Simplifying Algebraic Fractions

To simplify an algebraic fraction you generally follow these steps:

1. Factor the polynomials in both the numerator and denominator into irreducible form using factorization.
2. Determine the conditions under which the algebraic fraction is defined (C.E.).
3. Apply the invariant property: divide numerator and denominator by any common factors.
4. The result is the simplified fraction.

Note. You can only simplify by canceling common factors - never by canceling terms. For example, this fraction can be simplified by the factor “a”: $$ \frac{a}{ac} = \frac{\frac{a}{a}}{ \frac{ac}{a}} = \frac{1}{c} $$ But this one cannot be simplified by “a”: $$ \frac{a}{a+b} $$

A worked example

Consider the algebraic fraction:

$$ \frac{6x^2-12x+6}{x^2-1} $$

The condition of existence is:

$$ C.E. \ = \ x \ne 1 \ ∧ \ x \ne -1$$

Now let’s factor the polynomials into irreducible form:

$$ \frac{6x^2-12x+6}{x^2-1} $$

$$ \frac{3 \cdot 2 \cdot x^2- 3 \cdot 2^2 \cdot x+ 3 \cdot 2}{x^2-1} $$

The denominator is a difference of squares:

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

Now factor the numerator by grouping:

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

The numerator is a perfect square trinomial:

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

Both numerator and denominator share the factor (x-1).

By the invariant property of fractions, divide numerator and denominator by the common factor (x-1):

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

Now simplify:

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

So the simplified fraction is:

$$ \frac{6 \cdot (x-1)}{(x+1)} $$

This fraction is equivalent to the original expression.

And the same method applies in general.