Chapter 6. Rational Expressions. Rational Functions and Multiplying and Dividing Rational Expressions. § 6.1. Rational Expressions. Rational expressions can be written in the form where P and Q are both polynomials and Q 0. Examples of Rational Expressions.
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Rational expressions can be written in the form where P and Q are both polynomials and Q 0.
Examples of Rational Expressions
To evaluate a rational expression for a particular value(s), substitute the replacement value(s) into the rational expression and simplify the result.
Example:
Evaluate the following expression for y = 2.
In the previous example, what would happen if we tried to evaluate the rational expression for y = 5?
This expression is undefined!
We have to be able to determine when a rational expression is undefined.
A rational expression is undefined when the denominator is equal to zero.
The numerator being equal to zero is okay (the rational expression simply equals zero).
Find any real numbers that make the following rational expression undefined.
Example:
The expression is undefined when 15x + 45 = 0.
So the expression is undefined when x = 3.
or, simply,
Simplifying Rational Expressions
Simplifying a rational expression means writing it in lowest terms or simplest form.
Fundamental Principle of Rational Expressions
Simplifying a Rational Expression
1) Completely factor the numerator and denominator of the rational expression.
2) Divide out factors common to the numerator and denominator. (This is the same thing as “removing the factor of 1.”)
Warning!
Only common FACTORS can be eliminated from the numerator and denominator. Make sure any expression you eliminate is a factor.
Multiplying Rational Expressions
The rule for multiplying rational expressions is
as long as Q 0 and S 0.
1) Completely factor each numerator and denominator.
2) Use the rule above and multiply the numerators and denominators.
3) Simplify the product by dividing the numerator and denominator by their common factors.
Dividing Rational Expressions
The rule for dividing rational expressions is
as long as Q 0 and S 0 and R 0.
To divide by a rational expression, use the rule above and multiply by its reciprocal. Then simplify if possible.
Adding and Subtracting Rational Expressions
The rule for multiplying rational expressions is
Adding or Subtracting Rational Expressions with Common Denominators
are rational expression, then
This involves finding the least common denominator of the two original rational expressions.
Least Common Denominators1) Factor each denominator completely,
2) The LCD is the product of all unique factors each raised to a power equal to the greatest number of times that the factor appears in any factored denominator.
Least Common DenominatorsExample:
Both of the denominators are already factored.
Since each is the opposite of the other, you can
use either x – 3 or 3 – x as the LCD.
Example:
Find the LCD of the rational expressions.
Write each rational expression as an equivalent rational expression whose denominator is the LCD found in Step 1.
Add or subtract numerators, and write the result over the denominator.
Simplify resulting rational expression, if possible.
Unlike DenominatorsSimplifying Complex Fractions
Complex rational expressions (complex fraction) are rational expressions whose numerator, denominator, or both contain one or more rational expressions.
There are two methods that can be used when simplifying complex fractions.
Simplify the numerator and the denominator of the complex fraction so that each is a single fraction.
Perform the indicated division by multiplying the numerator of the complex fraction by the reciprocal of the denominator of the complex fraction.
Simplify, if possible.
Simplifying Complex FractionsExample:
Multiply the numerator and the denominator of the complex fraction by the LCD of the fractions in both the numerator and denominator.
Simplify, if possible.
Simplifying Complex FractionsExample:
The resulting expression is a complex fraction, so use either method to simplify the expression.
Simplifying with Negative ExponentsDividing Polynomials: Long Division and Synthetic Division
Dividing a Polynomial by a Monomial
Divide each term in the polynomial by the monomial.
Example: Divide
Example: Divide
We then write our result as
Divide 43 into 72.
Multiply 1 times 43.
Subtract 43 from 72.
Bring down 5.
Divide 43 into 295.
Multiply 6 times 43.
Subtract 258 from 295.
Bring down 6.
Divide 43 into 376.
Multiply 8 times 43.
Subtract 344 from 376.
Nothing to bring down.
As you can see from the previous example, there is a pattern in the long division technique.
Divide
Multiply
Subtract
Bring down
Then repeat these steps until you can’t bring down or divide any longer.
We will incorporate this same repeated technique with dividing polynomials.

35
x

15
Divide 7x into 28x2.
Multiply 4x times 7x+3.
Subtract 28x2 + 12x from 28x2 – 23x.
Bring down – 15.
Divide 7x into –35x.
Multiply – 5 times 7x+3.
Subtract –35x–15 from –35x–15.
Nothing to bring down.
So our answer is 4x – 5.

10
10
2
2
x
x
+

+
2
2
x
7
4
x
6
x
8

20
x


20
x
70
+
8
2
+
14
x
4
x
78
+
We write our final answer as
+
(
2
x
7
)
Dividing Polynomials
Divide 2x into 4x2.
Multiply 2x times 2x+7.
Subtract 4x2 + 14x from 4x2 – 6x.
Bring down 8.
Divide 2x into –20x.
Multiply 10 times 2x+7.
Subtract –20x–70 from –20x+8.
78
Nothing to bring down.
Example: Divide:
1. Divide the leading term of the dividend, c2, by the first term of the divisor, x.
2. Multiply c by c + 1.
3. Subtract c2 + c from c2 + 3c – 2.
Continued.
2
c
Remainder
Example continued:
Bring down the next term to obtain a new polynomial.
4. Repeat the process until the degree of the remainder is less than the degree of the binomial divisor.
5. Check by verifying that (Quotient)(Divisor) + Remainder = Dividend.
No remainder
Example: Divide: (y2 – 5y + 6) ÷ (y – 2)
y
– 3
y2 – 2y
+ 6
– 3y
– 3y + 6
0
Check : (y – 2)(y – 3)
= y2 – 5y + 6
(y2 – 5y + 6)÷ (y – 2) = y – 3
To find the quotient and remainder when a polynomial of degree 1 or higher is divided by x – c, a shortened version of long division called synthetic division may be used.
Long Division
Synthetic Division
Subtraction signs are not necessary.
The original terms in red are not needed because they are the same as the term directly above.
Continued.
The variables are not necessary.
Continued.
The “boxed” numbers can be aligned horizontally.
Continued.
The first two numbers in the last row are the coefficients of the quotient, the last number is the remainder.
To simplify further, the top row can be removed and instead of subtracting, we can change the sign of each entry and add.
The leading coefficient of the dividend can be brought down.
The x + 1 is replaced with a – 1.
Quotient
Remainder
Continued.
Solving Equations Containing Rational Expressions
First note that an equation contains an equal sign and an expression does not.
To solve EQUATIONS containing rational expressions, clear the fractions by multiplying both sides of the equation by the LCD of all the fractions.
Then solve as in previous sections.
Note: this works for equations only, not simplifying expressions.
Solving Equations
Example continued:
Substitute the value for x into the original equation, to check the solution.
true
Solving Equations
Example continued:
Substitute the value for x into the original equation, to check the solution.
true
Example continued:
Substitute the value for x into the original equation, to check the solution.
true
So the solution is x = 3.
Example continued:
Substitute the value for x into the original equation, to check the solution.
Since substituting the suggested value of a into the equation produced undefined expressions, the solution is .
Rational Equations and Problem Solving
Clear the equation of fractions or rational expressions by multiplying each side of the equation by the LCD of the denominators in the equation.
Use the distributive property to remove grouping symbols such as parentheses.
Combine like terms on each side of the equation.
Use the addition property of equality to rewrite the equation as an equivalent equation with terms containing the specified variable on one side and all other terms on the other side.
Use the distributive property and the multiplication property of equality to get the specified variable alone.
Solving Equations with Multiple Variables
Read and reread the problem. If we let
n = the number, then
= the reciprocal of the number
Example:
The quotient of a number and 9 times its reciprocal is 1. Find the number.
1.) Understand
Continued
is
1
a number
and 9 times its reciprocal
n
=
1
Finding an Unknown Number
Example continued:
2.) Translate
Continued
Example continued:
4.) Interpret
Check: We substitute the values we found from the equation back into the problem. Note that nothing in the problem indicates that we are restricted to positive values.
true
true
State: The missing number is 3 or –3.
Ratio is the quotient of two numbers or two quantities.
The ratio of the numbers a and b can also be written as a:b, or .
The units associated with the ratio are important.
The units should match.
If the units do not match, it is called a rate, rather than a ratio.
We can rewrite the proportion by multiplying by the LCD, bd.
This simplifies the proportion to ad = bc.
This is commonly referred to as the cross product.
Solving Proportions
Example continued:
Substitute the value for x into the original equation, to check the solution.
true
Example:
10 ounces for $0.99
16 ounces for $1.69
30 ounces for $3.29
Solving ProportionsExample:
Continued.
Example continued:
Size Price Unit Price
10 ounces $0.99 $0.99/10 = $0.099
16 ounces $1.69 $1.69/16 = $0.105625
30 ounces $3.29 $3.29/30 $0.10967
The 10 ounce size has the lower unit price, so it is the best buy.
Read and reread the problem. By using the times for each roofer to complete the job alone, we can figure out their corresponding work rates in portion of the job done per hour.
Time in hrs
Portion job/hr
Experienced roofer 26 1/26
Beginner roofer 39 /39
Together t 1/t
Example:
An experienced roofer can roof a house in 26 hours. A beginner needs 39 hours to do the same job. How long will it take if the two roofers work together?
1.) Understand
Continued
Example continued:
2.) Translate
Since the rate of the two roofers working together would be equal to the sum of the rates of the two roofers working independently,
Continued
Example continued:
4.) Interpret
Check: We substitute the value we found from the proportion calculation back into the problem.
true
State: The roofers would take 15.6 hours working together to finish the job.
Read and reread the problem. By using the formula d=rt, we can rewrite the formula to find that t = d/r.
We note that the rate of the boat downstream would be the rate in still water + the water current and the rate of the boat upstream would be the rate in still water – the water current.
Distance rate time = d/r
Down 20 r + 5 20/(r + 5)
Up 10 r – 5 10/(r – 5)
Example:
The speed of Lazy River’s current is 5 mph. A boat travels 20 miles downstream in the same time as traveling 10 miles upstream. Find the speed of the boat in still water.
1.) Understand
Continued
Example continued:
2.) Translate
Since the problem states that the time to travel downstairs was the same as the time to travel upstairs, we get the equation
Continued
Example continued:
4.) Interpret
Check: We substitute the value we found from the proportion calculation back into the problem.
true
State: The speed of the boat in still water is 15 mph.
Variation and Problem Solving
y varies directly as x, or y is directly proportional to x, if there is a nonzero constant k such that y = kx.
The family of equations of the form y = kx are referred to as direct variation equations.
The number k is called the constant of variation or the constant of proportionality.
y = kx
5 = k·30
k = 1/6
Direct VariationSo the direct variation equation is
1
6
y =
x
y = kx
48 = k·6
8 = k
So the equation is y = 8x.
y = 8·15
y = 120
Direct VariationExample:
Example:
Continued.
Example continued:
We substitute our given value for the elevation into the equation.
So our equation is
The family of equations of the form y = k/x are referred to as inverse variation equations.
The number k is still called the constant ofvariation or the constant of proportionality.
Inverse Variationy = k/x
63 = k/3
k = 63·3
k = 189
Inverse Variation189
x
So the inverse variation equation is
y =
Example:
y varies directly as a power of x if there is a nonzero constant k and a natural number n such that y = kxn
Powers of xy varies inversely as a power of x if there is a nonzero constant k and a natural number n such that
If an 8foot column can hold 2 tons, find how much weight a 10foot column can hold.
Powers of xExample:
Continued.
Example continued:
We substitute our given value for the height of the column into the equation.
So our equation is
k is STILL the constant of variation or the constant of proportionality.
We can also use combinations of direct, inverse, and joint variation. These variations are referred to as combined variations.
Combined Variationa varies jointly as b and c
a = kbc
P varies jointly as R and the square of S
P = kRS2
The weight (w) varies jointly with the width (d) and the square of the height (h) and inversely with the length (l)
Combined VariationExample: