Rational Expressions: Simplifying, Multiplying, Adding

Fractions with polynomials in them

A rational expression is a fraction of polynomials:

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

Rational expressions behave the way rational numbers do. Adding, subtracting, multiplying or dividing two of them gives another, as long as nothing divides by zero. So every rule you know for fractions applies unchanged, and the only new skill is factoring polynomials instead of numbers.

Simplifying: cancel factors, never terms

A fraction simplifies by dividing top and bottom by a common factor: $\tfrac{12}{18} = \tfrac{2 \cdot 6}{3 \cdot 6} = \tfrac{2}{3}$.

Rational expressions work the same way once they are factored:

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

The shared factor $x + 3$ divides out.

Only factors cancel. In $\tfrac{x + 3}{x}$, the $x$ on top is a term being added, not a factor of the whole numerator, so nothing cancels. Test it: at $x = 2$, $\tfrac{2 + 3}{2} = 2.5$, while “canceling” the $x$ would give $3$.

Excluded values

The original denominator $x^2 + 5x + 6 = (x + 2)(x + 3)$ is zero at $x = -2$ and $x = -3$. Those values are excluded — the expression has no value there.

After canceling, $\tfrac{x - 3}{x + 2}$ looks as though it works at $x = -3$. It does not, because the original expression is undefined there, and simplifying cannot create a value that did not exist.

$$\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{x - 3}{x + 2}, \qquad x \ne -2, \; x \ne -3$$

Multiplying and dividing

Multiply straight across, then cancel. Factoring first keeps the numbers small:

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

Divide by multiplying by the reciprocal, exactly as with fractions:

$$\frac{x + 1}{x} \div \frac{x + 1}{x - 2} = \frac{x + 1}{x} \cdot \frac{x - 2}{x + 1} = \frac{x - 2}{x}$$

Adding and subtracting over a common denominator

Fractions add only over a common denominator, and rational expressions are the same. The least common denominator is built from the factors of each denominator, each taken as many times as it appears in any one of them.

Add $\tfrac{2}{x} + \tfrac{3}{x + 1}$.

The common denominator is $x(x + 1)$. Rewrite each fraction over it:

$$\frac{2(x + 1)}{x(x + 1)} + \frac{3x}{x(x + 1)} = \frac{2x + 2 + 3x}{x(x + 1)} = \frac{5x + 2}{x(x + 1)}$$

Subtract $\tfrac{1}{x - 1} - \tfrac{1}{x + 1}$.

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

The parentheses around $(x - 1)$ matter. The subtraction applies to the whole second numerator, so its $-1$ becomes $+1$.

Why the common denominator comes from factors

Multiplying the denominators together always works, but it can build a denominator bigger than needed. Factoring shows what is shared.

$$\frac{4}{x^2 - x} + \frac{2}{x} = \frac{4}{x(x - 1)} + \frac{2}{x}$$

Both denominators contain $x$. The common denominator is $x(x - 1)$, not $x \cdot x(x - 1)$:

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

The smallest common denominator leaves less to simplify at the end, which is the same reason the least common denominator is preferred for numbers.

Worked examples

Common mistakes

Practice problems