Simplifying Rational Expressions

Learning goals

Take the domain from the original denominator, before cancelling
Cancel common factors only, never common terms
Attach the restriction to a simplified expression
Pull a $-1$ out of opposite factors
Recognize lowest terms as sharing no non-constant 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 we may write it as $\frac{x^2+4x}{1}$ . Every polynomial is a rational expression, exactly as every integer $n$ is the fraction $\frac{n}{1}$ . Something like $\frac{|x|+1}{x}$ is not a rational expression, because $|x|$ is not a polynomial.

The condition on $q$ deserves care, because it is easy to misread. Saying ” $q$ is not the zero polynomial” is a statement about the polynomial itself: not all of its coefficients are $0$ . It is a much weaker statement than ” $q(x)$ is never $0$ ”. The polynomial $x-1$ is certainly not the zero polynomial, yet it evaluates to $0$ at $x=1$ . That gap between a polynomial that is not zero and a polynomial that is not zero here 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, that is, the solutions of $q(x)=0$ . Nothing the numerator does can rescue them. If $q(x_0)=0$ and $p(x_0) \neq 0$ , the quotient would have to be a number that multiplies $0$ into something nonzero, and no such number exists. If $q(x_0)=0$ and $p(x_0)=0$ as well, the quotient is no better off: every number multiplies $0$ into $0$ , so nothing pins the value down. Both cases are simply 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, and here is the reason for it. Cancelling erases factors from the page, and a factor takes its restriction with it when it goes, even though the restriction still binds.

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 cancelling works

Reducing $\frac{6}{8}$ to $\frac{3}{4}$ is not a matter of the two $2$ s looking alike and being crossed off. It is a theorem about quotients, and it is worth seeing the theorem in full, because its hypotheses are the whole story of this lesson.

Cancelling a common factor: $\dfrac{ac}{bc} = \dfrac{a}{b}$ whenever $b \neq 0$ and $c \neq 0$ #

Start from what a quotient means. For real numbers $u$ and $v$ with $v \neq 0$ , the quotient $\frac{u}{v}$ is the unique number $t$ satisfying $tv = u$ . It is worth checking that “unique” is earned: if $t_1v = u$ and $t_2v = u$ , then subtracting gives $(t_1-t_2)v = 0$ , and since $v \neq 0$ we must have $t_1 - t_2 = 0$ , so $t_1 = t_2$ . Exactly one number does the job, which is what licenses us to give it a name.

Now let $a$ , $b$ , $c$ be real numbers with $b \neq 0$ and $c \neq 0$ , and set $t = \frac{a}{b}$ . By the definition just given, $tb = a$ . Multiply both sides of that equation by $c$ and regroup the product:

t(bc) = (tb)c = ac.

Since $b \neq 0$ and $c \neq 0$ , their product $bc$ is also nonzero, so $\frac{ac}{bc}$ is defined and is the unique number whose product with $bc$ equals $ac$ . The line above says that $t$ is such a number. Uniqueness then forces

\frac{ac}{bc} = t = \frac{a}{b},

which is the claim.

Look at what the argument actually consumed. It needed $b \neq 0$ , so that $\frac{a}{b}$ names something, and it needed $c \neq 0$ , so that $bc \neq 0$ and the cancelled factor was not secretly zero. Neither hypothesis is decoration, and when $a$ , $b$ and $c$ are polynomials evaluated at some number, both of them turn into restrictions on $x$ .

Transplanting the theorem to polynomials is immediate. 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)}.

The equation is asserted at those values of $x$ and at no others. That is not fine print, and you will see it bite within the page.

The theorem cancels factors, and only factors. The $c$ in $\frac{ac}{bc}$ multiplies all of the numerator and all of the denominator. Appearing as one term of a sum does not give a quantity that power, which is why the most popular cancellation in algebra is also the most wrong. Look at $\frac{x+2}{x}$ . The $x$ upstairs is added to $2$ ; it does not multiply the $2$ . Dividing the numerator by $x$ divides every term of the numerator, 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 stroll away untouched; it turns into $\frac{2}{x}$ . Crossing out the $x$ and reporting $2$ is not a slip of the pen, it is a false statement, and a single 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. Cancelling 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 cancelling 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. Two things cannot be equal “for all $x$ ” when one of them is not even a number at $x=1$ . The honest statement is an equality that carries a condition, and the condition is part of the statement, not a footnote to it.

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:

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

The factor you divide out carries its own restriction away with it. Nothing left on the page remembers that $x=1$ was ever forbidden, which is exactly why you must write the restriction down yourself, at the moment you cancel.

Every point of the original expression sits on the line y = x + 1, with the single point at x = 1 punched out. Cancelling the common factor cannot put that point back, and the graphing lesson later in this chapter will draw it as an open circle, a hole.

Not every restriction hides. If a factor of the denominator survives the cancelling, its restriction is still visible in the answer and takes care of itself. The dangerous ones are the restrictions belonging to factors you cancelled away, because 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 its numerator and denominator share no common non-constant factor. Constants are deliberately left out of that definition, because every nonzero constant divides every polynomial: if constants counted, 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. 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 cancelling. 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.

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 needs the Factor Theorem. Since $2^3-8=0$ , the number $2$ is a root of $x^3-8$ , so $x-2$ is a factor of it. Dividing $x^3-8$ by $x-2$ (synthetic division with the coefficients $1, 0, 0, -8$ works nicely) gives the quotient $x^2+2x+4$ with remainder $0$ :

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).

The result is in lowest terms: $x^2+2x+4$ has discriminant $2^2-4(1)(4) = -12 < 0$ , so it has no real zeros. By the Factor Theorem, $x^2+2x+4$ therefore has no factor of the form $x-r$ with $r$ real. In particular $x+2$ is not a factor of it. 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 by the Factor Theorem it has no factor $x-3$ and no factor $x+3$ . 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 cancelling the $x^2$ s would predict, leaving $\frac{-9}{9}$ ) nor $1$ (what cancelling 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.

Common mistakes

Cancelling terms instead of factors. $\frac{x+2}{x}$ is not $2$ ; at $x=3$ it is $\frac{5}{3}$ . The cancellation theorem needs a factor that multiplies all of the top and all of the bottom. 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 about $x=1$ : 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 $1$ .
Cancelling before factoring. Nothing can be divided out until the numerator and the denominator are written as products. Crossing symbols out of $\frac{x^2-1}{x-1}$ as it stands is guesswork; factoring it into $\frac{(x-1)(x+1)}{x-1}$ makes the common factor a fact.

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 cancelling.
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.
Cancelling 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 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 cancelling works

The hole that cancelling leaves behind

Simplifying, step by step

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)