dfrac{x^2 - 1}{x - 1} = x + 1,\ x \ne 1$.

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

Solution

Factor out $x$ first, then keep factoring:

$$\frac{x(x^2 - 4)}{x(x^2 + x - 6)} = \frac{x(x - 2)(x + 2)}{x(x + 3)(x - 2)} = \frac{x + 2}{x + 3}$$

Restrictions from the original denominator: $x \ne 0, 2, -3$.