Simplifying Rational Expressions

Factor, cancel, and always say what x cannot be.

A student council sold candy bars to raise money for prom. They collected a total of $x^2 - 9$ dollars, split evenly among $x + 3$ tables. To find how much each table earned, you divide: $\frac$ . If you try to evaluate that fraction at $x = 5$ , you get KaTeX can only parse string typed expression. But something interesting happens when you factor the numerator — the expression collapses into something much simpler. That process is called simplifying a rational expression.

A rational expression is a fraction where the numerator and denominator are polynomials. The same rule that lets you cancel KaTeX can only parse string typed expression by dividing top and bottom by 2 works here too — you can cancel any factor that appears in both the numerator and denominator. The key is that you can only cancel factors, never terms. Factors are things multiplied together. Terms are things added or subtracted. That distinction matters more than almost anything else in this topic.

The general process has two steps. First, factor the numerator and denominator completely. Second, cancel any factors that are identical on top and bottom.

\frac, = \frac, = x - 3

Every time you cancel a factor from the denominator, you create a restriction. The original denominator $x + 3$ cannot equal zero, so $x \neq -3$ . You must state that restriction even after simplifying, because the simplified expression $x - 3$ looks perfectly fine at $x = -3$ , but the original fraction is undefined there. The restriction comes from the original denominator, always.

Here is a straightforward example with a monomial factor.

Simplifying a rational expression with a common monomial

\frac{6x^2 + 9x}{3x}

Start by factoring the numerator. Look for what both terms share.

\frac{3x(2x + 3)}{3x}

3x divides into both 6x² and 9x, so it factors out front.

2x + 3, \quad x \neq 0

The 3x on top and bottom cancel. The denominator was 3x, so x = 0 is the restriction.

Now a harder one where both the numerator and denominator need full factoring.

Simplifying when both numerator and denominator factor as trinomials

\frac{x^2 + x - 6}{x^2 - 4}

Neither piece is factored yet. Work on numerator and denominator separately.

\frac{(x+3)(x-2)}{(x+2)(x-2)}

The numerator factors as a trinomial — find two numbers that multiply to -6 and add to 1. That's 3 and -2. The denominator is a difference of squares.

\frac{x+3}{x+2}, \quad x \neq 2, \quad x \neq -2

The (x-2) factor cancels. Both x = 2 and x = -2 made the original denominator zero, so both are restrictions.

One situation that trips people up is a factor that looks almost the same on top and bottom but has opposite signs. When the denominator contains $(a - x)$ and the numerator contains $(x - a)$ , those are not the same factor — but they are negatives of each other. You can factor out $-1$ from one of them to reveal the cancellation.

Simplifying when a factor appears with opposite signs

\frac{x - 5}{25 - x^2}

The denominator is a difference of squares, but it's written in the less common order.

\frac{x - 5}{(5+x)(5-x)}

Factor the denominator as a difference of squares: 25 - x² = (5+x)(5-x).

\frac{x - 5}{-(x+5)(x-5)}

Rewrite (5-x) as -(x-5) so the top and bottom share the exact same factor.

\frac{-1}{x+5}, \quad x \neq 5, \quad x \neq -5

The (x-5) cancels, leaving the -1 in the numerator. Both values from the original denominator are restrictions.

Here is the graph of the original expression and the simplified expression together. They look identical — but the original has a hole at $x = -2$ where the canceled factor made the denominator zero.

Interactive graph — scroll to zoom, drag to pan

The red curve is the original. The blue curve is the simplified form. They overlap everywhere except at $x = 2$ , where the original is undefined and the simplified form is not. That gap is the hole — a single missing point. The restriction you write is the algebraic way of marking that hole.

Simplify each expression completely and state all restrictions.

Practice Questions

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

\frac{x^2 - 7x + 12}{x^2 - 9}

\frac{2x^2 + 7x - 4}{2 - x - x^2 + x^3 - 2x^2 + x}

Let me give you cleaner third questions. Here are three solid practice problems.

Practice Questions

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

\frac{x^2 - 7x + 12}{x^2 - 9}

\frac{3x - 15}{x^2 - 10x + 25}

Regents Corner

On Part II and Part III of the Algebra II Regents, simplifying rational expressions almost always carries a requirement to state the domain restrictions. A fully simplified expression with no restrictions stated will lose credit — the grader is specifically looking for them. Write restrictions as $x \neq __$ and identify every value that made the original denominator zero, not just the factors you canceled.

A student sees (x-3) in both the numerator and denominator of a sum like

\frac

and cancels it, getting KaTeX can only parse string typed expression. That is wrong. You can only cancel a factor — something multiplied across the entire numerator. Here, (x-3) is one term added to 2, not a factor of the whole numerator. The fraction does not simplify. Factor completely first, and only cancel what multiplies everything on top.

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Polynomial Long Division

Solving Rational Equations