Lesson Objectives

- Demonstrate an understanding of how to factor polynomials
- Demonstrate an understanding of how to simplify fractions
- Learn how to simplify a rational expression

## How to Simplify a Rational Expression

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 1: 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 2: 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.

Example 1: 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 2: 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.

#### Skills Check:

Example #1

Simplify each. $$\frac{7x^4 - 49x^3 - 56x^2}{7x - 56}$$

Please choose the best answer.

A

$$x^2$$

B

$$x^2(x + 1)$$

C

$$\frac{x(3x - 10)}{7}$$

D

$$\frac{3x^2(x + 2)}{7x - 2}$$

E

$$\frac{7x - 2}{3x^2(x + 2)}$$

Example #2

Simplify each. $$\frac{6x^2 - 24}{6x^2 - 6x - 36}$$

Please choose the best answer.

A

$$\frac{3(7x + 5)}{2x(x + 3)}$$

B

$$\frac{5x - 3}{6x(x - 2)}$$

C

$$\frac{x - 2}{x - 3}$$

D

$$\frac{3x - 8}{7x + 9}$$

E

$$2x^2 - 1$$

Example #3

Simplify each. $$\frac{-10x^3 + 70x^2 + 80x}{5x^2 - 30x - 80}$$

Please choose the best answer.

A

$$-\frac{2x(x + 1)}{x + 2}$$

B

$$\frac{5(x - 1)}{2(7x + 3)}$$

C

$$\frac{7x - 2}{(x - 2)(5x + 7)}$$

D

$$\frac{4}{5(x - 2)}$$

E

$$3x - 1$$

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