Simplifying Rational Expressions: Videos & Practice Problems

Rational expressions are quotients of polynomials, similar to rational numbers being quotients of integers. To simplify them, factor the numerator and denominator completely, then cancel common factors. Remember, the denominator cannot be zero, as this makes the expression undefined. For example, setting the denominator equal to zero helps find values where the expression is undefined. Simplifying involves factoring terms like monomials and binomials, then reducing the quotient by canceling shared factors. This process deepens understanding of polynomials, coefficients, and exponents, reinforcing algebraic manipulation skills essential for working with multivariable polynomials and expressions in standard form.

Intro To Rational Expressions

Rational expressions are algebraic fractions where both the numerator and denominator are polynomials. Similar to how rational numbers are quotients of integers, rational expressions represent the quotient of two polynomials, typically written as \(\frac{p}{q}\), where \(p\) and \(q\) are polynomials. For example, the expression \(\frac{4x}{x - 2}\) is a rational expression because both the numerator \$4x\( and the denominator \)x - 2\( are polynomials.

One crucial rule when working with rational expressions is that the denominator cannot be zero, as division by zero is undefined. This is analogous to rational numbers, where the denominator cannot be zero. To determine where a rational expression is undefined, you set the denominator equal to zero and solve for the variable. For instance, in the expression \)\frac{4x}{x - 2}\(, setting the denominator equal to zero gives \)x - 2 = 0\(, which means \)x = 2\( is not allowed because it would make the denominator zero.

Consider the rational expression \)\frac{x - 1}{2x - 6}\(. To find where this expression is undefined, set the denominator equal to zero:

\[2x - 6 = 0\]

Add 6 to both sides:

\[2x = 6\]

Divide both sides by 2:

\[x = 3\]

This means the expression is undefined at \)x = 3\(.

To evaluate the expression at a specific value, such as \)x = 2\(, substitute 2 into both the numerator and denominator:

\[\frac{2 - 1}{2(2) - 6} = \frac{1}{4 - 6} = \frac{1}{-2} = -\frac{1}{2}\]

Thus, the value of the expression at \)x = 2\( is \)-\frac{1}{2}$.

Understanding rational expressions involves recognizing that they are fractions with polynomials in the numerator and denominator, and applying the fundamental rule that the denominator cannot be zero. This ensures the expression is defined and allows for proper evaluation and simplification. Mastery of these concepts is essential for working confidently with rational expressions in algebra.

For which of the following values of $x$ is the rational expression undefined?
$\frac{5}{x}$

• A. $0$
• B. $5$
• C. $1$
• D. 2

Answer: (A)

For which of the following values of $x$ is the rational expression undefined?
$\frac{7}{x-3}$

• A. $0$
• B. $-3$
• C. $3$
• D. $7$

Answer: (C)

For which of the following values of $x$ is the rational expression undefined?
$\frac{x+1}{x^2-9}$

• A. $0$
• B. $\pm 3$
• C. $\pm 9$
• D. $\pm 1$

Answer: (B)

Evaluate the rational expression below at $x=-1$
$\frac{12}{x+3}$

• A. $4$
• B. $3$
• C. $12$
• D. $6$

Answer: (D)

Evaluate the rational expression below at $x=2$
$\frac{x^2-4}{x^2-x-6}$

• A. $\infty$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer: (B)

Evaluate the rational expression below at $x=-2$
$\frac{3x}{x^2+x-6}$

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\infty$
• D. $\frac{3}{4}$

Answer: (A)

Simplifying Rational Expressions

Rational expressions are fractions where the numerator and denominator are polynomials. Simplifying rational expressions involves factoring both the numerator and denominator completely and then canceling any common factors, much like simplifying numerical fractions. For example, to simplify a fraction like \(\frac{28}{35}\), you factor both numbers: \$28 = 7 \times 4\( and \)35 = 7 \times 5\(. Since 7 is a common factor, it can be canceled, leaving \)\frac{4}{5}\( as the simplified form.

When simplifying rational expressions with variables, the process is similar. Consider the expression \)\frac{28x^3}{35x^5}\(. First, factor the constants and variables: \)28 = 7 \times 4\(, \)35 = 7 \times 5\(, \)x^3 = x \times x \times x\(, and \)x^5 = x \times x \times x \times x \times x\(. Cancel the common factors of 7 and three \)x\( terms from numerator and denominator. This leaves \)\frac{4}{5x^2}\(, since two \)x\( terms remain in the denominator, which can be written as \)x^2\(.

In cases where expressions are already factored, simplification is straightforward by canceling common binomial factors. For instance, in \)\frac{(x+2)(x-2)}{(x+6)(x+2)}\(, the common factor \)(x+2)\( cancels, resulting in \)\frac{x-2}{x+6}\(.

When expressions are not fully factored, factoring out the greatest common factor (GCF) is essential. For example, to simplify \)\frac{x-5}{3x^2 - 15x}\(, factor the denominator by pulling out the GCF: \)3x^2 - 15x = 3x(x - 5)\(. Now, the expression becomes \)\frac{x-5}{3x(x-5)}\(. Canceling the common factor \)(x-5)\( leaves \)\frac{1}{3x}$ as the simplified expression.

Mastering the simplification of rational expressions requires recognizing factoring patterns, identifying common factors, and applying cancellation rules. This skill is fundamental in algebra and prepares you for more advanced topics involving rational functions and equations.

Simplify the rational expressions below:
$\frac{6x}{12x}$

• A. $2x$
• B. $2$
• C. $\frac{1}{2}x$
• D. $\frac{1}{2}$

Answer: (D)

Simplify the rational expressions below:
$\frac{x^2-9}{x^2-3x}$

• A. $3$
• B. $\frac{x}{x+3}$
• C. $\frac{x+3}{x}$
• D. $\frac{x-3}{x}$

Answer: (C)

Simplify the rational expressions below:
$\frac{x^2-4x}{x^2-2x-8x}$

• A. $\frac{x-4}{x-10}$
• B. $\frac{x+4}{x+10}$
• C. $\frac{x-2}{x-5}$
• D. $\frac{x+2}{x+5}$

Answer: (A)

Simplifying Rational Expressions Example 1

Understanding how to write equivalent forms of rational expressions is a fundamental skill in algebra that enhances problem-solving flexibility. Consider the rational expression $$\backslash (\backslash frac\{-x^2\; +\; 4x\}\{x^2\; -\; 2x\}\backslash )$$. One effective method to find equivalent expressions is through factoring both the numerator and denominator.

Start by factoring a negative $\backslash (x\backslash )$ from the numerator. Since $\backslash (-x\; \backslash times\; x\; =\; -x^2\backslash )$ and $\backslash (-x\; \backslash times\; (-4)\; =\; 4x\backslash )$, the numerator factors to $\backslash (-x(x\; -\; 4)\backslash )$. Similarly, factor a negative $\backslash (x\backslash )$ from the denominator: $\backslash (-x\; \backslash times\; (-x)\; =\; x^2\backslash )$ and $\backslash (-x\; \backslash times\; 2\; =\; -2x\backslash )$, so the denominator becomes $\backslash (-x(-x\; +\; 2)\backslash )$. Canceling the common factor $\backslash (-x\backslash )$ leaves the equivalent expression $\backslash (\backslash frac\{x\; -\; 4\}\{-x\; +\; 2\}\backslash )$, which can be rewritten as $\backslash (\backslash frac\{x\; -\; 4\}\{2\; -\; x\}\backslash )$ by rearranging the denominator terms.

Another approach is to factor a positive $\backslash (x\backslash )$ from the denominator instead. Keeping the numerator as $\backslash (-x(x\; -\; 4)\backslash )$, factor the denominator as $\backslash (x(x\; -\; 2)\backslash )$. Canceling the common factor $\backslash (x\backslash )$ results in $\backslash (\backslash frac\{-\; (x\; -\; 4)\}\{x\; -\; 2\}\backslash )$, which simplifies to $\backslash (\backslash frac\{4\; -\; x\}\{x\; -\; 2\}\backslash )$ by distributing the negative sign in the numerator. This form highlights how factoring choices affect the appearance of equivalent expressions.

A third method involves multiplying the entire rational expression by a form of one, such as $\backslash (\backslash frac\{2\}\{2\}\backslash )$. Multiplying numerator and denominator by 2 yields $\backslash (\backslash frac\{2(-x^2\; +\; 4x)\}\{2(x^2\; -\; 2x)\}\; =\; \backslash frac\{-2x^2\; +\; 8x\}\{2x^2\; -\; 4x\}\backslash )$. This expression remains equivalent because multiplying by one does not change the value, but it provides a different form that might be useful in certain contexts.

These techniques—factoring with different signs and multiplying by constants—demonstrate the versatility in expressing rational expressions equivalently. Recognizing that multiplying numerator and denominator by the same nonzero constant preserves equivalence allows for infinite variations. Mastery of these methods deepens understanding of algebraic manipulation and prepares students for more complex problem-solving scenarios.

Simplifying Rational Expressions Example 2

Understanding how to write equivalent forms of rational expressions is a fundamental skill in algebra that enhances flexibility in problem-solving. Consider the rational expression $$\backslash (\backslash frac\{-x^2\; +\; 4x\}\{x^2\; -\; 2x\}\backslash )$$. One effective method to find equivalent expressions is through factoring both the numerator and denominator.

Start by factoring a negative $\backslash (x\backslash )$ from the numerator. Since $\backslash (-x\; \backslash times\; x\; =\; -x^2\backslash )$ and $\backslash (-x\; \backslash times\; (-4)\; =\; 4x\backslash )$, the numerator factors to $\backslash (-x(x\; -\; 4)\backslash )$. Similarly, factor a negative $\backslash (x\backslash )$ from the denominator: $\backslash (-x\; \backslash times\; (-x)\; =\; x^2\backslash )$ and $\backslash (-x\; \backslash times\; 2\; =\; -2x\backslash )$, so the denominator becomes $\backslash (-x(-x\; +\; 2)\backslash )$. Canceling the common factor $\backslash (-x\backslash )$ leaves the equivalent expression $\backslash (\backslash frac\{x\; -\; 4\}\{2\; -\; x\}\backslash )$, noting that $\backslash (-x\; +\; 2\backslash )$ is rewritten as $\backslash \$2\; -\; x\backslash ($ for clarity.

Another approach involves factoring a positive $\backslash )x\backslash ($ from the denominator instead. Keeping the numerator as $\backslash )-x(x\; -\; 4)\backslash ($, factor $\backslash )x\backslash ($ from the denominator to get $\backslash )x(x\; -\; 2)\backslash ($. Canceling the common factor $\backslash )x\backslash ($ results in $\backslash )\backslash frac\{-\; (x\; -\; 4)\}\{x\; -\; 2\}\backslash ($, which simplifies to $\backslash )\backslash frac\{4\; -\; x\}\{x\; -\; 2\}\backslash ($ by distributing the negative sign in the numerator. This form highlights how factoring choices affect the appearance but not the value of the expression.

A third method to generate an equivalent expression is by multiplying both numerator and denominator by the same nonzero constant, such as 2. Multiplying yields $\backslash )\backslash frac\{2(-x^2\; +\; 4x)\}\{2(x^2\; -\; 2x)\}\; =\; \backslash frac\{-2x^2\; +\; 8x\}\{2x^2\; -\; 4x\}\backslash ($. This operation preserves equivalence because multiplying by $\backslash )\backslash frac\{2\}\{2\}\$$ is effectively multiplying by 1. This technique demonstrates that scaling both parts of a rational expression by the same factor produces infinitely many equivalent forms.

These strategies—factoring with different signs and multiplying by constants—illustrate the versatility in expressing rational expressions equivalently. Mastery of these methods deepens understanding of algebraic manipulation and prepares students for more complex problem-solving scenarios involving rational functions.

Simplify the rational expressions below:
$\frac{x^2+5x+6}{x^2+7x+10}$

• A. $\frac{x+3}{x+2}$
• B. $\frac{x+3}{x+5}$
• C. $\frac{x+2}{x+2}$
• D. $1$

Answer: (B)

Here’s what students ask on this topic:

What is a rational expression and how is it different from a rational number?

A rational expression is a quotient of two polynomials, similar to how a rational number is a quotient of two integers. For example, a rational expression can be written as $\frac{p}{q}$, where $p$ and $q$ are polynomials. In contrast, a rational number is a fraction where both numerator and denominator are integers. The key difference is that rational expressions involve variables and polynomials, while rational numbers involve only integers. Also, just like rational numbers, the denominator of a rational expression cannot be zero because division by zero is undefined. Understanding this helps in identifying where the expression is undefined by setting the denominator equal to zero and solving for the variable.

How do you find the values for which a rational expression is undefined?

To find where a rational expression is undefined, you need to identify the values of the variable that make the denominator zero. Since division by zero is undefined, these values are excluded from the domain. The process involves setting the denominator equal to zero and solving for the variable. For example, if the denominator is $\mathrm{2x\; -\; 6}$, set $\mathrm{2x\; -\; 6\; =\; 0}$. Solving this gives $\mathrm{x\; =\; 3}$. Therefore, the rational expression is undefined at $\mathrm{x\; =\; 3}$. This step is crucial before simplifying or evaluating the expression to avoid invalid operations.

What are the steps to simplify a rational expression?

Simplifying a rational expression involves three main steps. First, factor both the numerator and denominator completely into their prime factors or simpler polynomials. For example, factor numbers into primes and expressions like $x^3$ into $\mathrm{x\; imes\; x\; imes\; x}$. Second, identify and cancel out any common factors that appear in both numerator and denominator. For instance, if both have a factor of $7$ or $x$, cancel them. Third, rewrite the expression with the remaining factors. This process reduces the expression to its simplest form, making it easier to work with in further calculations or evaluations.

How do you simplify rational expressions with polynomials that are not initially factored?

When simplifying rational expressions with polynomials that are not factored, start by factoring each polynomial completely. This may involve factoring out the greatest common factor (GCF) or using special factoring techniques like difference of squares or factoring trinomials. For example, if the denominator is $\mathrm{3x^2\; -\; 15x}$, factor out the GCF $\mathrm{3x}$ to get $\mathrm{3x(x\; -\; 5)}$. After factoring, look for common factors in numerator and denominator to cancel. This step is essential because canceling common factors simplifies the expression. Always ensure the expression is fully factored before canceling to avoid mistakes.

Why can't the denominator of a rational expression be zero?

The denominator of a rational expression cannot be zero because division by zero is undefined in mathematics. If the denominator equals zero for some value of the variable, the expression does not have a valid value at that point. This is why we say the expression is undefined there. For example, in the rational expression $\frac{\mathrm{4x}}{\mathrm{x\; -\; 2}}$, if $\mathrm{x\; =\; 2}$, the denominator becomes zero, making the expression undefined. Identifying these values is important to understand the domain and behavior of the expression.

How do you evaluate a rational expression at a specific value of x?

To evaluate a rational expression at a specific value of $x$, substitute that value into both the numerator and denominator. Then, perform the arithmetic operations to simplify the expression. For example, given $\frac{\mathrm{x\; -\; 1}}{\mathrm{2x\; -\; 6}}$ and $\mathrm{x\; =\; 2}$, substitute to get $\frac{\mathrm{2\; -\; 1}}{\mathrm{2\; imes\; 2\; -\; 6}}=\frac{1}{\mathrm{4\; -\; 6}}=\frac{1}{-2}=\; -rac\{1\}\{2\}$. This method allows you to find the value of the expression at any allowed point in its domain.