7.1 Multiply and Divide Rational Expressions

Rational Expressions

Rational expressions are fractions where the numerator and denominator are polynomials. Multiplying and dividing them follows the same logic as multiplying and dividing numeric fractions, but with an extra step: you need to factor first so you can cancel common factors. This section covers how to simplify, multiply, and divide rational expressions, plus how to handle rational functions.

Undefined Values in Rational Expressions

A rational expression is undefined whenever its denominator equals zero (division by zero isn't allowed). To find the undefined values, set the denominator equal to zero and solve.

Example: Find the undefined value of $\frac{x+1}{x-2}$

1. Set the denominator equal to zero: $x - 2 = 0$

2. Solve: $x = 2$

3. The expression is undefined when $x = 2$

The domain of a rational expression is all real numbers except the values that make the denominator zero. Always check for undefined values before simplifying, because canceling factors can hide them.

Simplifying Rational Expressions

Simplifying a rational expression means reducing it to lowest terms, just like simplifying a numeric fraction. The key is factoring.

1. Factor the numerator and denominator completely
2. Cancel any factors that appear in both the numerator and denominator
3. Write what remains

Example: Simplify $\frac{x^2-4}{x^2+3x-10}$

1. Factor the numerator: $x^2 - 4 = (x+2)(x-2)$

2. Factor the denominator: $x^2 + 3x - 10 = (x+5)(x-2)$

3. Cancel the common factor $(x-2)$:

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

Note that $x \neq 2$ still, even though $(x-2)$ was canceled. The original expression was undefined there, and simplifying doesn't change the domain.

Multiplying Rational Expressions

Multiplication works the same way as with numeric fractions: multiply straight across. But you should factor before multiplying so you can cancel common factors and keep things manageable.

1. Factor all numerators and denominators completely
2. Cancel any common factors between any numerator and any denominator
3. Multiply the remaining numerators together and the remaining denominators together

Example: Multiply $\frac{2x+6}{x-1} \cdot \frac{x+4}{3x+9}$

1. Factor everything:

   • $2x + 6 = 2(x+3)$
   • $3x + 9 = 3(x+3)$
   • $x+4$ and $x-1$ don't factor further

2. Rewrite with factors: $\frac{2(x+3)}{x-1} \cdot \frac{x+4}{3(x+3)}$

3. Cancel the common factor $(x+3)$: $\frac{2}{x-1} \cdot \frac{x+4}{3}$

4. Multiply across: $\frac{2(x+4)}{3(x-1)}$

This is much cleaner than multiplying the unfactored expressions and then trying to simplify afterward.

Dividing Rational Expressions

Division of rational expressions follows the same rule as numeric fractions: multiply by the reciprocal of the divisor. After flipping, it becomes a multiplication problem.

1. Rewrite the division as multiplication by flipping the second fraction (swap its numerator and denominator)
2. Factor all numerators and denominators
3. Cancel common factors, then multiply across

Example: Divide $\frac{3x^2-12}{2x+4} \div \frac{x+1}{x-2}$

1. Rewrite as multiplication: $\frac{3x^2-12}{2x+4} \cdot \frac{x-2}{x+1}$

2. Factor:

   • $3x^2 - 12 = 3(x^2 - 4) = 3(x+2)(x-2)$
   • $2x + 4 = 2(x+2)$

3. Substitute the factors: $\frac{3(x+2)(x-2)}{2(x+2)} \cdot \frac{x-2}{x+1}$

4. Cancel $(x+2)$: $\frac{3(x-2)}{2} \cdot \frac{x-2}{x+1}$

5. Multiply across: $\frac{3(x-2)^2}{2(x+1)}$

Rational Function Operations

A rational function is a function written as the ratio of two polynomials, like $f(x) = \frac{2x-1}{x+3}$. Multiplying and dividing rational functions uses the exact same steps as rational expressions. The one extra thing to track is the domain: you need to exclude every $x$-value that makes any denominator zero in the original problem, even if that factor cancels.

Example: Given $f(x) = \frac{2x-1}{x+3}$ and $g(x) = \frac{x^2+4x+3}{x-1}$, find $f(x) \cdot g(x)$.

1. Set up the multiplication: $\frac{2x-1}{x+3} \cdot \frac{x^2+4x+3}{x-1}$
2. Factor: $x^2 + 4x + 3 = (x+1)(x+3)$
3. Rewrite: $\frac{2x-1}{x+3} \cdot \frac{(x+1)(x+3)}{x-1}$
4. Cancel $(x+3)$: $\frac{(2x-1)(x+1)}{x-1}$

So $f(x) \cdot g(x) = \frac{(2x-1)(x+1)}{x-1}$, with $x \neq -3$ and $x \neq 1$.

Notice that $x \neq -3$ must still be excluded from the domain even though $(x+3)$ canceled. That restriction came from the original function $f(x)$.

Applying Rational Expressions