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A rational number is a number that can be written as a fraction of two integers. Similarly, a rational expression is an expression written as a fraction of two **polynomials**. Because of the similarity with rational numbers, many of the same rules apply.

A rational expression is a fraction where both the numerator and the denominator are polynomials. An example is $\dfrac{x^2-7}{x^3+5x}.$

Sometimes the numerator and denominator can be named using function notation. For example, suppose $p(x) = x^2 -7$ and $q(x) = x^3 + 5x.$ Rational expressions are undefined for values of $x$ that make the denominator equal $0$. Therefore, $q(x)$ cannot equal $0.$
To simplify a rational expression, there must be a common factor in the numerator and the denominator. If there is, then the expression can be simplified by canceling out that factor. Common factors can be found through various ways of factoring polynomials. For example, the rational expression
$\dfrac{9x - x^3}{x^2 - 6x + 9}$
can be simplified using the following checklist.
### 1

Check first if either the numerator or denominator (or both) can be factored by GCF. In this example, the factor $x$ is in both terms in the numerator.
### 2

See if it's possible to factor either the numerator or the denominator using the difference of squares. Look for an expression in the form $a^2 - b^2,$ which can be factored as $(a + b)(a - b).$ In the example, $(9 - x)^2$ can be factored using this rule.
### 3

See if either the numerator or denominator is a perfect square trinomial. Look for an expression in the form $a^2 + 2ab + b^2$ or $a^2 - 2ab + b^2,$ which can be factored as $(a + b)^2$ and $(a - b)^2,$ respectively. Here, the denominator is a perfect square trinomial.
### 4

In some cases, a negative sign can be factored out for one expression to have the same form as another factor. Here, the factor $3 - x$ in the numerator is almost identical with the factor $x - 3$ in the denominator.
### 5

When the numerator and the denominator are completely factored, it's possible to identify common factors and cancel them. Here, the factor $(x - 3)$ is in both the numerator and the denominator.
The rational expression is now written in its simplest form. In this case, it's possible to distribute the negative sign in denominator:
$\dfrac{x (3 + x)}{\text{-}(x - 3)}=\dfrac{x(3 + x)}{\text{-} x+3}=\dfrac{x(3 + x)}{3-x}.$

Are there any common single-term factors?

Is it possible to factor the expression using the difference of squares?

Is it possible to factor using perfect square trinomials?

Is it helpful to factor out a negative sign?

Cancel out factors

$\dfrac{\text{-} (x - 3)\cdot x \cdot(3 + x)}{(x - 3)^2}$

$\dfrac{\text{-} x (3 + x)}{x - 3}$

Write the rational expression in its simplest form. $\dfrac{3x^2 - 75}{3x (x + 5)}$

To begin, we'll factor the numerator completely. Notice that the denominator is already written in factored form. The numerator contains the expression $3x^2-75.$ Since $3$ is a factor of $75,$ we can factor $3$ out of the entire expression.
Notice that $x^2-25$ is a difference of squares, which means it can be factored to be $(x+5)(x-5).$ Thus, the given expression can be written as $\dfrac{3(x+5)(x-5)}{3x(x+5)}$
We can see that the common factors between the numerator and denominator are $3$ and $(x+5).$ Thus, we can cancel them out. Writing the expression in its simplest form gives $\dfrac{x-5}{x}.$

Since rational expressions are essentially fractions, it's possible to add, subtract, multiply and divide them. When creating a multiple of a rational expression, $\frac{p(x)}{q(x)},$ by multiplying by the factor $k$ in both the denominator and the numerator, the equality still holds true.

$\dfrac{p(x)}{q(x)}=\dfrac{p(x)\cdot k}{q(x)\cdot k}$

Canceling out a factor $k,$ yields the same equality. In both cases, the factor $k$ can take all values except $0.$

$\dfrac{p(x)}{q(x)}=\dfrac{p(x)/ k}{q(x)/ k}$

It is also possible to create a multiple or cancel out a factor by using a more complex polynomial: $\dfrac{(x - 1)}{x (x - 1)} = \dfrac{1}{x}.$ Consider the domain. The first expression is undefined for $x=1,$ but the second expression is not. It looks like the domain has been expanded when $(x-1)$ was canceled out, but this is not the case. For the equality to hold true, all $x$-values must give the same value on both sides. Taking this into account gives

$\dfrac{x - 1}{x (x - 1)} = \dfrac{1}{x}, \quad x\neq 1,0.$Multiplying rational expressions works the same as multiplying fractions. The numerators and denominators are multiplied separately.

$\dfrac{p(x)}{q(x)} \cdot \dfrac{h(x)}{g(x)}=\dfrac{p(x)\cdot h(x)}{q(x)\cdot g(x)}$

To divide two rational expressions, the first step is to invert the expression in the denominator, and then multiply, similar to dividing fractions.

$\left.{\dfrac{p(x)}{q(x)}}\middle/{\dfrac{h(x)}{g(x)}}\right. = \dfrac{p(x)}{q(x)} \cdot \dfrac{g(x)}{h(x)}$

When rewriting a division of a rational expression as multiplication, it might appear to change the domain. For example, $r(x)=\left.\dfrac{x+1}{x-3}\middle/\dfrac{x-9}{x+7}\right.$ is undefined for the $x$-values $3,$ $\text{-}7$ and $9.$ Each of these $x$-values result in the expression in any of the denominators being equal to $0$. The rewritten expression $q(x)=\dfrac{(x+1)(x+7)}{(x-3)(x-9)}$

is, however, undefined only for the $x$-values $3$ and $9.$ In order to be able to have an equality between the expressions theyDetermine the product and quotient of the rational expressions. Simplify completely. $\dfrac{x^2-1}{x+2} \quad \text{and} \quad \dfrac{2x}{x+1}$

Let's start with multiplying the expressions. The numerators and denominators are multiplied separately.
Since we want to simplify the expression, let's factor the numerator. $(x^2-1)$ is a difference of squares. We can factor it, and then cancel out any common factors between the numerator and the denominator.
The product is $\frac{2x(x-1)}{x+2}.$ However, since we divided out the common factor $x+1$ we have to take that into account:
$\dfrac{\left(x^2-1\right)\cdot 2x}{(x+2)\cdot (x+1)}=\dfrac{2x(x-1)}{x+2}, \quad x\neq \text{-} 2, \text{-} 1.$
Now, we divide the expressions.
From the multiplication, we know that $x^2-1$ can be written as the factors $x+1$ and $x-1.$ Therefore, the numerator and denominator do not have any common factors. The simplest form of the quotient is $\dfrac{(x^2-1)(x+1)}{2x(x+2)}.$

$\dfrac{x^2-1}{x+2} \cdot \dfrac{2x}{x+1}$

$\dfrac{\left(x^2-1\right)\cdot 2x}{(x+2)\cdot (x+1)}$

$\dfrac{\left(x^2-1\right)\cdot 2x}{(x+2)\cdot (x+1)}$

$\dfrac{\left(x^2-1^2\right)\cdot 2x}{(x+2)\cdot (x+1)}$

$\dfrac{(x+1)(x-1)\cdot 2x}{(x+2)\cdot (x+1)}$

$\dfrac{(x-1)\cdot 2x}{x+2}$

$\left.\dfrac{x^2-1}{x+2}\middle/\dfrac{2x}{x+1}\right.$

$\dfrac{x^2-1}{x+2}\cdot \dfrac{x+1}{2x}$

$\dfrac{(x^2-1)(x+1)}{(x+2)\cdot 2x}$

When adding and subtracting rational expressions, the same rules apply as when adding and subtracting fractions. If they share a denominator, the numerators can be added or subtracted directly.

$\dfrac{p(x)}{q(x)} + \dfrac{h(x)}{q(x)}=\dfrac{p(x)+h(x)}{q(x)}$

$\dfrac{p(x)}{q(x)} - \dfrac{h(x)}{q(x)}=\dfrac{p(x)-h(x)}{q(x)}$

Determine the sum of the rational expressions. $\dfrac{x+1}{x} \quad \text{and} \quad \dfrac{2x}{x-1}$

Similar to fractions, in order to add rational expressions they must have common denominators. Here, the denominators are not the same. This means we need to manipulate the expressions before adding them. We multiply the numerator and denominator of each expression by the denominator in the other expression. $\dfrac{x+1}{x} \cdot \dfrac{x-1}{x-1} \quad \text{and} \quad \dfrac{2x}{x-1} \cdot \dfrac{x}{x}$
We'll multiply the first set of fractions.
Next, we'll multiply the second set of fractions.
Now that the expressions have the same denominator, we can add them.
Thus, the sum of the rational expressions is
$\dfrac{3x^2-1}{x^2-x}.$

$\dfrac{x+1}{x}$

$\dfrac{(x+1)(x-1)}{x(x-1)}$

$\dfrac{x^2-1^2}{x(x-1)}$

$\dfrac{x^2-1}{x(x-1)}$

$\dfrac{x^2-1}{x^2-x}$

$\dfrac{2x}{x-1}$

$\dfrac{2x\cdot x}{(x-1)\cdot x}$

$\dfrac{2x^2}{(x-1)\cdot x}$

$\dfrac{2x^2}{x^2-x}$

$\dfrac{x^2-1}{x^2-x}+\dfrac{2x^2}{x^2-x}$

$\dfrac{x^2-1+2x^2}{x^2-x}$

$\dfrac{3x^2-1}{x^2-x}$

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