Example

Subtracting 2xx241x+2\frac{2x}{x^2-4} - \frac{1}{x+2}

Subtract two rational expressions where one denominator is a difference of squares that contains the other denominator as a factor:

2xx241x+2\frac{2x}{x^2 - 4} - \frac{1}{x + 2}

Step 1 — Find the LCD and rewrite each fraction. Factor the first denominator as a difference of squares: x24=(x2)(x+2)x^2 - 4 = (x - 2)(x + 2). The second denominator (x+2)(x + 2) is already one of these factors. The LCD is (x2)(x+2)(x - 2)(x + 2).

The first fraction already has the LCD. The second fraction is missing the factor (x2)(x - 2); multiply its numerator and denominator by (x2)(x - 2):

2x(x2)(x+2)1(x2)(x2)(x+2)\frac{2x}{(x - 2)(x + 2)} - \frac{1(x - 2)}{(x - 2)(x + 2)}

Distribute in the second numerator: 1(x2)=x21(x - 2) = x - 2.

Step 2 — Subtract the rational expressions. Subtract the numerators over the common denominator, placing the second numerator in parentheses:

2x(x2)(x2)(x+2)\frac{2x - (x - 2)}{(x - 2)(x + 2)}

Distribute the negative sign:

2xx+2(x2)(x+2)\frac{2x - x + 2}{(x - 2)(x + 2)}

Combine like terms:

x+2(x2)(x+2)\frac{x + 2}{(x - 2)(x + 2)}

Step 3 — Simplify, if possible. The numerator (x+2)(x + 2) shares a common factor with the denominator. Cancel the shared factor (x+2)(x + 2):

x+2(x2)(x+2)=1x2\frac{x + 2}{(x - 2)(x + 2)} = \frac{1}{x - 2}

This example demonstrates the subtraction process where finding the LCD involves factoring a difference of squares. After converting to a common denominator and distributing the negative sign across the second numerator, the resulting combined numerator shares a factor with the denominator, which cancels out to produce the simplified result.

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Updated 2026-04-30

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Ch.7 Rational Expressions and Functions - Intermediate Algebra @ OpenStax

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