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As with rational numbers, it is possible to write two polynomial expressions as a ratio, so long as the divisor is not zero. This lesson will explore various aspects of this type of algebraic expression, including how mathematical operations such as multiplication and division can be applied to it. ### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

The distance that a vehicle can travel per one gallon of fuel is measured as its mile per gallon (mpg) fuel economy. Each car has two fuel economy numbers, one measuring its efficiency for city driving and the other for highway driving. The combined fuel economy $C$ for $x$ mpg in the city and $y$ mpg on the highway is computed by the following formula.

$C=21 (x1 +y1 )1 $

a If a car travels $28$ miles per gallon in the city and $36$ mpg on the highway, what is the combined fuel economy for the car?

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b Let $x$ be the city fuel economy for a new car model. If the highway fuel economy in terms of $x$ is $x−2010x ,$ write a simplified expression for the combined fuel economy for such a car.

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A rational expression is a fraction where both the numerator and the denominator are polynomials.

$q(x)p(x) $

$x_{3}+5xx_{2}−7 $

A rational expression is said to be written in its Rational Expressions | |
---|---|

Not in Simplest Form | In Simplest Form |

$x(x−3)(y+2)xy $ | $x+1x−1 $ |

$x_{2}+1x_{4}+x_{2} $ | $x_{2}−x−6x_{3}+7 $ |

Notice that for some of the expressions shown in the table, there are some $x-$values that make the denominator $0.$ For example, the denominator of $x+1x−1 $ is $0$ when $x=-1.$ Any value of a variable for which a rational expression is undefined is called an excluded value.

Expression | Restriction | Excluded Value(s) |
---|---|---|

$x+1x−1 $ | $x+1 =0$ | $x =-1$ |

$x_{2}−x−6x_{3}+7 $ | $x_{2}−x−6 =0$ | $x =-2$ and $x =3$ |

$x(x−3)(y+2)xy $ | $x(x−3)(y+2) =0$ | $x =0,$ $x =3,$ and $y =-2$ |

$x_{2}+1x_{4}+x_{2} $ | There is no real number that makes $x_{2}+1$ zero | None |

Simplifying a rational expression can remove some of the excluded values that appear in the original expression. A rational expression and its simplified form must have the same domain in order for them to be equivalent expressions. This means that the excluded values that are no longer visible in the simplified expression must still be declared.

Equivalent Expressions | |
---|---|

Rational Expression | Simplified Form |

$(x+2)(x−3)x−3 ,x =-2,3$ | $x+21 ,x =-2,3$ |

$x_{2}−1x_{2}+2x+1 ,x =-1,1$ | $x−1x+1 ,x =-1,1$ |

$x_{2}x_{3}−2x_{2}+x ,x =0$ | $xx_{2}−2x+1 ,x =0$ |

A rational expression is undefined when its denominator is $0.$ The values that make the denominator of a rational expression equal to $0$ are called excluded values because they are excluded from its domain. Determine the excluded values for the indicated rational expressions.

A rational expression is written in simplest form if the numerator and denominator have no factors in common. Rational expressions that are not in simplest form can be simplified by canceling out the common factors. There are many various methods of factoring polynomials to find any common factors. Consider an example expression.
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$9x−x_{3}x_{2}−6x+9 $

Rational expressions can be simplified in three steps.
1

Factor the Numerator and the Denominator

Check first if either the numerator or the denominator — or both — can be factored by the greatest common factor. In this example, the factor $x$ is in both terms of the denominator.
Determine if it is possible to factor either the numerator or the denominator by using a 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.
See if either the numerator or the denominator is a perfect square trinomial. Look for an expression in the form $a_{2}±2ab+b_{2},$ which can be factored as $(a±b)_{2}.$ Here, the numerator is a perfect square trinomial.
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 denominator is almost identical to the factor $x−3$ in the numerator.
Now both the numerator and denominator are completely factored.

$9x−x_{3}x_{2}−6x+9 $

SplitIntoFactors

Split into factors

$x⋅9−x⋅x_{2}x_{2}−6x+9 $

FactorOut

Factor out $x$

$x(9−x_{2})x_{2}−6x+9 $

$x(9−x_{2})x_{2}−6x+9 $

WritePow

Write as a power

$x(3_{2}−x_{2})x_{2}−6x+9 $

FacDiffSquares

$a_{2}−b_{2}=(a+b)(a−b)$

$x(3+x)(3−x)x_{2}−6x+9 $

$x(3+x)(3−x)x_{2}−6x+9 $

SplitIntoFactors

Split into factors

$x(3+x)(3−x)x_{2}−2x(3)+9 $

WritePow

Write as a power

$x(3+x)(3−x)x_{2}−2⋅x⋅3+3_{2} $

FacNegPerfectSquare

$a_{2}−2ab+b_{2}=(a−b)_{2}$

$x(3+x)(3−x)(x−3)_{2} $

$x(3+x)(3−x)(x−3)_{2} $

CommutativePropMult

Commutative Property of Multiplication

$(3−x)x(3+x)(x−3)_{2} $

FactorOutNegSignSwap

$a−b=-(b−a)$

$-(x−3)x(3+x)(x−3)_{2} $

2

List Restricted Values

Before simplifying the common factors, check if there are any restrictions on $x.$

$-(x−3)x(3+x)(x−3)_{2} $

Note that the expression is undefined when $x=-3,$ $x=0,$ or $x=3.$ 3

Cancel Out Common Factors

In this expression, $(x−3)$ is the common factor of the numerator and the denominator and can be eliminated.
The rational expression is now written in its simplest form. In order for the given expression and the simplified expression to be equivalent, the domain of the simplified expression should be restricted by excluding $x=3.$

$-x(3+x)x−3 ,x =3 $

Since the restriction $x =3$ cannot be seen from the simplified expression, it should be written. The other restrictions are evident from the simplified expression.
Kevin and Zosia are asked to find the values that make the following rational expression undefined.
### Hint

### Solution

$x_{2}+2x−35x_{2}+12x+35 $

They disagree about the domain of the rational expression.
Who is correct? {"type":"choice","form":{"alts":["Kevin","Zosia"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

A rational expression is undefined for values that make its denominator zero. Therefore, those values should be excluded from the domain.

A rational expression is undefined when the $denominator$ equals $0.$
Now the Zero Product Property will be applied to find all excluded values.
The values $x=-7$ and $x=5$ make the denominator equal to $0.$ As a result, they make the rational expression undefined. Therefore, Zosia is correct in saying that the domain is all real numbers except $-7$ and $5.$ Kevin may have simplified the expression before identifying the excluded values. This is a common mistake!
Note that $x=5$ is the value that makes the denominator of the *simplified* expression equal to $0.$ However, Kevin should have noted that $-7$ makes the *original* expression undefined. Since the original and simplified expressions are equivalent, both should have the same domain.

$x_{2}+2x−35x_{2}+12x+35 $

To find the domain of the rational expression, the expression in the denominator is set equal to $0.$
$x_{2}+2x−35=0 $

To solve the resulting equation for $x,$ start by factoring the expression on the left-hand side.
$x_{2}+2x−35=0$

$(x+7)(x−5)=0$

$(x+7)(x−5)=0$

Solve using the Zero Product Property

ZeroProdProp

Use the Zero Product Property

$x+7=0x−5=0 (I)(II) $

SubEqn

$(I):$ $LHS−7=RHS−7$

$x=-7x−5=0 $

AddEqn

$(II):$ $LHS+5=RHS+5$

$x=-7x=5 $

$(x+7)(x−5)x_{2}+12x+35 $

$(x+7)(x−5)(x+7)(x+5) $

Simplify

$(x−5)(x+5) $

Operations with rational numbers and rational expressions are similar.

Multiplying rational expressions works the same way as multiplying fractions. The numerators and denominators are multiplied separately.

$Q(x)P(x) ⋅G(x)H(x) =Q(x)⋅G(x)P(x)⋅H(x) $

$7x+42x_{2}−2x−15 ⋅x_{2}−11x+30x_{2}+9x+18 $

Three steps can be followed to multiply the rational expressions.
1

Factor Each Numerator and Denominator

Before multiplying, it is helpful to factor the numerators and denominators, if possible. Start with the first rational expression.
Now the numerator and denominator of the second expression will be factored.

$x_{2}−11+30x_{2}+9x+18 $

$x_{2}−11x+30(x+3)(x+6) $

$(x−5)(x−6)(x+3)(x+6) $

2

Multiply the Numerators and Multiply the Denominators

Now that both expressions have been factored, the numerators and denominators can be multiplied.
The product is undefined when $x=-6,$ $x=5,$ and when $x=6.$

$7(x+6)(x+3)(x−5) ⋅(x−5)(x−6)(x+3)(x+6) $

MultFrac

Multiply fractions

$7(x+6)(x−5)(x−6)(x+3)(x−5)(x+3)(x+6) $

4

Cancel Out Common Factors

Note that $(x−5)$ and $(x+6)$ are common factors of the numerator and denominator. Finally, the product is simplified by canceling out common factors.
Considering the denominator of the simplified expression and any other denominator used, the values $x=-6,$ $x=5,$ and $x=6$ must be excluded from the domain of the simplified expression. In this expression it is clear that $x=6$ makes the denominator $0.$ Therefore, the fact that $x =6$ will not be expressed.

$7(x+6)(x−5)(x−6)(x+3)(x−5)(x+3)(x+6) $

CancelCommonFac

Cancel out common factors

$7(x+6) (x−5) (x−6)(x+3)(x−5) (x+3)(x+6) $

SimpQuot

Simplify quotient

$7(x−6)(x+3)(x+3) $

ProdToPowTwoFac

$a⋅a=a_{2}$

$7(x−6)(x+3)_{2} $

$7(x−6)(x+3)_{2} ,x =-6,x =5 $

Dividing two rational expressions is the same as multiplying the first expression by the reciprocal of the second expression.

$x_{2}+3x5−x ÷x_{2}+10x+21x_{2}−25 $

The process of dividing rationals expression can be completed in four steps.
1

Rewrite the Division as the Product of the Dividend and the Reciprocal of the Divisor

To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second.

$x_{2}+3x5−x ÷x_{2}+10x+21x_{2}−25 ⇕x_{2}+3x5−x ⋅x_{2}−25x_{2}+10x+21 $

Once the quotient is expressed as a product, the remaining steps are the same as those for multiplying rational expressions. 2

Factor Each Numerator and Denominator

Factor the numerators and the denominators, if possible. The first expression will be factored first.
Now the numerator and the denominator of the second expression will be factored.

$x_{2}−25x_{2}+10x+21 $

$x_{2}−25(x+3)(x+7) $

Factor the denominator

$(x+5)(x−5)(x+3)(x+7) $

3

Multiply the Numerators and Multiply the Denominators

Next, the numerators and denominators can be multiplied.

$x(x+3)5−x ⋅(x+5)(x−5)(x+3)(x+7) ⇕x(x+3)(x+5)(x−5)(5−x)(x+3)(x+7) $

This product is undefined when $x=-5,$ $x=-3,$ $x=0,$ and $x=5.$ Also, the values that make the divisor's denominator in the original quotient expression equal $0$ should be excluded. These values are $x=-7$ and $x=-3.$
$Excluded Values-7,-5,-3,0,5 $

4

Cancel Out Common Factors

Finally, the product is simplified by canceling out common factors.
Considering the denominator of the simplified expression and any other denominator used in this process, the values $x=-7,$ $x=-5,$ $x=-3,$ $x=0,$ and $x=5$ must be excluded from the domain of the expression. The simplified expression shows that $x=0$ and $x=-5$ must be excluded, so these values are not mentioned.

$x(x+3)(x+5)(x−5)(5−x)(x+3)(x+7) $

FactorOut

Factor out $-1$

$x(x+3)(x+5)(x−5)-1(x−5)(x+3)(x+7) $

CancelCommonFac

Cancel out common factors

$x(x+3) (x+5)(x−5) -1(x−5) (x+3) (x+7) $

SimpQuot

Simplify quotient

$x(x+5)-1(x+7) $

Distr

Distribute $-1$

$x(x+5)-x−7 $

$x(x+5)-x−7 ,x =-7,x =-3,x =5 $

Ramsha drew the plan of her house and labeled the sides, measured in meters, as shown.

a Write a simplified rational expression for the area of the house.

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b State any restrictions on $x.$

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a Start by factoring the numerator and the denominator of each rational expression.

b The denominator of a rational expression *cannot* be zero.

a Start by considering the formula for the area of a rectangle.

$A=ℓw $

According to the diagram, the length $ℓ$ of the house is represented by $x_{2}+18x+812x_{2}−6x $ and the width $w$ by $x_{2}−99x+81 .$
To write an expression for the area, these two expressions will be substituted into the formula for the area of a rectangle. Both factors will then be factored.
$A=ℓw$

SubstituteII

$ℓ=x_{2}+18x+812x_{2}−6x $, $w=x_{2}−99x+81 $

$A=x_{2}+18x+812x_{2}−6x ⋅x_{2}−99x+81 $

Factor the first expression

FactorOut

Factor out $2x$

$A=x_{2}+18x+812x(x−3) ⋅x_{2}−99x+81 $

SplitIntoFactors

Split into factors

$A=x_{2}+2(9)x+812x(x−3) ⋅x_{2}−99x+81 $

CommutativePropMult

Commutative Property of Multiplication

$A=x_{2}+2x(9)+812x(x−3) ⋅x_{2}−99x+81 $

WritePow

Write as a power

$A=x_{2}+2x(9)+9_{2}2x(x−3) ⋅x_{2}−99x+81 $

FacPosPerfectSquare

$a_{2}+2ab+b_{2}=(a+b)_{2}$

$A=(x+9)_{2}2x(x−3) ⋅x_{2}−99x+81 $

Factor second expression

FactorOut

Factor out $9$

$A=(x+9)_{2}2x(x−3) ⋅x_{2}−99(x+9) $

WritePow

Write as a power

$A=(x+9)_{2}2x(x−3) ⋅x_{2}−3_{2}9(x+9) $

FacDiffSquares

$a_{2}−b_{2}=(a+b)(a−b)$

$A=(x+9)_{2}2x(x−3) ⋅(x+3)(x−3)9(x+9) $

$(x+9)_{2}2x(x−3) ⋅(x+3)(x−3)9(x+9) $

Simplify

MultFrac

Multiply fractions

$(x+9)_{2}⋅(x+3)(x−3)2x(x−3)⋅9(x+9) $

PowToProdTwoFac

$a_{2}=a⋅a$

$(x+9)(x+9)⋅(x+3)(x−3)2x(x−3)⋅9(x+9) $

CancelCommonFac

Cancel out common factors

$(x+9) (x+9)⋅(x+3)(x−3) 2x(x−3) ⋅9(x+9) $

SimpQuot

Simplify quotient

$(x+9)⋅(x+3)2x⋅9 $

Multiply

Multiply

$(x+9)(x+3)18x $

b To identify the restrictions, the denominator of the simplified expression and *any other denominator* used in this process will be analyzed. For simplicity, their factored forms will be used.

Denominator | Restrictions on the Denominator | Restrictions on the Variable |
---|---|---|

$(x+9)_{2}$ | $(x+9)_{2} =0$ | $x =-9$ |

$(x+3)(x$ |