## Section14.4Simplifying Rational Expressions

Recall that multiplying fractions entails multiplying the numerators with one another and multiplying the denominators with one another. For example:

\begin{align*} \frac{2}{7} \cdot \frac{10}{3}\amp=\frac{2 \cdot 10}{7 \cdot 3}\\ \amp=\frac{20}{21} \end{align*}

The process of simplifying a rational expression involves employing the multiplication process in reverse, but in a strategic way. What we want to do is completely factor both the numerator and the denominator of the expression and then look for factors that occur in both the numerator and the denominator. We can then factor the common factors from the larger expression as fractions that reduce to one, thereby simplifying the original expression. This will make more sense when seen it in action.

Let's simplify the expression $\frac{x^2+6x-55}{x^2+x-30}\text{.}$ We begin by factoring both the numerator and the denominator.

\begin{equation*} \frac{x^2+6x-55}{x^2+x-30}=\frac{(x+11)(x-5)}{(x+6)(x-5)} \end{equation*}

We now see that there is indeed a factor that is common to both the numerator and the denominator, specifically $(x-5)\text{.}$ We can isolate these factors from the other factors in the form of a fraction that reduces to one. Picking up where we left off:

\begin{align*} \frac{x^2+6x-55}{x^2+x-30}\amp=\frac{(x+11)(x-5)}{(x+6)(x-5)}\\ \amp=\frac{x-5}{x-5} \cdot \frac{x+11}{x+6} \end{align*}

So long as the value of $x$ is not $5\text{,}$ the expression $\frac{x-5}{x-5}$ simplifies to $1\text{.}$ Let's go ahead and replace $\frac{x-5}{x-5}$ with $1\text{,}$ but when we do so we have to state a domain restriction, that the value of $x$ is not allowed to be $5\text{.}$

\begin{align*} \frac{x^2+6x-55}{x^2+x-30}\amp=\frac{(x+11)(x-5)}{(x+6)(x-5)}\\ \amp=\frac{x-5}{x-5} \cdot \frac{x+11}{x+6}\\ \amp=1 \cdot \frac{x+11}{x+6}, x \neq 5\\ \amp=\frac{x+11}{x+6}, x \neq 5 \end{align*}

As a rule we leave out the line with the explicit factor of $1\text{.}$ So a complete simplification would be as follows.

\begin{align*} \frac{x^2+6x-55}{x^2+x-30}\amp=\frac{(x+11)(x-5)}{(x+6)(x-5)}\\ \amp=\frac{x-5}{x-5} \cdot \frac{x+11}{x+6}\\ \amp=\frac{x+11}{x+6}, x \neq 5 \end{align*}

Several examples follow.

###### Example14.4.1.

Simplify the given expression and make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{t^2+16t+64}{t^2+14t+48} \end{equation*}
Solution
\begin{align*} \frac{t^2+16t+64}{t^2+14t+48}\amp=\frac{(t+8)(t+8)}{(t+8)(t+6)}\\ \amp=\frac{t+8}{t+8} \cdot \frac{t+8}{t+6}\\ \amp=\frac{t+8}{t+6}, t \neq -8 \end{align*}
###### Example14.4.2.

Simplify the given expression and make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{x^2-7x}{x^2-15x+56} \end{equation*}
Solution
\begin{align*} \frac{x^2-7x}{x^2-15x+56}\amp=\frac{x(x-7)}{(x-8)(x-7)}\\ \amp=\frac{x-7}{x-7} \cdot \frac{x}{x-8}\\ \amp=\frac{x}{x-8}, x \neq 7 \end{align*}
###### Example14.4.3.

Simplify the given expression and make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{y^3-36y}{y^3+36y} \end{equation*}
Solution
\begin{align*} \frac{y^3-36y}{y^3+36y}\amp=\frac{y(y^2-36)}{y(y^2+36)}\\ \amp=\frac{y(y+6)(y-6)}{y(y^2+36)}\\ \amp=\frac{y}{y} \cdot \frac{(y+6)(y-6)}{y^2+36}\\ \amp=\frac{(y+6)(y-6)}{y^2+36}, y \neq 0 \end{align*}
###### Example14.4.4.

Simplify the given expression and make sure that you state any necessary domain restrictions.

\begin{equation*} \frac{w^3-1}{w^2-1} \end{equation*}
Solution
\begin{align*} \frac{w^3-1}{w^2-1}\amp=\frac{(w-1)(w^2+w+1)}{(w-1)(w+1)}\\ \amp=\frac{w-1}{w-1} \cdot \frac{w^2+w+1}{w+1}\\ \amp=\frac{w^2+w+1}{w+1},w \neq 1 \end{align*}

### ExercisesExercises

Simplify each expression. Make sure that you state any necessary domain restrictions.

###### 1.

$\frac{x-3}{x^2+x-12}$

Solution

\begin{aligned}[t] \frac{x-3}{x^2+x-12}\amp=\frac{x-3}{(x-3)(x+4)}\\ \amp=\frac{x-3}{x-3} \cdot \frac{1}{x+4}\\ \amp=\frac{1}{x+4}, x \neq 3 \end{aligned}

###### 2.

$\frac{x^2-5x+4}{x^2+6x-40}$

Solution

\begin{aligned}[t] \frac{x^2-5x+4}{x^2+6x-40}\amp=\frac{(x-4)(x-1)}{(x-4)(x+10)}\\ \amp=\frac{x-4}{x-4} \cdot \frac{x-1}{x+10}\\ \amp=\frac{x-1}{x+10}, x \neq 4 \end{aligned}

###### 3.

$\frac{x^2+7x}{x^2-7x}$

Solution

\begin{aligned}[t] \frac{x^2+7x}{x^2-7x}\amp=\frac{x(x+7)}{x(x-7)}\\ \amp=\frac{x}{x} \cdot \frac{x+7}{x-7}\\ \amp=\frac{x+7}{x-7}, x \neq 0 \end{aligned}

###### 4.

$\frac{x^4+21x^2}{5x^3+x^7}$

Solution

\begin{aligned}[t] \frac{x^4+21x^2}{5x^3+x^7}\amp=\frac{x^2(x^2+21)}{x^3(5+x^4)}\\ \amp=\frac{x^2}{x^2} \cdot \frac{x^2+21}{x(5+x^4)}\\ \amp=\frac{x^2+21}{x(5+x^4)} \end{aligned}

###### 5.

$\frac{4t^2-9}{2t-3}$

Solution

\begin{aligned}[t] \frac{4t^2-9}{2t-3}\amp=\frac{(2t-3)(2t+3)}{2t-3}\\ \amp=\frac{2t-3}{2t-3} \cdot \frac{2t+3}{1}\\ \amp=2t+3, t \neq \frac{3}{2} \end{aligned}

###### 6.

$\frac{t^2-16}{t^2+16}$

Solution

\begin{aligned}[t] \frac{t^2-16}{t^2+16}=\frac{(t-4)(t+4)}{t^2+16} \end{aligned}