Simplifying Rational Expressions

Learning goals

Find the domain from the original denominator, before canceling
Cancel common factors only, never common terms
Attach the restriction to a simplified expression
Pull a $-1$ out of opposite factors
Recognize lowest terms, including reducing any common numerical factor

What a rational expression is

A rational expression is a quotient

\frac{p(x)}{q(x)},

where $p$ and $q$ are polynomials and $q$ is not the zero polynomial. The name is the same word as in rational number, and for the same reason: both are ratios. A rational number is a ratio of two integers, and a rational expression is a ratio of two polynomials.

These all qualify:

\frac{x+2}{x^2-9}, \qquad \frac{5}{x}, \qquad \frac{3x^3-x}{7}, \qquad x^2+4x.

The last two are worth a second look. In $\frac{3x^3-x}{7}$ the denominator is the constant polynomial $7$ , which is a perfectly good polynomial. And $x^2+4x$ is a rational expression because it can be written as $\frac{x^2+4x}{1}$ . Every polynomial is a rational expression, exactly as every integer $n$ is the fraction $\frac{n}{1}$ .

The condition on $q$ needs care, because it is easy to misread. Saying ” $q$ is not the zero polynomial” means not all of its coefficients are $0$ . It does not mean ” $q(x)$ is never $0$ ”. The polynomial $x-1$ is a perfectly good denominator, yet it evaluates to $0$ at $x=1$ . That gap, between a polynomial that is never the zero polynomial and a polynomial that can still equal zero at a particular input, is the entire subject of the next section.

The domain comes first

The domain of a rational expression is the set of real numbers you are allowed to substitute for $x$ . Polynomials themselves cause no trouble: a polynomial can be evaluated at every real number. The only thing that can go wrong is division by zero. So:

\text{the domain of } \frac{p(x)}{q(x)} \text{ is every real number } x \text{ with } q(x) \neq 0.

The excluded values are exactly the real zeros of the denominator, the solutions of $q(x)=0$ . Nothing the numerator does can rescue them. If $q(x_0)=0$ and $p(x_0) \neq 0$ , no number times $0$ can give something nonzero, so no quotient exists there. If $p(x_0)=0$ as well, the quotient is no better off: every number times $0$ gives $0$ , so nothing pins a single value down. Either way, $x_0$ is excluded.

Take $\frac{x+5}{x^2-4x+3}$ . Factor the denominator:

x^2-4x+3 = (x-1)(x-3),

which is zero exactly when $x=1$ or $x=3$ . The domain is every real number except $1$ and $3$ . Notice that $x=-5$ , the zero of the numerator, is not excluded at all; there the expression is a legitimate $\frac{0}{48}=0$ .

Find the excluded values from the original denominator, before you cancel anything. This is the most important habit in the chapter. Canceling can erase a factor from the page, but it cannot erase the restriction that factor caused. The value that made it zero is still excluded, even though nothing left in the simplified expression shows it.

Check your understanding

Which values must be excluded from the domain of $\dfrac{x+5}{x^2-x-6}$ ?

x = -5

only B

x = 3

and

x = -2

x = -3

and

x = 2

x = 6

and

x = -1

Why canceling works

Reducing $\frac{6}{8}$ to $\frac{3}{4}$ works because dividing the top and the bottom of a fraction by the same nonzero number never changes its value:

\frac{6}{8} = \frac{2\cdot3}{2\cdot4} = \frac{3}{4}.

The two $2$ s divide out because they are a shared factor, not because they merely look alike. Written for any numbers, that is

\frac{ac}{bc} = \frac{a}{b} \qquad \text{whenever } b \neq 0 \text{ and } c \neq 0.

Both conditions matter. You need $b \neq 0$ so that $\frac{a}{b}$ names a number at all, and you need $c \neq 0$ so that the factor you are canceling was never secretly zero.

Move to polynomials and nothing about the idea changes, only the objects being multiplied. If $p$ , $q$ and $d$ are polynomials, then at any real number $x$ for which $q(x) \neq 0$ and $d(x) \neq 0$ ,

\frac{p(x)\,d(x)}{q(x)\,d(x)} = \frac{p(x)}{q(x)}.

That equation holds at those values of $x$ , and nowhere else. Keep the condition in view. You will see it bite within the page.

The rule cancels factors, and only factors. The $c$ in $\frac{ac}{bc}$ multiplies all of the numerator and all of the denominator. A term in a sum does not have that power, which is why $\frac{x+2}{x}$ is the single most common cancellation mistake in algebra. The $x$ upstairs is added to $2$ ; it does not multiply the $2$ . Dividing the numerator by $x$ divides every term of it, so what really happens is

\frac{x+2}{x} = \frac{x}{x} + \frac{2}{x} = 1 + \frac{2}{x} \qquad (x \neq 0).

The $2$ does not vanish; it turns into $\frac{2}{x}$ . Crossing out the $x$ and reporting $2$ is a false statement, and one input settles it. At $x=3$ ,

\frac{3+2}{3} = \frac{5}{3}, \qquad \text{but the crossing-out answer says } 2.

So $\frac{x+2}{x}$ is already in lowest terms; there is nothing there to cancel. Canceling is never “erase a symbol you can see twice”. It is “divide the top and the bottom by the same nonzero quantity”, and dividing the top means dividing all of it.

This is why the first step of every simplification is to factor. Until the numerator and denominator are written as products, the theorem has nothing to grip. Written as they stand, $x^2-1$ and $x-1$ share no visible common factor; written as $(x-1)(x+1)$ and $(x-1)$ , they obviously do.

Check your understanding

Which statement about $\dfrac{x^2+3x}{x}$ is correct?

A It equals

x+3

for every real number

x

. B It equals

x^2+3

, since the

x

cancels the

x

3x

. C It equals

x+3

for every

x \neq 0

. D It cannot be simplified, because

x

is a term of the numerator.

The hole that canceling leaves behind

Here is the example the rest of the chapter leans on. Take

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

Read the domain off that denominator before touching anything: it is zero exactly at $x=1$ , so $x=1$ is excluded. For every other $x$ the factor $x-1$ is nonzero, so the cancellation theorem applies with $c = x-1$ , and it gives

\frac{x^2-1}{x-1} = x+1 \qquad (x \neq 1).

Now look hard at what that line does and does not say. The expression $\frac{x^2-1}{x-1}$ and the polynomial $x+1$ are not the same object. The polynomial $x+1$ is perfectly happy at $x=1$ , where it takes the value $2$ . The original expression is undefined at $x=1$ , because its denominator is $0$ there. The table shows precisely how far the agreement goes.

$x$	$\dfrac{x^2-1}{x-1}$	$x+1$
$-2$	$\dfrac{3}{-3} = -1$	$-1$
$0$	$\dfrac{-1}{-1} = 1$	$1$
$\tfrac{1}{2}$	$\dfrac{-3/4}{-1/2} = \tfrac{3}{2}$	$\tfrac{3}{2}$
$1$	$\dfrac{0}{0}$ , undefined	$2$
$3$	$\dfrac{8}{2} = 4$	$4$

Every row agrees except one, and in that row the left column has nothing at all to report. So here is the principle, and it is the intellectual heart of the lesson:

Canceling preserves the value wherever both sides are defined. It does not preserve the domain.

Every point of the original expression sits on the line y = x + 1, except at x = 1: the original denominator excludes that point, and canceling only erases the visible evidence of the exclusion, not the exclusion itself. The graphing lesson later in this chapter marks a gap like this with an open circle.

Not every restriction hides. If a factor of the denominator survives the canceling, its restriction is still visible in the answer. The dangerous ones are the restrictions that belonged to factors you canceled away: the simplified expression is defined there and gives no hint that the original was not.

Check your understanding

For which values of $x$ is the statement $\dfrac{x^2-4}{x-2} = x+2$ true?

A For every real number

x

. B For no real number

x

, since the two sides are different expressions. C Only for

x = 2

. D For every real number

x

except

x = 2

Simplifying, step by step

A rational expression is in lowest terms when there is nothing left to divide out: any common numerical factor between the coefficients has been reduced, the way $\frac{6}{8}$ reduces to $\frac{3}{4}$ , and the numerator and denominator share no common non-constant factor, one that actually contains $x$ . Bare constants are left out of that second half on purpose, because every nonzero constant divides every polynomial; if a shared constant alone disqualified an expression, nothing would ever be in lowest terms. To get there, follow four steps, and never skip the second.

Factor the numerator and the denominator completely.
Read the excluded values off the original denominator. Every zero of it is excluded, whether or not its factor survives step 3.
Divide out every common factor, numerical and polynomial. Each cancellation is legal precisely because that factor is nonzero everywhere on the domain from step 2.
Write the simplified expression together with its restrictions.

Worked example 1 Simplify $\dfrac{3x^2+6x}{x^2-4}$ and state its domain

Factor the top and the bottom completely. The numerator has a common factor of $3x$ , and the denominator is a difference of squares:

\frac{3x^2+6x}{x^2-4} = \frac{3x(x+2)}{(x-2)(x+2)}.

Now read the excluded values from that denominator, before canceling. It is zero when $x=2$ and when $x=-2$ , so the domain is every real number except $2$ and $-2$ .

On that domain both $x-2$ and $x+2$ are nonzero, so the common factor $x+2$ may be divided out:

\frac{3x(x+2)}{(x-2)(x+2)} = \frac{3x}{x-2} \qquad (x \neq 2 \text{ and } x \neq -2).

Watch what just happened to the two restrictions. The restriction $x \neq 2$ is still plain to see in the answer, because $x-2$ is still in the denominator. The restriction $x \neq -2$ has vanished from sight: $\frac{3x}{x-2}$ is perfectly well defined at $x=-2$ , where it equals $\frac{-6}{-4} = \frac{3}{2}$ . The original is not defined there. That is why the restriction is written beside the answer, and it is why step 2 comes before step 3.

Worked example 2 Simplify $\dfrac{9-x^2}{x^2-2x-3}$

Factor both parts. The numerator is a difference of squares, and the denominator is a trinomial whose factors must multiply to $-3$ and add to $-2$ :

\frac{9-x^2}{x^2-2x-3} = \frac{(3-x)(3+x)}{(x-3)(x+1)}.

The denominator is zero at $x=3$ and $x=-1$ , so those two values are excluded.

The top and bottom look as though they share nothing, but $3-x$ and $x-3$ are opposites, and one is $-1$ times the other:

3-x = -(x-3).

Rewrite the numerator using that, and the common factor appears:

\frac{-(x-3)(x+3)}{(x-3)(x+1)} = \frac{-(x+3)}{x+1} \qquad (x \neq 3 \text{ and } x \neq -1).

A quick check at $x=0$ , which is in the domain: the original gives $\frac{9-0}{0-0-3} = \frac{9}{-3} = -3$ , and the simplified form gives $\frac{-(0+3)}{0+1} = -3$ . They agree, as they must.

Check your understanding

Simplify $\dfrac{x^2-16}{4-x}$ and state the result with its domain.

x+4

, for

x \neq 4

-(x+4)

, for

x \neq 4

x+4

, for every real number

x

-(x+4)

, for

x \neq -4

Worked example 3 Simplify $\dfrac{x^3-8}{x^2-4}$

The denominator factors on sight as $(x-2)(x+2)$ , so the excluded values are $x=2$ and $x=-2$ .

The numerator is a difference of cubes, $x^3-8=x^3-2^3$ , and the difference-of-cubes identity factors it directly:

x^3-8 = (x-2)(x^2+2x+4).

Now assemble and cancel the common factor $x-2$ , which is nonzero everywhere on the domain:

\frac{x^3-8}{x^2-4} = \frac{(x-2)(x^2+2x+4)}{(x-2)(x+2)} = \frac{x^2+2x+4}{x+2} \qquad (x \neq 2 \text{ and } x \neq -2).

Is that in lowest terms? Check whether $x+2$ could still be a factor of $x^2+2x+4$ : if it were, then $x=-2$ would make $x^2+2x+4$ equal $0$ . It does not, since $(-2)^2+2(-2)+4=4$ , so $x+2$ is not a factor and nothing more cancels. Once again the restriction $x \neq 2$ is invisible in the answer and must be carried along in writing.

Worked example 4 Simplify $\dfrac{x^2-9}{x^2+9}$ and state its domain

Start, as always, with the denominator. The equation $x^2+9=0$ would need $x^2=-9$ , and no real number has a negative square, so the denominator is never zero. This rational expression has no excluded values at all: its domain is every real number.

Factoring gives

\frac{x^2-9}{x^2+9} = \frac{(x-3)(x+3)}{x^2+9},

and $x^2+9$ has no real zeros, so it cannot have a factor of the form $x-3$ or $x+3$ : if it did, plugging in $x=3$ or $x=-3$ would have to give $0$ , and it does not (both give $18$ ). There is no common non-constant factor, and the expression is already in lowest terms.

The tempting moves here are all illegal. The $9$ s are terms, not factors, so they do not cancel, and neither do the $x^2$ s. One input kills both ideas at once. At $x=1$ the expression is

\frac{1-9}{1+9} = \frac{-8}{10} = -\frac{4}{5},

which is neither $-1$ (what canceling the $x^2$ s would predict, leaving $\frac{-9}{9}$ ) nor $1$ (what canceling the $9$ s would predict, leaving $\frac{x^2}{x^2}$ ). The instruction “simplify” never promises that something will cancel; sometimes the honest answer is that the expression is already in lowest terms.

Check your understanding

Simplify $\dfrac{2x^2-8}{4x+8}$ and state the result in lowest terms with its domain.

\dfrac{x-2}{2}

, for every real number

x

\dfrac{2x-4}{4}

, for

x \neq -2

\dfrac{x-2}{2}

, for

x \neq -2

\dfrac{x-2}{2}

, for

x \neq 2

Common mistakes

Canceling terms instead of factors, or canceling before factoring. $\frac{x+2}{x}$ is not $2$ ; at $x=3$ it is $\frac{5}{3}$ . Nothing can be divided out until the numerator and denominator are written as products: $\frac{x^2-1}{x-1}$ shows no common factor as it stands, but $\frac{(x-1)(x+1)}{x-1}$ does. Same trap in $\frac{x^2+9}{x+3}$ : the $9$ and the $3$ are not factors, and $x+3$ is not a factor of $x^2+9$ , since $(-3)^2+9 = 18 \neq 0$ .
Reading the domain off the simplified form. $\frac{3x^2+6x}{x^2-4}$ becomes $\frac{3x}{x-2}$ , and the restriction $x \neq -2$ is nowhere to be seen in that answer. It still binds. Excluded values come from the original denominator, always.
Writing an unconditional equality. " $\frac{x^2-1}{x-1} = x+1$ " is false as stated, because the two sides disagree at $x=1$ : one is undefined and the other is $2$ . Write ” $\frac{x^2-1}{x-1} = x+1$ for $x \neq 1$ ” and the statement becomes true.
Losing the sign on opposite factors. $3-x$ and $x-3$ are not the same factor; $3-x = -(x-3)$ . So $\frac{3-x}{x-3} = -1$ for $x \neq 3$ , not $1$ .

Key takeaways

A rational expression is $\frac{p(x)}{q(x)}$ with $p$ and $q$ polynomials and $q$ not the zero polynomial. Every polynomial is one, with denominator $1$ .
Its domain is every real number except the zeros of the denominator. Find those zeros from the original denominator, before any canceling.
Cancellation law: $\frac{ac}{bc} = \frac{a}{b}$ whenever $b \neq 0$ and $c \neq 0$ . It removes common factors, never common terms, which is why every simplification starts by factoring.
Canceling preserves the value wherever both sides are defined, but it does not preserve the domain. The honest way to record a simplification is with its restriction attached, as in $\frac{x^2-1}{x-1} = x+1$ for $x \neq 1$ .
Opposite factors differ by a sign: $3-x = -(x-3)$ , so a factor of $-1$ comes out rather than nothing.
An expression is in lowest terms when any numerical coefficients are reduced and the numerator and denominator share no common non-constant factor. Some expressions, like $\frac{x^2-9}{x^2+9}$ , are there already, and some have no excluded values at all.

Simplifying Rational Expressions

Learning goals

What a rational expression is

The domain comes first

Why canceling works

The hole that canceling leaves behind

Simplifying, step by step

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice