Rational Expressions & Functions

A2.E.1Simplify01/04

Simplify rational expressions.

Step01of 04

A rational expression is a fraction of polynomials. To simplify: factor numerator and denominator, then cancel common factors.

Always factor first, never cancel terms. $\frac{x + 3}{x}$ does NOT simplify to $3$ — the $x$ in the numerator isn't a factor.

A2.E.1 — Simplify rational expressions.

Step 01 of 04

A rational expression is a fraction of polynomials. To simplify: factor numerator and denominator, then cancel common factors.

Always factor first, never cancel terms. $\frac{x + 3}{x}$ does NOT simplify to $3$ — the $x$ in the numerator isn't a factor.

Step 02 of 04

Worked example.

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

Factor both:

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

The $(x + 3)$ factor cancels.

Step 03 of 04

Domain restrictions persist. The original expression was undefined at $x = - 2$ AND $x = - 3$ (from the denominator). After cancelling, the simplified version is still undefined at both — even though $x = - 3$ doesn't appear anywhere in the simplified form.

State restrictions explicitly: $x \neq = - 2, x \neq = - 3$ .

Step 04 of 04

Tricky factor — opposite binomials.

\frac{x - 5}{5 - x} = \frac{x - 5}{- ( x - 5 )} = - 1

Whenever you see $a - b$ over $b - a$ , factor out a $- 1$ from one of them; the rest cancels and you're left with just $- 1$ .

Key insight

Factor everything → cancel common factors → state any restrictions from the ORIGINAL denominator. Never cancel terms across a sum or difference; only factors come out.

A2.E.2 — Multiply and divide rational expressions.

Step 01 of 04

Multiplication. Factor everything, then cancel any factor that appears in any numerator with the same factor in any denominator. Multiply what's left.

\frac{x ^{2} - 4}{x ^{2} + 6 x + 9} \cdot \frac{x + 3}{x + 2}

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

Step 02 of 04

Division — flip the second fraction and multiply. Same as dividing numeric fractions.

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

Worked example:

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

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

Step 03 of 04

Don't forget restrictions. Every original denominator's zero stays excluded — and so does any factor you flipped (since it was originally a denominator).

In the division example above: $x \neq = 0$ (both denominators) AND $x \neq = - 1$ (after flipping).

Step 04 of 04

Coefficient handling. Pull out numerical GCFs early to make the algebra cleaner.

\frac{6 x ^{2}}{4 x} = \frac{3 x}{2}

Same logic on more complex expressions: factor out constants, factor the polynomial, cancel.

Key insight

Multiplication: factor → cancel factors → multiply. Division: flip the second fraction → multiply (then same procedure). Track restrictions from every original denominator, AND from any flipped numerator.

A2.E.3 — Add and subtract rational expressions.

Step 01 of 04

Same denominator — combine numerators directly.

\frac{3 x}{x + 2} + \frac{6}{x + 2} = \frac{3 x + 6}{x + 2} = \frac{3 ( x + 2 )}{x + 2} = 3

Result simplifies because $(x + 2)$ cancels.

Step 02 of 04

Different denominators — find the LCD (least common denominator), then build each fraction up to it.

The LCD is the product of EVERY distinct factor, each raised to its highest power across all denominators.

Step 03 of 04

Worked example. Add $\frac{2}{x - 3} + \frac{5}{x + 1}$ .

Denominators are different and have no common factor. LCD = $(x - 3) (x + 1)$ .

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

= \frac{2 x + 2 + 5 x - 15}{( x - 3 ) ( x + 1 )} = \frac{7 x - 13}{( x - 3 ) ( x + 1 )}

Step 04 of 04

Subtraction needs the same minus-distribution care as polynomial subtraction.

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

LCD = $x^{2}$ :

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

Notice the parentheses around $x + 2$ — without them the minus only flips the first term and you'd get $3 x - x + 2$ , which is wrong.

Key insight

Find LCD = product of all distinct factors at highest power. Build each fraction up to LCD. Combine numerators carefully (parens around everything you subtract!). Simplify the result.

A2.E.4 — Solve rational equations — watch for extraneous solutions.

Step 01 of 04

Multiply both sides by the LCD to clear all fractions in one step. Then solve the resulting polynomial equation.

Always check final answers against the original — multiplying by the LCD can introduce extraneous solutions if a candidate makes any original denominator zero.

Step 02 of 04

Worked example. Solve $\frac{1}{x} + \frac{1}{x + 2} = \frac{3}{4}$ .

LCD = $4 x (x + 2)$ . Multiply every term by it:

4 (x + 2) + 4 x = 3 x (x + 2)

4 x + 8 + 4 x = 3 x^{2} + 6 x

8 x + 8 = 3 x^{2} + 6 x

0 = 3 x^{2} - 2 x - 8

Step 03 of 04

Solve the quadratic.

3 x^{2} - 2 x - 8 = (3 x + 4) (x - 2) = 0

x = - \frac{4}{3} or x = 2

Check both against the original. Neither makes $x = 0$ or $x = - 2$ , so both are valid.

Step 04 of 04

Extraneous-solution example. Solve $\frac{x}{x - 2} = \frac{2}{x - 2} + 3$ . Multiply by $(x - 2)$ :

x = 2 + 3 (x - 2) \Rightarrow x = 2 + 3 x - 6 \Rightarrow x = 2

But $x = 2$ makes the original denominators zero — extraneous. No solution.

Key insight

Multiply both sides by the LCD to wipe out fractions. Solve the resulting polynomial. Reject any answer that makes the ORIGINAL denominators zero — that's an extraneous solution.

A2.E.5 — Asymptotes of rational functions.

Step 01 of 04

A rational function $f (x) = \frac{p ( x )}{q ( x )}$ has two kinds of asymptotes that shape its graph.

Vertical asymptote	→where the denominator equals zero (and numerator does NOT)
Horizontal asymptote	→depends on degree comparison of numerator and denominator

Step 02 of 04

Vertical asymptotes. Set the (factored) denominator equal to zero. Each zero gives a vertical asymptote — UNLESS that factor also cancels with the numerator (then it's a "hole" instead).

f (x) = \frac{x + 1}{( x - 3 ) ( x + 2 )} \Rightarrow VA at x = 3, x = - 2

Step 03 of 04

Horizontal asymptote — three cases based on degree comparison.

Degree(num) < Degree(denom)	→ $y = 0$
Degree(num) = Degree(denom)	→ $y = \frac{leading coef of num}{leading coef of denom}$
Degree(num) > Degree(denom)	→no horizontal asymptote (slant or curve at the ends)

Step 04 of 04

Worked example. Find all asymptotes of $f (x) = \frac{2 x ^{2} - 8}{x ^{2} - 4 x + 3}$ .

Factor: numerator $2 (x^{2} - 4) = 2 (x - 2) (x + 2)$ ; denominator $(x - 1) (x - 3)$ . Nothing cancels.

Vertical asymptotes: $x = 1, x = 3$ . Same degree (both 2), so horizontal asymptote: $y = 2/1 = 2$ .

Two vertical asymptotes, one horizontal asymptote.

Key insight

Vertical asymptotes from the denominator's zeros (after cancellation). Horizontal asymptote from comparing degrees: smaller top → 0; equal → ratio of leading coefficients; bigger top → none.

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