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Algebra and Precalculus A Practical Guide

Cameron Geisler

Section 7.1 Intro to Rational Expressions

Recall that a fraction, or rational number, is a number that can be written by dividing two integers (like \(\frac{3}{4}\text{,}\) or \(-\frac{5}{8}\)). A rational expression is similar, except with variables.
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Subsection 7.1.1 Rational Expressions

Definition 7.1.1.

A rational expression is an algebraic expression of the form,
\begin{gather*} \frac{\text{polynomial}}{\text{polynomial}} \end{gather*}
In other words, it is the quotient of two polynomials.
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  • It’s also called an algebraic fraction, as opposed to a “regular” fraction which only involves numbers.
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Example 7.1.2. Identifying Rational Expressions.

For example, these are all rational expressions,
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  • \(\displaystyle \dfrac{3x + 4}{2x^2 - 1}\)
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  • \(\displaystyle \dfrac{x^3 - x - 1}{x^2 + 2x + 5}\)
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  • \(\dfrac{5}{x}\) (\(5\text{,}\) being just a constant, is still considered a polynomial)
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  • \(\dfrac{3xy}{x^2 - 6}\) (polynomials can have multiple different variables)
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However, these are not rational expressions:
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  • \(\dfrac{\sqrt{x} + 1}{x}\text{,}\) because \(\sqrt{x} + 1\) is not a polynomial.
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  • \(\dfrac{2^x + 3}{4x - 2}\text{,}\) because \(2^x + 3\) is not a polynomial.
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Example 7.1.3. Polynomials as Rational Expressions.

All polynomials are technically also considered rational expressions, because any polynomial can be written as itself divided by 1 (and 1 is a polynomial also). For example,
\begin{gather*} \underbrace{x^2 + 2x - 3}_{\text{polynomial}} = \underbrace{\frac{x^2 + 2x - 3}{1}}_{\text{rational expression}} \end{gather*}
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Subsection 7.1.2 Non-Permissible Values

A rational expression is not defined if their denominator is 0, because division by zero is undefined. Values of \(x\) (or whatever the variable is) that make the denominator 0 are called non-permissible values.
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Definition 7.1.4.

The non-permissible values (NPVs) of a rational expressions are the values of the variable(s) that make the denominator 0.
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  • Also called restrictions, or excluded values.
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To determine the non-permissible values of a rational expression, figure out what values of \(x\) will make the denominator 0.
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Example 7.1.5. Basic Non-Permissible Values.

Find the non-permissible values of the rational expression,
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\begin{gather*} \frac{5}{x-3} \end{gather*}
This is when \(x-3\) is equal to 0. You may be able to observe that this is when \(x = 3\text{.}\) Or, you can set \(x-3\) equal to 0 and solve for \(x\text{.}\)
\begin{align*} x-3 \amp= 0\\ x \amp= 3 \amp\amp \text{adding 3 to both sides} \end{align*}
Therefore, the NPV is \(x \neq 3\text{.}\)
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The purpose of NPVs is to communicate what values of the variable are not allowed to be plugged into the expression.
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Example 7.1.6. NPVs with Difference of Squares.

Find the non-permissible values of,
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\begin{gather*} \frac{x+1}{x^2-4} \end{gather*}
The denominator is \(x^2-4\text{,}\) so the NPVs are the values that make \(x^2-4 = 0\text{.}\) To solve this, one way is to factor it as a difference of squares, then set each factor equal to 0 and solve for \(x\text{.}\)
\begin{align*} x^2 - 4 \amp= 0\\ (x - 2)(x + 2) \amp= 0 \amp\amp \text{factoring as a difference of squares}\\ x - 2 = 0 \qquad x + 2 \amp= 0\\ x = 2 \qquad x \amp= -2 \end{align*}
Therefore, the NPVs are \(x \neq 2,-2\text{.}\) Alternatively, you could take square roots,
\begin{align*} x^2 - 4 \amp = 0\\ x^2 \amp = 4\\ x \amp = \pm 2 \amp\amp \text{taking square roots of both sides} \end{align*}
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Example 7.1.7. No Non-Permissible Values.

Find the non-permissible values of,
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\begin{gather*} \frac{2x^2}{x^2+4} \end{gather*}
The denominator is \(x^2+4\text{,}\) so the NPVs are when,
\begin{gather*} x^2 + 4 = 0 \end{gather*}
To solve this, you may recognize that the left side \(x^2+4\) is always positive, so it is never equal to 0, and so there is no solution. Or, isolate for \(x\text{,}\)
\begin{gather*} x^2 = -4 \end{gather*}
Then, there is no number \(x\) such that \(x^2 = -4\) (in other words, the equation has no solution), so the denominator is never 0. Therefore, there are no NPVs.
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Example 7.1.8. NPVs with Trinomial Factoring.

Find the non-permissible values of,
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\begin{gather*} \frac{x^2-1}{x^2-3x+2} \end{gather*}
The NPVs are when,
\begin{gather*} x^2 - 3x + 2 = 0 \end{gather*}
We can factor, then set each factor equal to 0 and solve.
\begin{align*} (x - 1)(x - 2) \amp= 0\\ x - 1 = 0 \qquad x - 2 \amp= 0\\ x = 1 \qquad x \amp= 2 \end{align*}
Therefore, the NPVs are \(x \neq 1,2\text{.}\)
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Example 7.1.9. NPVs with Multiple Variables.

Find the non-permissible values of,
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\begin{gather*} \frac{2x+3y}{xy} \end{gather*}
The denominator is \(xy\text{,}\) so the NPVs are when,
\begin{align*} xy \amp= 0 \end{align*}
This is a product of two variables, so it is equal to 0 precisely when either \(x = 0\) or \(y = 0\text{.}\) Both of these make the denominator 0, so both must be excluded. Therefore, the NPVs are \(x \neq 0\) and \(y \neq 0\text{.}\)
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Example 7.1.10. NPVs with Common Factors.

Find the non-permissible values of,
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\begin{gather*} \frac{1}{x^2 + 3x} \end{gather*}
The NPVs are when,
\begin{gather*} x^2 + 3x = 0 \end{gather*}
You can factor out a common factor of \(x\text{,}\) then set each factor equal to 0 and solve for \(x\text{.}\)
\begin{align*} x(x + 3) \amp= 0\\ x = 0 \qquad x + 3 \amp= 0\\ x = 0 \qquad x \amp= -3 \end{align*}
Therefore, the NPVs are \(x \neq 0,-3\text{.}\)
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Example 7.1.11. NPVs Using Quadratic Formula.

Find the non-permissible values of,
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\begin{gather*} \frac{x+1}{4x^2 - 2x - 3} \end{gather*}
The NPVs are when,
\begin{gather*} 4x^2 - 2x - 3 = 0 \end{gather*}
In fact, this quadratic does not factor, so we must use the quadratic formula.
\begin{align*} x \amp= \frac{-(-2) \pm \sqrt{(-2)^2 - 4(4)(-3)}}{2(4)}\\ x \amp= \frac{2 \pm \sqrt{4 + 48}}{8}\\ x \amp= \frac{2 \pm \sqrt{52}}{8}\\ x \amp= \frac{2 \pm 2\sqrt{13}}{8}\\ x \amp= \frac{1 \pm \sqrt{13}}{4} \end{align*}
Therefore, the NPVs are \(x \neq \frac{1 + \sqrt{13}}{4}, \frac{1 - \sqrt{13}}{4}\text{.}\)
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In summary, to find non-permissible values, find the zeros of the denominator, by setting it equal to 0 and solving for \(x\text{.}\)
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Note: the non-permissible values do not depend on what the numerator is at all, only on what the denominator is.
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Exercise Group 7.1.1. Practice Exercises.

Find all non-permissible values (if any) for each rational expression.
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(a)
\(\dfrac{3}{x}\)
Answer.
\(x\neq 0\)
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(b)
\(\dfrac{x+2}{x-1}\)
Answer.
\(x\neq 1\)
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(c)
\(\dfrac{4x^2-12x}{x^2-5x+6}\)
Answer.
\(x\neq 2,3\)
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(e)
\(\dfrac{x^2-x-6}{x^2-4x}\)
Answer.
\(x\neq 0,4\)
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(f)
\(\dfrac{4x^2-9}{2x+3}\)
Answer.
\(x\neq -\dfrac{3}{2}\)
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(g)
\(\dfrac{x^2-x-6}{x^2-4x+4}\)
Answer.
\(x\neq 2\)
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(h)
\(\dfrac{x^2+4}{x^2-3x+2}\)
Answer.
\(x\neq 1,2\)
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(i)
\(\dfrac{x^2-5x+6}{9-x^2}\)
Answer.
\(x\neq \pm 3\)
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(j)
\(\dfrac{3a}{ab}\)
Answer.
\(a\neq 0\text{,}\) \(b\neq 0\)
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(k)
\(\dfrac{3}{x^2+4x+3}\)
Answer.
\(x \neq -1, -3\)
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(l)
\(\dfrac{x^2+3x-18}{x+6}\)
Answer.
\(x\neq -6\)
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(n)
\(\dfrac{x+2}{x^2-x-6}\)
Answer.
\(x\neq 3,-2\)
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(o)
\(\dfrac{5x-1}{x^2-4x+1}\)
Hint.
the quadratic does not factor
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Answer.
\(x \neq 2 \pm \sqrt{3}\)
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(p)
\(\dfrac{x^2-4}{x^4-1}\)
Hint.
difference of squares
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Answer.
\(x\neq -1,1\)
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