FLC Ch 6

Page 1 of 14

Math 120 Intermediate Algebra

Sec 6.1: Rational Expressions and Functions: Multiplying and Dividing

Ex 1 Simplify. State any restrictions (if necessary). a) b)

10𝑦𝑧4

40𝑦2𝑧9

3𝑦2 + 10𝑦 + 3

3𝑦2 − 14𝑦 − 5

Ex 2 Simplify each. List all restrictions on the domain. Next, graph the function 𝑓.

𝑔(𝑥) =8𝑥 − 16

𝑥2 − 4 𝑓(𝑥) =

(12 𝑥 + 1) (𝑥 − 2)

𝑥 − 2

Ex 3 Simplify. State any restrictions (if necessary).

a) b) c)

𝑦2 + 10𝑦 + 25

𝑦2 − 9∙

𝑦2 + 3𝑦

𝑦 + 5

𝑎3 − 𝑏3

3𝑎2 + 9𝑎𝑏 + 6𝑏2∙

𝑎2 + 2𝑎𝑏 + 𝑏2

𝑎2 − 𝑏2

4𝑎2 − 1

𝑎2 − 4÷

2𝑎 − 1

2 − 𝑎

Defn A polynomial divided by another nonzero polynomial is called a rational expression. If 𝑃 and

𝑄 are polynomials in 𝑥 with 𝑄 ≠ 0, then 𝑟(𝑥) =𝑃

𝑄 is called a rational function.

FLC Ch 6

Page 2 of 14

d) (As time permits)

5𝑥2 − 5𝑦2

27𝑥3 + 8𝑦3÷

𝑥2 − 2𝑥𝑦 + 𝑦2

9𝑥2 − 6𝑥𝑦 + 4𝑦2∙

6𝑥 + 4𝑦

10𝑥 − 15𝑦

Ans: 𝟐(𝒙 + 𝒚)/(𝒙 − 𝒚)(𝟐𝒙 − 𝟑𝒚) 𝒙 ≠ 𝒚, −𝟐𝒚/𝟑 Do problems from 𝒚 =

𝟏

𝒙 Handout

Intro to Sec 6.2: Rational Expressions and Functions: Adding and Subtracting

Ex 4 Find the LCD.

1

23 ∙ 52 ∙ 𝑥2 𝑎𝑛𝑑

1

23 ∙ 3 ∙ 5 ∙ 𝑥2

1

56(𝑥 − 1)2(𝑥 + 1) 𝑎𝑛𝑑

1

24(𝑥 − 1)(𝑥 + 1)(𝑥 + 2)

1

𝑥,

1

𝑥 + 1

1

𝑥 + 𝑦,1

𝑥, 𝑎𝑛𝑑

1

𝑦

1

4𝑥,

1

4𝑥 − 8

1

9𝑥,

1

2𝑥 + 4

Sec 6.3: Complex Rational Expressions

Ex 5 Simplify using each method. Method TWO Method ONE

13 −

110

215

+13

13 −

110

215

+13

*Method 1: LCD Method - Avoid Multiplying by Reciprocal Method 2: Multiply by Reciprocal *generally preferred EC?: Determine which method is more efficient.

Good Exercises:

𝑓(𝑥) =1

2(𝑥 + 1)3

𝑔(𝑥) =3

4|𝑥 + 2|

ℎ(𝑥) = −1

3(𝑥 + 1)2 − 2

FLC Ch 6

Page 3 of 14

Ex 6 Simplify. a) b)

3𝑥𝑦−1 − 𝑥

2𝑦 − 𝑦𝑥−1

𝑥2 − 𝑥 − 12𝑥2 − 2𝑥 − 15𝑥2 + 8𝑥 + 12𝑥2 − 5𝑥 − 14

c) d) PP

𝑦𝑦2 − 1

−3

1 − 𝑦2

𝑦3

𝑦2 − 1−

271 − 𝑦2

6𝑥2 − 25

−1

𝑥 − 524

𝑥2 − 25+

6𝑥 + 5

e) f) In-Class Prob: Ans: −

𝟐𝒚−𝟕

𝟐𝒚+𝟏𝟏

2𝑥 + 33

2𝑥 + 6 −7𝑥

𝑦𝑦2 − 1

−3𝑦

𝑦2 + 5𝑦 + 43𝑦

𝑦2 − 1−

𝑦𝑦2 + 3𝑦 − 4

FLC Ch 6

Page 4 of 14

g) DO h) PP −2𝑥

𝑥2 − 7𝑥 + 12+

8𝑥 − 4

𝑥 + 4

105𝑥 − 1

− 10

105𝑥 − 1

+ 10

Ans:

𝟐−𝟓𝒙

𝟓𝒙

Complete Problems from Function Worksheet IV (Graphs Review A)

Sec 6.6: Division of Polynomials

Ex 7 Divide and check. a) b) Fractions

−25𝑥3 + 20𝑥2 − 3𝑥 + 5

−10𝑥 (3𝑥3 − 5𝑥2 − 3𝑥 − 2) ÷ (3𝑥 − 2)

c) (3𝑥4 + 2𝑥3 − 11𝑥2 − 2𝑥 + 5) ÷ (𝑥2 − 2)

Type I (easier): Dividing by a monomial. “Break up numerator.” Type II (more difficult): Dividing by a binomial (or polynomial with more than 1 term). Use Long Division (LD) OR Synthetic Division (SD) (sec 6.7)

FLC Ch 6

Page 5 of 14

d) e) 𝑥3 − 𝑥 + 6

𝑥 + 2

𝑥3 − 8

𝑥 − 2

*answer only f) Practice Problem 𝑥4 + 4𝑥2 + 6

𝑥2 + 1

Answer: 𝒙𝟐 + 𝟑 +𝟑

𝒙𝟐+𝟏

g) (As time permits) Find a simplified expression for 𝐹(𝑥) if 𝐹(𝑥) = (𝑓

𝑔) (𝑥). Be sure to list all

restrictions on the domain of 𝐹(𝑥). 𝑓(𝑥) = 𝑥4 − 3𝑥2 − 54; 𝑔(𝑥) = 𝑥2 − 9 Answer: 𝒙𝟐 + 𝟔, 𝒙 ≠ ±𝟑

Sec 6.7: Synthetic Division

Ex 8 Use synthetic division to divide.

a) (𝑥3 − 4𝑥2 + 5𝑥 − 6) ÷ (𝑥 − 3) b) (2𝑥3 − 3𝑥2 + 8) ÷ (𝑥 + 2) RT says: RT says:

A streamlined process of long division (synthetic division) can be used when the divisor has the form 𝒙 − 𝒄.

FLC Ch 6

Page 6 of 14

c) (8𝑥3 − 1 + 7𝑥 − 6𝑥2) ÷ (𝑥 −1

2) d) PP (𝑥5 − 243) ÷ (𝑥 − 3)

Answer: 𝒙𝟒 + 𝟑𝒙𝟑 + 𝟗𝒙𝟐 + 𝟐𝟕𝒙 + 𝟖𝟏

RT says: What does the Remainder Theorem tell us about the above examples? Ex 9 Use synthetic division to find 𝑓(−2) where 𝑓(𝑥) = 3𝑥4 + 8𝑥3 + 2𝑥2 − 7𝑥 − 4. Final?

The Remainder Theorem

The remainder obtained upon dividing a polynomial 𝑃(𝑥) by 𝑥 − 𝒄 is 𝑃(𝒄).

FLC Ch 6

Page 7 of 14

Show a check: Ex 10 By the Factor Theorem, 𝑝(𝑐) = 0 (where 𝑝 is a polynomial) if and only if (𝑥 − 𝑐) is a factor of 𝑝(𝑥). Use synthetic division to show that (𝑥 + 2) is a factor of (𝑥3 − 𝑥 + 6). Refer to example 7d. Show work and write conclusion using both the Remainder and Factor Theorems.

Sec 6.2: Rational Expressions and Functions: Adding and Subtracting & Sec 6.4: Rational Equations

READ AND STUDY:

Ex 11 Simplify OR solve. a) DO b) DO

9

𝑦−

5

𝑦 − 8

5𝑎𝑏

𝑎2 − 𝑏2+

𝑎 + 𝑏

𝑎 − 𝑏

ADDING/SUBTRACTING – *Do not factor/cancel until AFTER combining numerators* IF cancelling occurs (which means the final numerator factored), be sure to list the restrictions.

SOLVING – CLEAR FRACTIONS and LIST any bad values (restrictions) and CHECK for vacuous answers. Warning: On exams, when solving rational equations, if we fail to recognize to clear fractions, NO partial credit for incorrect answers.

FLC Ch 6

Page 8 of 14

c) DO d) PP

𝑥 + 2

𝑥2 − 1+

𝑥 − 3

𝑥2 + 2𝑥 − 3−

2𝑥 − 3

𝑥2 + 4𝑥 + 3

2

𝑥2 + 3𝑥 − 10−

5𝑥

𝑥2 − 6𝑥 + 8

Answer: −𝟓𝒙𝟐+𝟐𝟑𝒙+𝟖

(𝒙+𝟓)(𝒙−𝟐)(𝒙−𝟒)

e) f) DO

5𝑥

𝑥2 − 6𝑥 + 8−

3𝑥 + 1

𝑥2 − 𝑥 − 12

3

𝑥2 − 11𝑥 + 30−

1

𝑥 − 6=

1

9𝑥 − 45

g) h) DO

2𝑡

3 − 𝑡−

4

𝑡=

1

𝑡 − 3

𝑥 − 7

𝑥2 − 16−

𝑥 − 1

16 − 𝑥2

FLC Ch 6

Page 9 of 14

i) PP j) Start

𝑥 − 3

𝑥 − 2=

2𝑥2 − 8

𝑥2 + 𝑥 − 6−

𝑥 − 1

𝑥 + 3 𝑡 +

6

𝑡= −5

Ans: 𝒕 = −𝟐, −𝟑 Ans: 𝒙 = 𝟏/𝟑

k) DO l)

𝑥2 − 1

𝑥 + 2=

3

𝑥 + 2

2𝑥

𝑥2 − 6𝑥 + 8−

𝑥

𝑥2 − 𝑥 − 12

m) DO n) DO

2𝑥

𝑥2 − 𝑥 − 6+

𝑥 + 1

3 − 𝑥

6

𝑥 − 8+

6

𝑥= −

48

𝑥2 − 8𝑥

FLC Ch 6

Page 10 of 14

o) DO 𝑥 − 3

𝑥 − 2=

2𝑥2 − 9

𝑥2 + 𝑥 − 6−

𝑥 + 1

𝑥 + 3

Ex 12 a) Find all values for 𝑎 such that 𝑓(𝑎) = 𝑔(𝑎). b) Find 𝐻(𝑥) = 𝑓(𝑥) − 𝑔(𝑥).

𝑓(𝑥) =𝑥 − 4

𝑥 − 3; 𝑔(𝑥) =

2𝑥2 − 13

𝑥2 + 𝑥 − 12−

𝑥 + 2

𝑥 + 4 𝑓(𝑥) =

2

𝑥2 + 5𝑥 + 6; 𝑔(𝑥) =

1

𝑥 + 2

Ex 13 Find the simplified form for 𝑓(𝑥) and list all restrictions on the domain.

𝑓(𝑥) =3𝑥 − 2

𝑥2 + 5𝑥 − 24−

𝑥 − 3

𝑥2 − 9

FLC Ch 6

Page 11 of 14

Ex 14 PP (#62) Let 𝑓(𝑥) =2𝑥+5

𝑥2+4𝑥+3 and 𝑔(𝑥) =

𝑥+2

𝑥2−9+

𝑥−1

𝑥2−2𝑥−3. Find all values of 𝑎 for which

𝑓(𝑎) = 𝑔(𝑎). Answer: −𝟕

𝟑

PP

Solve. 𝑥 − 4

2𝑥 + 1−

24

2𝑥2 − 7𝑥 − 4=

𝑥 + 2

4 − 𝑥

Start “Which Fractions Reduce” handout. Finish-IC or hw assignment.

Sec 6.5: Solving Applications Using Rational Equations

Ex 15 (# 12) The reciprocal of the product of two consecutive integers is 1

30. Find the two integers.

-6, -5 and 5, 6 Ex 16 (#16) A community water tank can be filled in 18 hours by the town office well alone and in 22 hours by the high school well alone. How long will it take to fill the tank if both wells are working? 𝟗 𝟗

𝟏𝟎 hrs

FLC Ch 6

Page 12 of 14

Ex 17 (# 20) Kent can cut and split a cord of wood twice as fast as Brent can. When they work together, it takes them 4 hours. How long would it take each of them to do the job alone? B: 12 hrs K: 6 hrs Ex 18 (# 36) The A train goes 12 mph slower than the E train. The A train travels 230 miles in the same amount of time that the E train travels 290 miles. Find the speed of each train. A: 46mph; E: 58mph Ex 19 (# 44) Fiona’s Boston Whaler cruised 45 miles upstream and 45 miles back in a total of 8 hours. The speed of the river is 3 mph. Find the speed of the boat in still water. 12mph

FLC Ch 6

Page 13 of 14

Sec 6.8: Formulas, Applications, and Variation

Ex 20 Solve for each specified variable. a) (# 20) b) (# 24)

𝐾 =𝑟𝑡

𝑟 − 𝑡; 𝑡

1

𝑝+

1

𝑞=

1

𝑓; 𝑝

Ex 21 (# 46) Find the variation constant and an equation of variation if 𝑦 varies directly as 𝑥 and 𝑦 =2 when 𝑥 = 5. Ex 22 (# 52) Find the variation constant and an equation of variation in which 𝑦 varies inversely as 𝑥 and 𝑦 = 9 when 𝑥 = 10.

Direct Variation

The situation is modeled by a linear function of the form 𝒇(𝒙) = 𝒌𝒙, or 𝒚 = 𝒌𝒙, where 𝑘 is a nonzero constant.

We say there is direct variation, that 𝑦 varies directly as 𝑥, or that 𝑦 is proportional to 𝑥.

Inverse Variation

The situation is modeled by a rational function of the form 𝒇(𝒙) = 𝒌/𝒙, or 𝒚 = 𝒌/𝒙, where 𝑘 is a nonzero constant.

We say the there is inverse variation, that 𝑦 varies inversely as 𝑥, or that 𝑦 is inversely proportional to 𝑥.

Joint Variation When a variable varies directly with more than one variable, we say that there is a joint variation. 𝑦 varies jointly as 𝑥 and 𝑧 if for some nonzero constant 𝑘, 𝒚 = 𝒌𝒙𝒛

Note: When a variable varies directly and/or inversely at the same time with more than one other variable, there is a combined variation. Joint variation is a form of combined variation.

The number 𝑘 is called the variation constant, or constant of proportionality.

FLC Ch 6

Page 14 of 14

Ex 23 (# 76) Find an equation of variation in which 𝑦 varies directly as 𝑥 and inversely as 𝑤 and the square of 𝑧, and 𝑦 = 4.5 when 𝑥 = 15, 𝑤 = 5, and 𝑧 = 2. Ex 24 (# 80) The intensity 𝐼 of a television signal varies inversely as the square of the distance from the transmitter. If the intensity is 25 𝑊/𝑚2 at a distance of 2 𝑘𝑚, what is the intensity 6.25 𝑘𝑚 from the transmitter? Grade each problem. Identify any and all mistakes. Problem: Simplify or solve.

2

𝑥2 + 5𝑥 + 6−

𝑥 + 4

𝑥 + 2

Solution 1:

(𝑥 + 2)(𝑥 + 3)2

(𝑥 + 2)(𝑥 + 3)−

𝑥 + 4

𝑥 + 2(𝑥 + 2)(𝑥 + 3)

2 − 𝑥2 + 7𝑥 + 12

−𝑥2 + 7𝑥 + 14

Solution 2: 2 − (𝑥 + 4)(𝑥 + 3)

𝐿𝐶𝐷

2 − (𝑥2 + 7𝑥 + 12)

𝐿𝐶𝐷

2 − 𝑥2 − 7𝑥 − 12 −𝑥2 − 7𝑥 − 10

𝐿𝐶𝐷=

(𝑥 − 5)(𝑥 + 2)

(𝑥 + 2)(𝑥 + 3)

𝑥 − 5

𝑥 + 3

Solution 3: 2

(𝑥 + 2)(𝑥 + 3)−

𝑥 + 4

𝑥 + 2

2 − 𝑥2 + 7𝑥 + 12 𝑥2 + 7𝑥 + 10 (𝑥 + 5)(𝑥 + 2)

(𝑥 + 2)(𝑥 + 3)

𝑥 + 5

𝑥 + 3