# Simplifying Rational Expressions Examples

The rational expressions introduction page explained how a rational expression is a fraction or quotient that contains two polynomials.

For example a fraction such as     \frac{x^2 \space + \space 1}{2x \space {\text{--}} \space 5}     is a rational expression.

It was also touched on how like in the case of whole number fractions, rational expressions can often be simplified further from the original form they are given to you in.

The simplifying rational expressions examples on this page will show how such situations can be approached when you encounter them.
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But first we can have a brief recap on the introduction to simplifying rational expressions.

### Rational Expression Simplifying Recap

With   \frac{b^{\tt{3}}}{2b}.

The top and bottom can both be factored to give a common factor.

b^3 = \space b^2 \cdot b     ,     2b = \space 2 \cdot b

\frac{b^{\tt{3}}}{2b} \space {\scriptsize{=}} \space \frac{b^{\tt{2}} \cdot b}{2 \cdot b} \space {\scriptsize{=}} \space \frac{b^{\tt{2}} \cdot {\cancel{b}}}{2 \cdot {\cancel{b}}} \space {\scriptsize{=}} \space \frac{b^{\tt{2}}}{2}

The fraction  \frac{b^{\tt{2}}}{2}  is a simpler form of the original rational expression.

No more factoring can be done any further in search of common factors.

### Domain Restrictions:

It’s not always a requirement to state any values that cannot be in the domain if they were to result in a division by  0.

But when it is, it’s important to know what to write.

For a rational expression such as    \frac{(x \space + \space 2)(x \space + \space 1)}{(x \space + \space 2)(x \space {\text{–}} \space 4)}.

If  x  was either  -2  or  4,  the denominator below would give a result of  0.

What we can do is write our simplified expression as   \frac{(x \space + \space 1)}{(x \space {\text{–}} \space 4)}     x \space {\cancel{=}} \space {\text{-}}2 , 4.

Which takes into account the whole denominator before the simplified form.

## Simplifying Rational Expressions Examples

(1.1)

Simplify further the quotient     \frac{x^2 \space {\text{–}} \space 5x \space {\text{–}} \space 14}{x^3 + x^2 \space {\text{–}} \space 2x}.

Solution

\frac{x^2 \space {\text{–}} \space 5x \space {\text{–}} \space 14}{x^3 + x^2 \space {\text{–}} \space 2x} \space \space {\scriptsize{=}} \space \space \frac{(x \space + \space 2)(x \space {\text{–}} \space 7)}{x^3 + x^2 \space {\text{–}} \space 2x}

{\scriptsize{=}} \space \frac{(x \space + \space 2)(x \space {\text{–}} \space 7)}{x(x^{\tt{2}} + x \space {\text{–}} \space 2)}

Now we can factor further in the denominator, and have a common factor with the numerator above.

\frac{(x \space + \space 2)(x \space {\text{–}} \space 7)}{x(x^{\tt{2}} + x \space {\text{–}} \space 2)} \space \space {\scriptsize{=}} \space \space \frac{(x \space + \space 2)(x \space {\text{–}} \space 7)}{x(x + 2)(x \space {\text{–}} \space 1)}

=>    \frac{({\cancel{x \space + \space 2})}(x \space {\text{–}} \space 7)}{x({\cancel{x \space + \space 2}})(x \space {\text{–}} \space 1)} \space \space {\scriptsize{=}} \space \space \frac{(x \space {\text{–}} \space 7)}{x(x \space {\text{–}} \space 1)}

If we note the values that values that couldn’t be in the denominator,   x \space {\cancel{=}} \space {\text{-}}2 , 0 , 1.

(1.2)

Simplify further the rational expression     \frac{2x^2 \space + \space 5x \space {\text{–}} \space 3}{2x^2 \space + \space 9x \space {\text{–}} \space 5}.

Solution

\frac{2x^2 \space + \space 5x \space {\text{–}} \space 3}{2x^2 \space + \space 9x \space {\text{–}} \space 5} \space \space {\scriptsize{=}} \space \space \frac{(2x \space {\text{–}} \space 1)(x \space + \space 3)}{(2x \space {\text{–}} \space 1)(x \space + \space 5)}

Now:

\frac{({\cancel{2x \space {\text{–}} \space 1}})(x \space + \space 3)}{({\cancel{2x \space {\text{–}} \space 1}})(x \space + \space 5)} \space \space {\scriptsize{=}} \space \space \frac{x \space + \space 3}{x \space + \space 5}       ,     x {\cancel{=}} \space {\text{-}}5 \space , \space \frac{1}{2}

(1.3)

a)   Simplify further the rational expression     \frac{x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 12}{x \space {\text{–}} \space 6}.

Solution

\frac{x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 12}{x \space {\text{–}} \space 6} \space \space {\scriptsize{=}} \space \space \frac{(x \space + \space 2)(x \space {\text{–}} \space 6)}{(x \space {\text{–}} \space 6)}

Now:

\frac{(x \space + \space 2)({\cancel{x \space {\text{–}} \space 6}})}{({\cancel{x \space {\text{–}} \space 6}})} \space \space {\scriptsize{=}} \space \space \frac{x \space + \space 2}{1} \space \space \space {\scriptsize{=}} \space \space \space {\scriptsize{x + 2}}       ,     x {\cancel{=}} \space 6

b)   Simplify further the rational expression     \frac{x \space {\text{–}} \space 6}{x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 12}.

Solution

\frac{x \space {\text{–}} \space 6}{x^2 \space {\text{–}} \space 4x \space {\text{–}} \space 12} \space \space {\scriptsize{=}} \space \space \frac{(x \space {\text{–}} \space 6)}{(x \space + \space 2)(x \space {\text{–}} \space 6)}

Now:

\frac{({\cancel{x \space {\text{–}} \space 6}})}{(x \space + \space 2)({\cancel{x \space {\text{–}} \space 6}})} \space \space {\scriptsize{=}} \space \space \frac{1}{x \space + \space 2}       ,     x {\cancel{=}} \space {\text{-}}2 \space , \space 6

(1.4)

Simplify the rational expression     \frac{x^2 \space {\text{–}} \space 5x \space {\text{–}} \space 8}{4 \space {\text{–}} \space x}.

Solution

\frac{x^2 \space {\text{–}} \space 5x \space {\text{–}} \space 8}{4 \space {\text{–}} \space x} \space \space {\scriptsize{=}} \space \space \frac{(x \space {\text{–}} \space 4)(x \space + \space 2)}{4 \space {\text{–}} \space x}

From here we can rewrite the denominator to give us a common factor.

\frac{(x \space {\text{–}} \space 4)(x \space + \space 2)}{4 \space {\text{–}} \space x} \space \space {\scriptsize{=}} \space \space \frac{(x \space {\text{–}} \space 4)(x \space + \space 2)}{{\text{-}}1({\text{-}}4 \space + \space x)}

\space {\scriptsize{=}} \space \space \frac{(x \space {\text{–}} \space 4)(x \space + \space 2)}{{\text{-}}1(x \space {\text{–}} \space 4)}

Now:

\frac{({\cancel{x \space {\text{–}} \space 4}})(x \space + \space 2)}{{\text{-}}1({\cancel{x \space {\text{–}} \space 4}})} \space \space {\scriptsize{=}} \space \space \frac{x \space + \space 2}{{\text{-}}1} \space \space {\scriptsize{=}} \space \space {\scriptsize{{\text{-}}x \space {\text{–}} \space 2}}       ,     x {\cancel{=}} \space 4

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