Lesson Objectives

- Demonstrate an understanding of how to factor polynomials
- Demonstrate an understanding of how to simplify fractions
- Learn how to find the restricted values for a rational expression
- Learn how to simplify a rational expression

## Rational Expressions - Simplifying and Finding the Restricted Values

A rational number is any number that can be written as the quotient of two integers, with a non-zero denominator. Some examples:
$$\frac{2}{3}, \frac{-8}{7}, -7, 5$$
Since any integer can be written over 1, all integers are also rational numbers.

An algebraic fraction or rational expression is the quotient of two polynomials, with a non-zero denominator. In our textbook, we will see a rational expression defined as: $$\frac{P}{Q}, Q ≠ 0$$ Where P and Q are polynomials.

Since a rational expression is a fraction, we must make sure that we never divide by zero. One of the first tasks we face with rational expressions is finding where the rational expression is undefined. We can say that the domain (set of allowable x-values) of a rational expression/function includes all values that result in a non-zero denominator. In other words, to find the domain, find values for x that make the denominator equal to zero and exclude those values. Let's look at a few examples.

Example 1: Find the excluded values for each, state the domain $$f(x) = \frac{x^2 + 5x - 6}{x^2 + 16x + 60}$$ To find the restricted values, we want to set the denominator equal to zero and solve for x. The solutions will give us the values for x that result in a denominator of zero. These values will represent the restricted values for the rational expression/function. $$x^2 + 16x + 60 = 0$$ Factor the left side: $$(x + 6)(x + 10) = 0$$ Solve using the zero-product property: $$x + 6 = 0$$ $$x = -6$$ $$x + 10 = 0$$ $$x = -10$$ This tells us the rational function is not defined for an x-value of -6 or an x-value of -10.

domain: {x| x ≠ -6,-10}

Example 2: Find the excluded values for each, state the domain $$f(x) = \frac{7x^2 - 8x + 1}{14x^3 - 20x^2 + 6x}$$ To find the restricted values, we want to set the denominator equal to zero and solve for x. The solutions will give us the values for x that result in a denominator of zero. These values will represent the restricted values for the rational expression/function. $$14x^3 - 20x^2 + 6x = 0$$ Factor the left side: $$2x(7x^2 - 10x + 3) = 0$$ $$2x(7x - 3)(x - 1) = 0$$ Solve using the zero-product property: $$2x = 0$$ $$x = 0$$ $$7x - 3 = 0$$ $$x = \frac{3}{7}$$ $$x - 1 = 0$$ $$x = 1$$ This tells us the rational function is not defined for an x-value of 0, 1, or 3/7.

domain: {x | x ≠ 0, 1, 3/7}

Example 3: Simplify each and state the domain $$f(x) = \frac{7x^2 - 32x - 15}{10x^3 - 60x^2 + 50x}$$ First, we want to identify the restricted values. Let's set the denominator equal to zero and solve for x. $$10x^3 - 60x^2 + 50x = 0$$ Factor: $$10x(x^2 - 6x + 5) = 0$$ $$10x(x - 5)(x - 1) = 0$$ $$10x = 0$$ $$x = 0$$ $$x - 5 = 0$$ $$x = 5$$ $$x - 1 = 0$$ $$x = 1$$ domain: {x |x ≠ 0, 1, 5}

Now that we know the domain, we can simplify the rational function. Let's factor the numerator and denominator. $$f(x) = \frac{7x^2 - 32x - 15}{10x^3 - 60x^2 + 50x}$$ $$f(x) = \frac{(7x + 3)(x - 5)}{10x(x - 5)(x - 1)}$$ We can cancel a common factor of (x - 5) between the numerator and the denominator. $$\require{cancel}f(x) = \frac{(7x + 3)\cancel{(x - 5)}}{10x\cancel{(x - 5)}(x - 1)}$$ $$f(x) = \frac{7x + 3}{10x(x - 1)}$$ It is normal to leave a rational expression in factored form. It's worth noting that the simplified form does not have a restriction of x ≠ 5 in its domain. Since we started with a restriction of x ≠ 5, this carries over to the simplified form. This is why it is important to note any restrictions before any simplifying is done.

Example 4: Simplify each and state the domain $$f(x) = \frac{3x^2 + 2x - 8}{7x^2 - 28}$$ First, we want to identify the restricted values. Let's set the denominator equal to zero and solve for x. $$7x^2 - 28 = 0$$ Factor: $$7(x^2 - 4) = 0$$ $$7(x - 2)(x + 2) = 0$$ $$x - 2 = 0$$ $$x = 2$$ $$x + 2 = 0$$ $$x = -2$$ domain: {x |x ≠ -2, 2}

Now that we know the domain, we can simplify the rational function. Let's factor the numerator and denominator. $$f(x) = \frac{3x^2 + 2x - 8}{7x^2 - 28}$$ $$f(x) = \frac{(3x - 4)(x + 2)}{7(x - 2)(x + 2)}$$ We can cancel a common factor of (x + 2) between the numerator and the denominator. $$f(x) = \frac{(3x - 4)\cancel{(x + 2)}}{7(x - 2)\cancel{(x + 2)}}$$ $$f(x) = \frac{3x - 4}{7(x - 2)}$$ Again, we can see that the simplified form does not have the restriction of x ≠ -2 in its domain. This is why we take the domain from the original form.

An algebraic fraction or rational expression is the quotient of two polynomials, with a non-zero denominator. In our textbook, we will see a rational expression defined as: $$\frac{P}{Q}, Q ≠ 0$$ Where P and Q are polynomials.

Since a rational expression is a fraction, we must make sure that we never divide by zero. One of the first tasks we face with rational expressions is finding where the rational expression is undefined. We can say that the domain (set of allowable x-values) of a rational expression/function includes all values that result in a non-zero denominator. In other words, to find the domain, find values for x that make the denominator equal to zero and exclude those values. Let's look at a few examples.

Example 1: Find the excluded values for each, state the domain $$f(x) = \frac{x^2 + 5x - 6}{x^2 + 16x + 60}$$ To find the restricted values, we want to set the denominator equal to zero and solve for x. The solutions will give us the values for x that result in a denominator of zero. These values will represent the restricted values for the rational expression/function. $$x^2 + 16x + 60 = 0$$ Factor the left side: $$(x + 6)(x + 10) = 0$$ Solve using the zero-product property: $$x + 6 = 0$$ $$x = -6$$ $$x + 10 = 0$$ $$x = -10$$ This tells us the rational function is not defined for an x-value of -6 or an x-value of -10.

domain: {x| x ≠ -6,-10}

Example 2: Find the excluded values for each, state the domain $$f(x) = \frac{7x^2 - 8x + 1}{14x^3 - 20x^2 + 6x}$$ To find the restricted values, we want to set the denominator equal to zero and solve for x. The solutions will give us the values for x that result in a denominator of zero. These values will represent the restricted values for the rational expression/function. $$14x^3 - 20x^2 + 6x = 0$$ Factor the left side: $$2x(7x^2 - 10x + 3) = 0$$ $$2x(7x - 3)(x - 1) = 0$$ Solve using the zero-product property: $$2x = 0$$ $$x = 0$$ $$7x - 3 = 0$$ $$x = \frac{3}{7}$$ $$x - 1 = 0$$ $$x = 1$$ This tells us the rational function is not defined for an x-value of 0, 1, or 3/7.

domain: {x | x ≠ 0, 1, 3/7}

### Simplifying Rational Expressions/Functions

Since a rational expression/function is nothing more than a fraction, we use the same techniques we used when simplifying fractions. We will factor the numerator and denominator completely and then cancel any common factors other than 1. When we simplify a rational expression/function, we use the restricted values from the non-simplified form. In some cases, simplifying a rational expression will create a new rational expression with a broader domain. We must remember to carry over the old restrictions to the simplified form. Let's look at a few examples.Example 3: Simplify each and state the domain $$f(x) = \frac{7x^2 - 32x - 15}{10x^3 - 60x^2 + 50x}$$ First, we want to identify the restricted values. Let's set the denominator equal to zero and solve for x. $$10x^3 - 60x^2 + 50x = 0$$ Factor: $$10x(x^2 - 6x + 5) = 0$$ $$10x(x - 5)(x - 1) = 0$$ $$10x = 0$$ $$x = 0$$ $$x - 5 = 0$$ $$x = 5$$ $$x - 1 = 0$$ $$x = 1$$ domain: {x |x ≠ 0, 1, 5}

Now that we know the domain, we can simplify the rational function. Let's factor the numerator and denominator. $$f(x) = \frac{7x^2 - 32x - 15}{10x^3 - 60x^2 + 50x}$$ $$f(x) = \frac{(7x + 3)(x - 5)}{10x(x - 5)(x - 1)}$$ We can cancel a common factor of (x - 5) between the numerator and the denominator. $$\require{cancel}f(x) = \frac{(7x + 3)\cancel{(x - 5)}}{10x\cancel{(x - 5)}(x - 1)}$$ $$f(x) = \frac{7x + 3}{10x(x - 1)}$$ It is normal to leave a rational expression in factored form. It's worth noting that the simplified form does not have a restriction of x ≠ 5 in its domain. Since we started with a restriction of x ≠ 5, this carries over to the simplified form. This is why it is important to note any restrictions before any simplifying is done.

Example 4: Simplify each and state the domain $$f(x) = \frac{3x^2 + 2x - 8}{7x^2 - 28}$$ First, we want to identify the restricted values. Let's set the denominator equal to zero and solve for x. $$7x^2 - 28 = 0$$ Factor: $$7(x^2 - 4) = 0$$ $$7(x - 2)(x + 2) = 0$$ $$x - 2 = 0$$ $$x = 2$$ $$x + 2 = 0$$ $$x = -2$$ domain: {x |x ≠ -2, 2}

Now that we know the domain, we can simplify the rational function. Let's factor the numerator and denominator. $$f(x) = \frac{3x^2 + 2x - 8}{7x^2 - 28}$$ $$f(x) = \frac{(3x - 4)(x + 2)}{7(x - 2)(x + 2)}$$ We can cancel a common factor of (x + 2) between the numerator and the denominator. $$f(x) = \frac{(3x - 4)\cancel{(x + 2)}}{7(x - 2)\cancel{(x + 2)}}$$ $$f(x) = \frac{3x - 4}{7(x - 2)}$$ Again, we can see that the simplified form does not have the restriction of x ≠ -2 in its domain. This is why we take the domain from the original form.

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