Rational Expressions

A review of reduction, multiplication, division, addition and subtraction of rational expressions

The Essentials

When reducing a rational expression, first factor the numerator and denominator. Next, note the numbers that make the denominator equal to zero and cancel identical terms:

\[ \frac{4x^s + 16x - 20}{x^2 + 2x - 3} \] \[ \frac{4(x + 5)(x - 1)}{(x + 3)(x - 1)} \]

Here we note that x = 3 and x = 1 make it so that this expression doesn’t exist. Canceling gives us:

\[ \frac{4x + 20}{x + 3} \]

When multiplying rational expressions, the terms can just be combined and reduced as one fraction. When dividing rational expressions, just flip the divisor, combine the expressions, and reduce them:

\[ \frac{2x^2 - 4x - 30}{x^2 + 10x + 21} \div \frac{4x^2 - 16x - 20}{3x + 21} \]

Flipping and combining gives us:

\[ \frac{{(2x^2 - 4x - 30)(3x + 21)}}{{(x^2 + 10x + 21)(4x^2 - 16x - 20)}} \] \[ \frac{{6(x + 3)(x - 5)(x + 7)}}{{4(x + 7)(x + 3)(x - 5)(x + 1)}} \]

Here we note that \( x = 5,−1,−3,−7 \) make the expression non-existent.

\[ \frac{3}{2(x + 1)} \]

When adding or subtracting rational expressions, the denominators need to be factored. After they are factored it’s easy to see what is needed to make the denominators equal:

\[ \frac{3}{4x + 8} - \frac{x - 5}{x^2 + 4x + 4} \] \[ \frac{3}{4(x + 2)} - \frac{x - 5}{(x + 2)(x + 2)} \] \[ \frac{3(x + 2)}{4(x + 2)(x + 2)} - \frac{4(x - 5)}{4(x + 2)(x + 2)} \]

The numerators are distributed and combined while the denominator stays factored:

\[ \frac{3x + 6 + 20 - 4x}{4(x + 2)(x + 2)} \] \[ \frac{26 - x}{4(x + 2)(x + 2)} \]

Example

Find x in the following rational expression:

\[ \frac{3x^2 - 12x + 9}{x^2 + x - 20} = \frac{6x}{x^2 + x - 20} - \frac{6}{x + 5} \] \[ \frac{3(x - 1)(x - 3)}{(x + 5)(x - 4)} = \frac{6x}{(x + 5)(x - 4)} - \frac{6}{x + 5} \]

Here note that the equation is undefined at \( x = −5, 4 \)

\[ \frac{3(x - 1)(x - 3)}{(x + 5)(x - 4)} = \frac{6x}{(x + 5)(x - 4)} - \frac{6x - 24}{(x + 5)(x - 4)} \] \[ \frac{3(x - 1)(x - 3)}{(x + 5)(x - 4)} = \frac{24}{(x + 5)(x - 4)} \] \[ \frac{3(x - 1)(x - 3)}{(x + 5)(x - 4)} \div \frac{24}{(x + 5)(x - 4)} = \frac{24}{(x + 5)(x - 4)} \div \frac{24}{(x + 5)(x - 4)} \] \[ \frac{3(x - 1)(x - 3)}{(x + 5)(x - 4)} \cdot \frac{(x + 5)(x - 4)}{24} = \frac{24}{(x + 5)(x - 4)} \cdot \frac{(x + 5)(x - 4)}{24} \] \[ \frac{(x - 1)(x - 3)}{8} = 1 \] \[ (x - 1)(x - 3) = 8 \] \[ x^2 - 4x + 3 = 8 \] \[ x^2 + 4x - 5 = 0 \] \[ (x + 5)(x - 1) = 0 \]

The numbers -5 and 1 are solutions to this equation. However, as noted above -5 makes the equation undefined and is therefore an extraneous solution making 1 the only solution.

Practice

Evaluate this expression and determine if any of the solutions are extraneous:

\[ \frac{x + 1}{x - 3} = \frac{-x^2 + 8x + 2}{(x + 1)(x - 3)} \]

Solution:

\( x = 0, 3 \) where \( x = 3 \) is an extraneous solution

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