1Cademy - Rewriting $$\frac{8}{x^2-2x-3}$$ and $$\frac{3x}{x^2+4x+3}$$ with LCD $$(x+1)(x-3)(x+3)$$

Learn Before

Finding the LCD of $\frac{8}{x^2-2x-3}$ and $\frac{3x}{x^2+4x+3}$

Example

Rewriting $\frac{8}{x^2-2x-3}$ and $\frac{3x}{x^2+4x+3}$ with LCD $(x+1)(x-3)(x+3)$

Rewrite $\frac{8}{x^2 - 2x - 3}$ and $\frac{3x}{x^2 + 4x + 3}$ as equivalent rational expressions with denominator $(x + 1)(x - 3)(x + 3)$ .

Step 1 — Factor each denominator.

$x^2 - 2x - 3 = (x + 1)(x - 3)$

$x^2 + 4x + 3 = (x + 1)(x + 3)$

Step 2 — Find the LCD. From the factored denominators, the LCD is $(x + 1)(x - 3)(x + 3)$ .

Step 3 — Multiply each expression by the missing factor. Compare each factored denominator to the LCD to identify the factor that is absent.

For $\frac{8}{(x + 1)(x - 3)}$ : the denominator is missing $(x + 3)$ . Multiply both the numerator and denominator by $(x + 3)$ :

$\frac{8}{(x + 1)(x - 3)} \cdot \frac{(x + 3)}{(x + 3)} = \frac{8(x + 3)}{(x + 1)(x - 3)(x + 3)}$

For $\frac{3x}{(x + 1)(x + 3)}$ : the denominator is missing $(x - 3)$ . Multiply both the numerator and denominator by $(x - 3)$ :

$\frac{3x}{(x + 1)(x + 3)} \cdot \frac{(x - 3)}{(x - 3)} = \frac{3x(x - 3)}{(x + 1)(x + 3)(x - 3)}$

Step 4 — Simplify the numerators. Distribute in each numerator:

$\frac{8x + 24}{(x + 1)(x - 3)(x + 3)} \quad \text{and} \quad \frac{3x^2 - 9x}{(x + 1)(x + 3)(x - 3)}$

Both expressions now share the same denominator and are ready to be added or subtracted. This process mirrors the missing factors technique used for numerical fractions: just as multiplying $\frac{7}{12}$ by $\frac{3}{3}$ converts it to $\frac{21}{36}$ , multiplying a rational expression by a binomial factor over itself converts it to an equivalent expression with the LCD as its denominator.

Updated 2026-04-30

Contributors are:

Who are from:

References

Learn Before

Related

Learn After