Rational Expressions and Excluded Values, SAT math lesson

Type $\frac{x^2 - 9}{x - 3}$ into a calculator at $x = 3$ and it reports an error, yet the simplified form $x + 3$ happily gives $6$ . Both can’t be the same function unless something is said about $x = 3$ . That small gap is what this topic is about. Simplifying a fraction with variables is mostly factoring, but the values that were forbidden at the start stay forbidden at the end, and the graph shows them as a missing point.

See also Cancellation and Excluded Values and Rational Functions, Holes, and Asymptotes in Foundations for Calculus and Rational Functions and Domain Restrictions in Precalculus.

Key terms

Rational expression

Definition A rational expression is a quotient

\frac{P(x)}{Q(x)}

of two polynomials, where $Q$ is not the zero polynomial. It is defined exactly for the real numbers $x$ with $Q(x) \ne 0$ . The polynomial $Q(x)$ is the denominator.

In plain words It is a fraction whose top and bottom are polynomials. It works like a number fraction, with the one rule that the bottom may never be zero.

Example The expression

\frac{x + 1}{x^2 - 4}

is rational. At $x = 3$ it equals $\frac{4}{5}$ . At $x = 2$ the denominator is $0$ , so the expression has no value there.

Excluded values of the denominator

Definition An excluded value of a rational expression is a real number that makes its denominator equal to $0$ . To find them, set the denominator equal to $0$ and solve, usually by factoring. The domain of the expression is every real number except the excluded values. A zero numerator causes no trouble, since $\frac{0}{5} = 0$ is a perfectly good number.

In plain words Look only at the bottom. Whatever makes the bottom zero is off limits.

Example For

\frac{x + 5}{x^2 - x - 12}

the denominator factors as $(x - 4)(x + 3)$ , so the excluded values are $4$ and $-3$ . The value $-5$ makes the numerator zero, and there the expression is simply $0$ .

Cancellation

Definition For polynomials $A$ , $B$ and $C$ ,

\frac{A \cdot C}{B \cdot C} = \frac{A}{B}

for every $x$ with $B(x) \ne 0$ and $C(x) \ne 0$ . Cancellation removes a common factor of the entire numerator and the entire denominator. It never removes a term of a sum. The values where $C(x) = 0$ were excluded before the step and stay excluded after it.

In plain words Factor completely first, then cancel whole matching factors. If you are tempted to cancel something attached by a plus or minus sign, stop.

Example Factor both parts and cancel $x + 3$ ,

\begin{aligned} \frac{x^2 - 9}{x^2 + x - 6} &= \frac{(x - 3)(x + 3)}{(x + 3)(x - 2)} \\[0.4em] &= \frac{x - 3}{x - 2}, \end{aligned}

with $x \ne -3$ and $x \ne 2$ . The restriction $x \ne -3$ must be written, because the right side alone would allow it.

Here is the classic mistake. Writing $\frac{x + 6}{6} = x + 1$ cancels a term, not a factor. At $x = 6$ the left side is $\frac{12}{6} = 2$ and the right side is $7$ , so they are not equal.

Equivalent expressions

Definition Two expressions are equivalent on a set of real numbers when both are defined and have the same value at every number in that set. A rational expression and its simplified form are equivalent on the original domain, the set where every original denominator is nonzero. When the simplified form allows more inputs, the shared domain has to be stated, for example by writing “for $x \ne 3$ ”.

In plain words Equal formulas need equal domains. Simplifying is allowed to make a formula shorter, not to make it defined in new places.

Example Factoring $2x$ out of the numerator gives

\begin{aligned} \frac{2x^2 - 6x}{x - 3} &= \frac{2x(x - 3)}{x - 3} \\[0.4em] &= 2x \quad \text{for } x \ne 3. \end{aligned}

Check a value. At $x = 5$ the left side is $\frac{20}{2} = 10$ and the right side is $10$ . On the SAT a question like this often says “for $x > 3$ ” so that every excluded value is already left out.

Hole

Definition If a factor $x - c$ cancels completely from the denominator of a rational function, so that $c$ is no longer a zero of the simplified denominator, the graph has a hole (a removable discontinuity) at $x = c$ . The hole is the point $(c, L)$ , where $L$ is the value of the simplified expression at $c$ , and it is drawn as an open circle because the function is not defined there.

In plain words The graph looks like the graph of the simplified formula with one single point punched out.

Example The function

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

simplifies to $x + 3$ for $x \ne 3$ . Its graph is the line $y = x + 3$ with an open circle at $(3, 6)$ , since $3 + 3 = 6$ . The value $f(3)$ does not exist, even though the points around it approach height $6$ .

Operations with rational expressions

Definition For rational expressions $\frac{A}{B}$ and $\frac{C}{D}$ , wherever every denominator is nonzero,

\begin{aligned} \frac{A}{B} \cdot \frac{C}{D} &= \frac{AC}{BD}, \\[0.4em] \frac{A}{B} \pm \frac{C}{D} &= \frac{AD \pm BC}{BD}. \end{aligned}

For a sum it is usually cleaner to use the least common denominator, the product of each distinct factor of the denominators to the highest power in which it appears. The result keeps every exclusion of every original expression, even if a factor later cancels.

In plain words Multiply straight across. To add or subtract, rewrite both fractions over one common bottom first. Then factor and simplify, and carry along every forbidden value from the start.

Example The common denominator of $\frac{2}{x - 1}$ and $\frac{3}{x + 2}$ is $(x - 1)(x + 2)$ , so

\begin{aligned} &\frac{2}{x - 1} + \frac{3}{x + 2} \\[0.4em] &= \frac{2(x + 2) + 3(x - 1)}{(x - 1)(x + 2)} \\[0.4em] &= \frac{5x + 1}{(x - 1)(x + 2)}, \end{aligned}

with $x \ne 1$ and $x \ne -2$ . For a product, the factors cancel but the exclusions stay,

\begin{gathered} \frac{x^2 - 4}{x + 3} \cdot \frac{x + 3}{x - 2} = x + 2 \\[0.4em] \text{for } x \ne -3 \text{ and } x \ne 2. \end{gathered}

Division and the nonzero divisor

Definition Dividing by a rational expression means multiplying by its reciprocal,

\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C} = \frac{AD}{BC}.

This needs $B \ne 0$ and $D \ne 0$ so that both expressions are defined, and also $C \ne 0$ so that the divisor $\frac{C}{D}$ is not zero.

In plain words Flip the second fraction and multiply. The thing you divide by must exist and must not be zero, so a value that makes its numerator zero is excluded too.

Example Simplify

\frac{x + 1}{x - 2} \div \frac{x + 1}{x + 5}.

Flipping and multiplying gives $\frac{(x + 1)(x + 5)}{(x - 2)(x + 1)}$ , which simplifies to $\frac{x + 5}{x - 2}$ . The excluded values are $2$ from the first denominator, $-5$ from the divisor’s denominator, and $-1$ because there the divisor equals $0$ . The simplified result looks defined at $-1$ and $-5$ , so the restrictions $x \ne 2$ , $x \ne -5$ and $x \ne -1$ must be stated with it.

Shifted or scaled reciprocal

Definition A shifted or scaled reciprocal is a function of the form

f(x) = \frac{a}{x - h} + k,

with real constants $a \ne 0$ , $h$ and $k$ . Its only excluded value is $x = h$ . Its graph is the graph of $y = \frac{1}{x}$ stretched vertically by the factor $\lvert a \rvert$ (and reflected across the x-axis if $a < 0$ ), then moved $h$ units right and $k$ units up. It has the vertical asymptote $x = h$ and the horizontal asymptote $y = k$ , and the output $k$ is never reached, since $\frac{a}{x - h}$ is never $0$ .

In plain words It is the basic $\frac{1}{x}$ curve, made taller or flipped by $a$ and slid so that its two invisible guide lines cross at $(h, k)$ .

Example For

f(x) = \frac{3}{x - 2} + 1

the excluded input is $2$ and the asymptotes are $x = 2$ and $y = 1$ . At $x = 5$ the value is $\frac{3}{3} + 1 = 2$ , and at $x = -1$ it is $-1 + 1 = 0$ . Unlike a hole, the excluded input here does not hide a single missing point. Near $x = 2$ the outputs grow without bound in size.

Check yourself

Which of the following is equivalent to $\frac{x^2 - 4x}{x^2 - 16}$ for $x > 4$ ? (A) $\frac{x}{x + 4}$ (B) $\frac{x}{x - 4}$ (C) $\frac{x}{4}$ (D) $\frac{1}{x + 4}$

Show the answer

The answer is (A). Factor the top and bottom,

\begin{aligned} \frac{x^2 - 4x}{x^2 - 16} &= \frac{x(x - 4)}{(x - 4)(x + 4)} \\[0.4em] &= \frac{x}{x + 4}. \end{aligned}

The condition $x > 4$ keeps away both excluded values $4$ and $-4$ . Choice (C) comes from cancelling the $x^2$ terms and the minus signs, which cancels terms instead of factors. Choice (B) keeps the factor that should cancel and drops the one that should stay, and choice (D) loses the factor $x$ . A check at $x = 5$ settles it, since the original is $\frac{5}{9}$ and only (A) gives $\frac{5}{9}$ .

The graph of $f(x) = \frac{x^2 + 2x - 15}{x - 3}$ in the xy-plane has a hole at the point $(3, b)$ . What is the value of $b$ ?

Show the answer

The numerator factors as $(x - 3)(x + 5)$ , so $f(x) = x + 5$ for $x \ne 3$ . The hole is at height $3 + 5 = 8$ , so $b = 8$ .

Which of the following is equivalent to $\frac{1}{x - 1} - \frac{1}{x + 1}$ for $x > 1$ ? (A) $0$ (B) $\frac{2}{x^2 - 1}$ (C) $\frac{-2}{x^2 - 1}$ (D) $\frac{2x}{x^2 - 1}$

Show the answer

The answer is (B). Over the common denominator $(x - 1)(x + 1) = x^2 - 1$ ,

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

Choice (C) subtracts the numerators in the wrong order, and choice (D) adds the numerators instead of subtracting them. Choice (A) treats the two fractions as if they were equal. A quick check at $x = 3$ confirms (B), since $\frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ and $\frac{2}{8} = \frac{1}{4}$ .

Rational Expressions and Excluded Values

Theory, part 59

Practice, part 60

Key terms

Rational expression

Excluded values of the denominator

Cancellation

Equivalent expressions

Hole

Operations with rational expressions

Division and the nonzero divisor

Shifted or scaled reciprocal

Check yourself

How this lesson connects

Builds on

Used later in