Rational expressions — simplification, operations

Domain restrictions: The expression is undefined wherever the denominator equals zero. We must always identify and state these restrictions.

Step-by-step from first principles:

1. Factor numerator and denominator completely

   • Why? Because $\frac{ab}{ac} = \frac{b}{c}$ (cancel common factor $a$)
   • This only works when factors are multiplied, not added

2. Identify restrictions BEFORE canceling

   • Why? Canceling removes factors, but the original denominator zeros still matter
   • Example: $\frac{x-2}{(x-2)(x+3)}$ is undefined at $x=2$ AND $x=-3$, even after canceling

3. Cancel common factors

   • Rule: $\frac{A \cdot C}{B \cdot C} = \frac{A}{B}$ for $C \neq 0$
   • NEVER cancel terms that are added/subtracted

4. State simplified form with domain

General form: $\frac{P(x)}{Q(x)} = \frac{P(x)}{Q(x)} \quad \text{where } Q(x) \neq 0$

After factoring: $\frac{A(x) \cdot C(x)}{B(x) \cdot C(x)} = \frac{A(x)}{B(x)} \quad \text{where } B(x) \neq 0, C(x) \neq 0$

Solution with WHY at each step:

Step 1: Factor numerator $x^2 - 4 = (x-2)(x+2)$ Why difference of squares? Because $a^2 - b^2 = (a-b)(a+b)$

Step 2: Factor denominator $x^2 + 5x + 6 = (x+2)(x+3)$ Why? Find two numbers that multiply to 6 and add to 5: those are 2 and 3

Step 3: Identify restrictions $x + 2 \neq 0 \Rightarrow x \neq -2$ $x + 3 \neq 0 \Rightarrow x \neq -3$ Why before canceling? These values made the ORIGINAL denominator zero

Step 4: Rewrite with factored forms $\frac{(x-2)(x+2)}{(x+2)(x+3)}$

Step 5: Cancel common factor $(x+2)$ $= \frac{x-2}{x+3}$ Why can we cancel? Because $\frac{a \cdot b}{a \cdot c} = \frac{b}{c}$ when $a \neq 0$

Final answer: $\frac{x-2}{x+3}$, where $x \neq -2, -3$

Derivation of multiplication rule: $\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}$ Why? Multiply numerators together, multiply denominators together (like numeric fractions)

Solution:

Step 1: Factor everything FIRST (before multiplying) $\frac{(x-1)(x+1)}{x(x+3)} \cdot \frac{x}{x-1}$ Why factor first? So we can cancel across fractions before multiplying

Step 2: Identify all restrictions

• From $x(x+3)$: $x \neq 0, -3$
• From $x-1$: $x \neq 1$

Step 3: Write as single fraction $\frac{(x-1)(x+1) \cdot x}{x(x+3) \cdot (x-1)}$

Step 4: Cancel common factors

• Cancel $x$ from numerator and denominator
• Cancel $(x-1)$ from numerator and denominator

$= \frac{x+1}{x+3}$

Final answer: $\frac{x+1}{x+3}$, where $x \neq 0, 1, -3$

Derivation of division rule: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$ Why? Dividing by a fraction = multiplying by its reciprocal Proof: $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \cdot \frac{d}{c}$ (multiply top and bottom by $\frac{d}{c}$)

Solution:

Step 1: Rewrite division as multiplication by reciprocal $\frac{x^2 + 2x}{x+3} \cdot \frac{x^2 - 9}{x}$

Step 2: Factor everything $\frac{x(x+2)}{x+3} \cdot \frac{(x-3)(x+3)}{x}$ Why $x^2 - 9$? Difference of squares: $x^2 - 9 = (x-3)(x+3)$

Step 3: Restrictions

• From denominators: $x \neq -3, 0$
• From original divisor's denominator: $x \neq 3$

Step 4: Combine and cancel $\frac{x(x+2)(x-3)(x+3)}{x(x+3)}$ Cancel $x$ and $(x+3)$: $= (x+2)(x-3)$ $= x^2 - x - 6$

Final answer: $x^2 - x - 6$, where $x \neq -3, 0, 3$

Rule for like denominators: $\frac{a}{c} + \frac{b}{c} = \frac{a+b}{c}$ Why? Just like $\frac{2}{7} + \frac{3}{7} = \frac{5}{7}$ — combine numerators, keep common denominator

Solution: $\frac{3x + (2x+1)}{x-5} = \frac{5x+1}{x-5}$ where $x \neq 5$

Derivation of addition rule: To add fractions with different denominators, we need a common denominator.

Why? You can't add $\frac{1}{2 + \frac{1}{3}}$ directly — you need $\frac{3}{6} + \frac{2}{6} = \frac{5}{6}$

Finding LCD (Least Common Denominator):

• Factor each denominator
• LCD = product of highest powers of all factors
• Here: $\text{LCD} = (x+1)(x-2)$

Solution:

Step 1: Write each fraction with LCD $\frac{2}{x+1} \cdot \frac{x-2}{x-2} + \frac{3}{x-2} \cdot \frac{x+1}{x+1}$ Why multiply by $\frac{x-2}{x-2}$? It equals1, so we're not changing the value

Step 2: Expand numerators $\frac{2(x-2)}{(x+1)(x-2)} + \frac{3(x+1)}{(x+1)(x-2)}$ $= \frac{2x-4}{(x+1)(x-2)} + \frac{3x+3}{(x+1)(x-2)}$

Step 3: Combine numerators $= \frac{2x-4+3x+3}{(x+1)(x-2)} = \frac{5x-1}{(x+1)(x-2)}$

Final answer: $\frac{5x-1}{(x+1)(x-2)}$, where $x \neq -1, 2$

Solution:

Step 1: Factor denominators $x^2 - 4 = (x-2)(x+2)$ $\frac{x}{(x-2)(x+2)} - \frac{2}{x+2}$

Step 2: Find LCD $\text{LCD} = (x-2)(x+2)$

Step 3: Rewrite with LCD $\frac{x}{(x-2)(x+2)} - \frac{2(x-2)}{(x+2)(x-2)}$ Why multiply second fraction by $\frac{x-2}{x-2}$? To get common denominator

Step 4: Combine $= \frac{x - 2(x-2)}{(x-2)(x+2)} = \frac{x - 2x + 4}{(x-2)(x+2)}$ $= \frac{-x+4}{(x-2)(x+2)}$

Final answer: $\frac{4-x}{(x-2)(x+2)}$, where $x \neq 2, -2$

Why it feels right: The $x$'s "look" the same in both parts.

Steel-man the mistake: You're pattern-matching to $\frac{xa}{xb} = \frac{a}{b}$, which IS correct for multiplication. The confusion is mixing up addition with multiplication.

The fix:

• You can only cancel factors (things that are multiplied)
• You CANNOT cancel terms (things that are added)
• $\frac{x \cdot 3}{x \cdot 5} = \frac{3}{5}$ ✓ (factors)
• $\frac{x+3}{x+5} \neq \frac{3}{5}$ ✗ (terms)

Test with numbers: Let $x=1$: $\frac{1+3}{1+5} = \frac{4}{6} = \frac{2}{3} \neq \frac{3}{5}$

Why it feels right: After canceling $(x-3)$, it's "gone" so why mention it?

Steel-man: You're thinking algebraically—once you've simplified, the restricted form is the "new" function.

The fix:

• The original expression was undefined at $x=3$ (makes denominator 0)
• Even though the simplified form $\frac{1}{x+2}$ would be defined at $x=3$, we maintain the original restriction
• Think: we're simplifying the expression, not changing its domain

Correct answer: $\frac{1}{x+2}$ where $x \neq 3, -2$

Visual test: Graph both functions—there's a "hole" at $x=3$ in the original, which persists.

Why it feels right: "Addition in fractions, addition in denominator"—pattern matching gone wrong.

Steel-man: You're overgeneralizing the rule "operations combine." You might be thinking $\frac{a}{b} + \frac{a}{c}$ should involve $(b+c)$ somehow.

The fix:

• LCD is the product $(x-2)(x+3)$, not the sum
• Why product? Each fraction needs to be a multiple of its original denominator
• $\frac{1}{x-2} = \frac{x+3}{(x-2)(x+3)}$ — we multiply to build up to LCD

Correct: $\frac{1}{x-2} + \frac{1}{x+3} = \frac{x+3}{(x-2)(x+3)} + \frac{x-2}{(x-2)(x+3)} = \frac{2x+1}{(x-2)(x+3)}$

Simplifying is like reducing $\frac{6}{8}$ to $\frac{3}{4}$ by canceling the common2. But here, we cancel common polynomial factors like $(x+2)$. The trick: you can ONLY cancel things that are multiplied together, not added.

Adding these expressions is like adding $\frac{1}{2} + \frac{1}{3}$—you need a common denominator first! With polynomials, you find the "least common multiple" of the denominators (factor them, then multiply together unique factors).

The big rule: Never divide by zero! So if your denominator has $(x-5)$, then $x$ can't be 5 (it would make the bottom zero, and $\frac{\text{anything}}{0}$ is undefined/breaks math). We write these as "restrictions."

Think of rational expressions as "fraction algebra"—every rule for numeric fractions (find common denominator, factor and cancel, multiply straight across) works the same way, but with polynomials!

For addition/subtraction: "LCD needs ALL factors each raised to HIGHEST power"

For canceling: "Only cancel FACTORS (×), never TERMS (+)"

Memory hook: "Can't FRaED without the RED" — Restrictions and Execution must be Done right.

Summary

Rational expressions are fractions with polynomial numerator and denominator. Core operations:

1. Simplification: Factor completely, identify restrictions, cancel common factors
2. Multiplication: Factor, multiply straight across, cancel
3. Division: Multiply by reciprocal, then follow multiplication rules
4. Addition/Subtraction: Find LCD, rewrite each fraction, combine numerators

Critical rule: State domain restrictions (values that make ANY denominator zero) with every final answer.

Connections

• Polynomial factoring — essential prerequisite for simplification
• Domain and range — restrictions define the domain
• Complex fractions — nested rational expressions
• Rational equations — solving when rational expressions are set equal
• Polynomial long division — for improper rational expressions
• Limits and continuity — calculus perspective on "holes" from canceled factors
• Partial fraction decomposition — breaking complex rational expressions apart

#flashcards/maths

What is a rational expression? :: A ratio of two polynomials $\frac{P(x)}{Q(x)}$ where $Q(x) \neq 0$

Why can't you cancel in $\frac{x+5}{x+3}$? :: Because $x$ is a TERM (added), not a FACTOR (multiplied). You can only cancel factors.

Concept Map

Hinglish (regional understanding)

Hinglish (regional understanding)