LCD of Monomials

Find the LCD of $\frac{5}{6x^2}$ and $\frac{7}{9x^3}$

Factor each denominator: $6x^2 = 2 \cdot 3 \cdot x^2$ and $9x^3 = 3^2 \cdot x^3$ = Prime factorizations found

List unique factors: Factors: $2$, $3$, and $x$ = Three unique factors

Take highest power of each: $2^1$ (from $6x^2$), $3^2$ (from $9x^3$), $x^3$ (from $9x^3$) = Highest powers identified

Multiply together: $\text{LCD} = 2 \cdot 3^2 \cdot x^3 = 2 \cdot 9 \cdot x^3$ = $18x^3$

Answer: $\text{LCD} = 18x^3$

LCD with Different Binomials

Find the LCD of $\frac{4}{x-3}$ and $\frac{2}{x+5}$

Examine each denominator: $(x-3)$ and $(x+5)$ are both linear and cannot be factored further = Already fully factored

Check for common factors: $(x-3)$ and $(x+5)$ share no common factors = No common factors

Multiply the distinct factors: $\text{LCD} = (x-3)(x+5)$ = Product of both denominators

Answer: $\text{LCD} = (x-3)(x+5)$

LCD with a Common Factor

Find the LCD of $\frac{3}{x^2-4}$ and $\frac{5}{x+2}$

Factor the first denominator: $x^2-4 = (x+2)(x-2)$ (difference of squares) = $(x+2)(x-2)$

Factor the second denominator: $(x+2)$ is already factored = $(x+2)$

Identify common and unique factors: Common: $(x+2)$. Unique to first: $(x-2)$ = $(x+2)$ and $(x-2)$

Build the LCD: Use $(x+2)$ once and $(x-2)$ once = $\text{LCD} = (x+2)(x-2)$

Answer: $\text{LCD} = (x+2)(x-2)$ or equivalently $x^2-4$

LCD with Repeated Factors

Find the LCD of $\frac{2}{(x-1)^2}$ and $\frac{7}{(x-1)(x+4)}$

List factors of first denominator: $(x-1)^2 = (x-1) \cdot (x-1)$ = $(x-1)$ appears twice

List factors of second denominator: $(x-1)(x+4)$: $(x-1)$ once, $(x+4)$ once = Two distinct factors

Take maximum occurrences: $(x-1)$: max is 2 times. $(x+4)$: max is 1 time = $(x-1)^2$ and $(x+4)$

Build the LCD: $\text{LCD} = (x-1)^2(x+4)$ = Complete LCD

Answer: $\text{LCD} = (x-1)^2(x+4)$

LCD with Quadratic Factoring

Find the LCD of $\frac{1}{x^2+5x+6}$ and $\frac{4}{x^2+2x}$

Factor the first denominator: $x^2+5x+6 = (x+2)(x+3)$ = $(x+2)(x+3)$

Factor the second denominator: $x^2+2x = x(x+2)$ = $x(x+2)$

Identify all unique factors: Factors: $x$, $(x+2)$, $(x+3)$ = Three unique factors

Build the LCD: Each factor appears at most once = $\text{LCD} = x(x+2)(x+3)$

Answer: $\text{LCD} = x(x+2)(x+3)$

Mistake: Multiplying denominators without factoring first

Why: This gives a common denominator, but not the LEAST common denominator. For $\frac{1}{x+2}$ and $\frac{1}{x^2-4}$, multiplying gives $(x+2)(x^2-4)$, but the LCD is just $x^2-4$.

Correct: Always factor first, then identify the minimum factors needed.

Mistake: Forgetting to use the highest power of repeated factors

Why: With $(x-1)^2$ and $(x-1)$, using $(x-1)$ only once won't work for the first expression.

Correct: Use each factor the MAXIMUM number of times it appears in ANY denominator.

Mistake: Not recognizing factoring patterns

Why: Expressions like $x^2-9$ look unfactorable but are $(x+3)(x-3)$.

Correct: Look for difference of squares, perfect square trinomials, and other patterns.

Mistake: Treating $(x-3)$ and $(3-x)$ as different factors

Why: $(3-x) = -(x-3)$, so they are the same factor (with a sign difference).

Correct: Factor out $-1$: $\frac{1}{3-x} = \frac{-1}{x-3}$

Combined Work Problems

When two machines or workers complete a job at different rates, you add their work rates, which are fractions.

Machine A completes a job in $x$ hours (rate: $\frac{1}{x}$), Machine B in $x+2$ hours (rate: $\frac{1}{x+2}$). Combined rate: $\frac{1}{x} + \frac{1}{x+2}$. LCD: $x(x+2)$.

Electrical Circuits

Parallel resistors combine using the formula $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}$, requiring LCD to solve.

With resistors of $x$ ohms and $(x+5)$ ohms: $\frac{1}{R_T} = \frac{1}{x} + \frac{1}{x+5}$. LCD: $x(x+5)$.

Q: What's the difference between LCD and LCM?

A: They're the same concept! LCM (Least Common Multiple) applies to numbers, while LCD (Least Common Denominator) is specifically used when working with fractions. The LCD of fractions IS the LCM of their denominators.

Q: Can I just multiply all denominators together?

A: Yes, that gives A common denominator, but not always the LEAST one. Using the LCD keeps numbers smaller and simplification easier. For $\frac{1}{4}$ and $\frac{1}{6}$, multiplying gives 24, but LCD is 12.

Q: What if denominators share no common factors?

A: Then the LCD is simply the product of all denominators. For $\frac{1}{x-1}$ and $\frac{1}{x+3}$, the LCD is $(x-1)(x+3)$.