1Cademy - Adding $$\frac{8}{x^2-2x-3} + \frac{3x}{x^2+4x+3}$$

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How to Add or Subtract Rational Expressions

Example

Adding $\frac{8}{x^2-2x-3} + \frac{3x}{x^2+4x+3}$

Carry out the complete addition of two rational expressions whose LCD and equivalent forms involve trinomial denominators with a shared binomial factor:

$\frac{8}{x^2 - 2x - 3} + \frac{3x}{x^2 + 4x + 3}$

Step 1 — Rewrite each fraction with the LCD. Factor the denominators: $x^2 - 2x - 3 = (x + 1)(x - 3)$ and $x^2 + 4x + 3 = (x + 1)(x + 3)$ . The LCD is $(x + 1)(x - 3)(x + 3)$ . Multiply each fraction by its missing factor:

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

Distribute in each numerator: $8(x + 3) = 8x + 24$ and $3x(x - 3) = 3x^2 - 9x$ .

Step 2 — Add the numerators over the common denominator. Combine and collect like terms:

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

Step 3 — Simplify, if possible. To check whether $3x^2 - x + 24$ factors, look for two integers whose product is $3 \cdot 24 = 72$ and whose sum is $-1$ . No such pair exists, so the trinomial is prime. Because no factor of the numerator matches any factor of the denominator, the result is already in simplified form:

$\frac{3x^2 - x + 24}{(x + 1)(x - 3)(x + 3)}$

This example completes the addition workflow that begins with finding the LCD of these two expressions and rewriting them as equivalent fractions. It illustrates a case where the combined numerator turns out to be a prime polynomial — after verifying that no factorization is possible, no further simplification can occur. The problem also demonstrates that when two trinomial denominators share a common binomial factor (here $(x + 1)$ ), the LCD contains three binomial factors rather than four.

Updated 2026-04-30

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