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6. Rational Functions
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Algebra II

6.03 Rational expressions

Lesson
Practice

Introduction

We will use our understanding of rational numbers to extend to rational expressions in this lesson. We will combine our knowledge of polynomial division, the remainder theorem and factoring polynomials from lessons  3.04 Polynomial division  and  3.05 Factoring polynomials  to rewrite rational expressions. Then we will rewrite rational functions in order to identify features of the graph, using the information we learned in  5.01 Rational functions. 

Rational expressions

Exploration

  1. Rewrite the following improper fractions as mixed numbers: \dfrac{7}{3}, \dfrac{19}{8}, \dfrac{11}{9}
  2. How do you think simplifying improper fractions is related to simplifying an expression such as \dfrac{2\left(x+2\right)+5}{\left(x+2\right)}?

Recall that improper fractions can be written as mixed numbers. Given integers a and b, where both a and b are nonzero integers and a>b, the fraction \dfrac{a}{b} can be rewritten as q + \dfrac{r}{b}, where q represents a quotient and r represents a remainder, for instance:\dfrac{7}{3} = 2 + \dfrac{1}{3}

A rational expression is a quotient of two polynomials with a non-zero denominator. Examples of rational expressions include \dfrac{x + 1}{x - 1}, \dfrac{2}{3a + 2b}, and \dfrac{12y^2 + 4y - 1}{y^3 + 8y}, but also expressions like \dfrac{\sqrt{3}x - 2}{x} and \dfrac{5x}{1} = 5x.

Rational expressions can sometimes be simplified by factoring the numerator and denominator and then simplifying any common factors between them. Similar to improper fractions, rational expressions can be written as polynomial division:{\dfrac{p\left(x\right)}{b\left(x\right)}=q\left(x\right)+\dfrac{r\left(x\right)}{b\left(x\right)}}Where p\left(x\right) and b\left(x\right) are polynomials, q\left(x\right) is the quotient and r\left(x\right) is the remainder.

Recall from the previous lesson that rational functions will have domain and range restrictions due to asymptotes and points of discontinuity (holes). When rewriting and simplifying rational expressions, we need to keep in mind the restrictions on the variables to avoid undefined values in the original expressions.

Examples

Example 1

Fully simplify each expression, so that the degrees of the polynomials in the numerator and denominator are as low as possible. Justify each step. State any restrictions on the variables.

a

\dfrac{2x^2 + 7x}{6x + 21}

Worked Solution
Create a strategy

Any variables that lead the denominators of the rational expressions to equal zero should be excluded.

Since there is more than one term in the denominator, it is not straightforward to see what can be simplified. So, first we want to factor the numerator and denominator.

Apply the idea

Before simplifying, determine the value(s) of x that will make the denominator equivalent to zero.

\displaystyle 6x+21\displaystyle =\displaystyle 0When the denominator equals 0, the expression is undefined
\displaystyle 6x\displaystyle =\displaystyle -21Subtract 21 from both sides of the equation
\displaystyle x\displaystyle =\displaystyle -\dfrac{7}{2}Divide both sides by 6 and simplify

When x= -\dfrac{7}{2}, the expression is undefined. We will exclude this value of x.

We can begin factoring a GCF. In this case, the numerator has a common factor of x and the denominator has a common factor of 3. Factoring these out gives:\dfrac{2x^2 + 7x}{6x + 21} = \dfrac{x\left(2x + 7\right)}{3\left(2x + 7\right)}We can now see the common factor of 2x + 7 which can be simplified from the numerator and denominator by the multiplicative inverse property of equality, since \dfrac{\left (2x + 7 \right)}{\left (2x+7 \right)} = 1 when x\neq -\dfrac{7}{2}:\dfrac{x\left(2x + 7\right)}{3\left(2x + 7\right)} = \dfrac{x}{3}

So the simplified form is \dfrac{x}{3}, x \neq -\dfrac{7}{2}.

Reflect and check

Note that if we were to graph the corresponding rational function y = \dfrac{2x^2 + 7x}{6x + 21} it would look like a linear graph of the form y = \dfrac{1}{3}x with a hole at x = -\dfrac{7}{2}.

b

\dfrac{y^2 + 5y - 24}{y^3 - 27}

Worked Solution
Create a strategy

Since there is more than one term in the denominator, it is not straightforward to see what can be simplified. We first want to factor the numerator and denominator, determine the values of y to exclude, then simplify.

Apply the idea

First, we will factor the denominator to determine restrictions on the variable. The denominator is a difference of two cubes, since 27 = 3^3, so we can factor it using that identity.\dfrac{y^2+5y-24}{y^3-27}=\dfrac{y^2+5y-24}{\left(y-3 \right) \left( y^2 + 3y + 9 \right)}

When the denominator is equal to zero, the rational expression is undefined. So, we need to determine when \left(y-3 \right) \left( y^2 + 3y + 9 \right)=0 by the zero product property.

\displaystyle y-3\displaystyle =\displaystyle 0
\displaystyle y\displaystyle =\displaystyle 3Add 3 to both sides

We can use the quadratic formula to determine when y^2+3y+9=0.

\displaystyle y\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle y\displaystyle =\displaystyle \dfrac{-3 \pm \sqrt{3^2-4\left(1 \right) \left(9 \right)}}{2 \left(1 \right)}Substitute a=1, b=3, and c=9
\displaystyle y\displaystyle =\displaystyle \dfrac{-3 \pm \sqrt{-27}}{2}Evaluate the multiplication and subtraction

Since the discriminant is negative, we can stop evaluating because the solution will be complex, and there will be no real values of y for which the rational expression is undefined in the simplified form other than y=3.

Second, we will factor the numerator.

In this case, the numerator is a quadratic. To factor it by grouping, we need to find two numbers that have a sum of b=5 and a product of ac=-24. The two numbers are 8 and -3, so

\displaystyle \dfrac{y^2+5y-24}{\left(y-3 \right) \left( y^2 + 3y + 9 \right)}\displaystyle =\displaystyle \dfrac{y^2+8y-3y-24}{\left(y-3 \right) \left( y^2 + 3y + 9 \right)}Rewrite the numerator
\displaystyle =\displaystyle \dfrac{y \left(y + 8 \right) - 3 \left(y + 8 \right)}{\left(y-3 \right) \left( y^2 + 3y + 9 \right)}Factor out each GCF
\displaystyle =\displaystyle \dfrac{\left(y + 8 \right) \left(y - 3 \right)}{\left(y-3 \right) \left( y^2 + 3y + 9 \right)}Factor out the common factor

Finally, we can now see the common factor of y - 3 which can be simplified from the numerator and denominator since \dfrac{\left(y - 3 \right)}{\left( y -3 \right)}=1 when y\neq 3:\dfrac{\left(y + 8\right)\left(y - 3\right)}{\left(y - 3\right)\left(y^2 + 3y + 9\right)} = \dfrac{y + 8}{y^2 + 3y + 9}

So the simplified form is \dfrac{y + 8}{y^2 + 3y + 9}, y \neq 3.

Reflect and check

In this case, the simplified form appears to have about the same "complexity" as the original expression, since it has the same number of terms between the numerator and denominator as the original expression did.

The reason that this form is simpler is that the exponents involved are smaller. The numerator and denominator of the original expression had degrees of 2 and 3 respectively, while the simplified expression has degrees of 1 and 2 respectively.

Note that we have to consider the full denominator when determining restricted values. If we had only examined the fully simplified fraction, we would have missed the domain restriction of y \neq 3.

Example 2

Consider the expression \dfrac{3x-4}{x-2}.

a

Rewrite the rational expression in the form q\left(x\right) + \dfrac{r\left(x\right)}{b\left(x\right)} where r\left(x\right) is the remainder. State any restrictions on the variables.

Worked Solution
Create a strategy

To rewrite the expression, we can use algebraic manipulation, long division, or synthetic division. To use algebraic manipulation, we need to rewrite the numerator so one of the terms contains a factor of the denominator.

Although the current numerator cannot be factored, it could have been factored if the constant term was -6. Then, one of the factors would be the same as the expression in the denominator. We can rewrite the numerator as 3x-6+2 since this is still equal to the original, 3x-4, then factor.

Apply the idea
\displaystyle \dfrac{3x-4}{x-2}\displaystyle =\displaystyle \dfrac{3x -6 + 2}{x-2}Rewrite the numerator
\displaystyle =\displaystyle \dfrac{3x-6}{x-2}+\dfrac{2}{x-2}Rewrite using \dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}
\displaystyle =\displaystyle \dfrac{3\left(x-2\right)}{x-2}+\dfrac{2}{x-2}Factor out 3
\displaystyle =\displaystyle 3+\dfrac{2}{x-2}Simplify since \dfrac{x-2}{x-2}=1, x\neq 2

If x=2, the expression \dfrac{x-2}{x-2} would be undefined. Therefore, we restricted the domain in the last step such that x\neq 2.

Reflect and check

To rewrite this in desired the form, \dfrac{p\left(x\right)}{b\left(x\right)}=q\left(x\right)+\dfrac{r\left(x\right)}{b\left(x\right)}, we divided 3x-4 by x-2. This means:

p\left(x\right)=3x-4 (the dividend)

b\left(x\right)=x-2 (the divisor)

q\left(x\right)=3 (the quotient)

r\left(x\right)=2 (the remainder)

b

Graph the corresponding function of the expression you found in part (a).

Worked Solution
Create a strategy

The corresponding function of the expression in part (b) is y=3+\dfrac{2}{x-2}. This is in the form of a transformed rational function. This form can help us identify key characteristics of its corresponding graph.

Apply the idea

When looking at the corresponding function, y=3+\dfrac{2}{x-2}, we can see that the parent function y = \dfrac{1}{x}, was stretched vertically by a factor of 2 and translated 2 units right and 3 units up.

This tells us that the restricted value of x = 2 is a vertical asymptote and that the horizontal asymptote will be y = 3.

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Idea summary

Rational expressions can sometimes be simplified by factoring the numerator and denominator and then simplifying any common factors between them.

Rational expressions can be written as polynomial division:{\dfrac{p\left(x\right)}{b\left(x\right)}=q\left(x\right)+\dfrac{r\left(x\right)}{b\left(x\right)}}Where p\left(x\right) and b\left(x\right) are polynomials, q\left(x\right) is the quotient and r\left(x\right) is the remainder.

Rewriting rational functions in different forms may help us identify key characteristics of their graphs.

Outcomes

AII.1a

Add, subtract, multiply, divide, and simplify rational algebraic expressions

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