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Polynomials and Polynomial Functions
One of the simplest types of algebraic expressions are polynomials. They are formed only by addition and multiplication of variables and constants. Since both addition and multiplication produce unique values for any given inputs, polynomials are in fact functions. One of the simplest polynomial functions are linear functions, such as ππ(π₯π₯) = 2π₯π₯ + 1, or quadratic functions, such as ππ(π₯π₯) = π₯π₯2 + π₯π₯ β 6. Due to the comparably simple form, polynomial functions appear in a wide variety of areas of mathematics and science, economics, business, and many other areas of life. Polynomial functions are often used to model various natural phenomena, such as the shape of a mountain, the distribution of temperature, the trajectory of projectiles, etc. The shape and properties of polynomial functions are helpful when constructing such structures as roller coasters or bridges, solving optimization problems, or even analysing stock market prices.
In this chapter, we will introduce polynomial terminology, perform operations on polynomials, and evaluate and compose polynomial functions.
P.1 Addition and Subtraction of Polynomials
Terminology of Polynomials
Recall that products of constants, variables, or expressions are called terms (see section R3, Definition 3.1). Terms that are products of only numbers and variables are called monomials. Examples of monomials are β2π₯π₯, π₯π₯π¦π¦2, 2
3π₯π₯3, etc.
Definition 1.1 A polynomial is a sum of monomials.
A polynomial in a single variable is the sum of terms of the form πππ₯π₯ππ, where ππ is a numerical coefficient, π₯π₯ is the variable, and ππ is a whole number.
An n-th degree polynomial in π₯π₯-variable has the form
ππππππππ + ππππβππππππβππ + β―+ ππππππππ + ππππππ + ππππ,
where ππππ,ππππβ1, β¦ ,ππ2,ππ1,ππ0 β β, ππππ β 0.
Note: A polynomial can always be considered as a sum of monomial terms even though there are negative signs when writing it. For example, polynomial π₯π₯2 β 3π₯π₯ β 1 can be seen as the sum of signed terms π₯π₯2 + β3π₯π₯ + β1
Definition 1.2 The degree of a monomial is the sum of exponents of all its variables.
For example, the degree of 5π₯π₯3π¦π¦ is 4, as the sum of the exponent of π₯π₯3, which is 3 and the exponent of π¦π¦, which is 1. To record this fact, we write deg(5π₯π₯3π¦π¦) = 4.
The degree of a polynomial is the highest degree out of all its terms.
For example, the degree of 2π₯π₯2π¦π¦3 + 3π₯π₯4 β 5π₯π₯3π¦π¦ + 7 is 5 because deg(2π₯π₯2π¦π¦3) = 5 and the degrees of the remaining terms are not greater than 5.
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Polynomials that are sums of two terms, such as π₯π₯2 β 1, are called binomials. Polynomials that are sums of three terms, such as π₯π₯2 + 5π₯π₯ β 6 are called trinomials.
The leading term of a polynomial is the highest degree term. The leading coefficient is the numerical coefficient of the leading term. So, the leading term of the polynomial 1 β π₯π₯ β π₯π₯2 is βπ₯π₯2, even though it is not the first term. The leading coefficient of the above polynomial is β1, as βπ₯π₯2 can be seen as (β1)π₯π₯2.
A first degree term is often referred to as a linear term. A second degree term can be referred to as a quadratic term. A zero degree term is often called a constant or a free term.
Below are the parts of an n-th degree polynomial in a single variable π₯π₯:
ππππ πππποΏ½οΏ½οΏ½ππππππππππππππππππππππ
+ ππππβππππππβππ +β―+ πππππππποΏ½οΏ½οΏ½ππππππππππππππππππ
ππππππππ
+ πππππποΏ½ππππππππππππππππππππ
+ πππποΏ½ππππππππππππππππ
(ππππππππ)ππππππππ
Note: Single variable polynomials are usually arranged in descending powers of the variable. Polynomials in more than one variable are arranged in decreasing degrees of terms. If two terms are of the same degree, they are arranged with respect to the descending powers of the variable that appers first in alphabetical order.
For example, polynomial π₯π₯2 + π₯π₯ β 3π₯π₯4 β 1 is customarily arranged as follows β3π₯π₯4 + π₯π₯2 + π₯π₯ β 1,
while polynomial 3π₯π₯3π¦π¦2 + 2π¦π¦6 β π¦π¦2 + 4 β π₯π₯2π¦π¦3 + 2π₯π₯π¦π¦ is usually arranged as below.
πππππποΏ½ππππππ
ππππππππππππππππππππ
+ππππππππππ β πππππππποΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππππ ππππππππππππ ππππππππππ
ππππππππππππππππ ππππππππ ππππππππππππππ ππππ ππ
+ππππππ β πππποΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππππ ππππππππππππ
ππππππππππ ππππππππππππππππππππππππ ππππππππππππππ ππππ ππ
+πποΏ½ππππππππ
ππππππππππππππππππππ
Writing Polynomials in Descending Order and Identifying Parts of a Polynomial
Suppose ππ = π₯π₯ β 6π₯π₯3 β π₯π₯6 + 4π₯π₯4 + 2 and ππ = 2π¦π¦ β 3π₯π₯π¦π¦π₯π₯ β 5π₯π₯2 + π₯π₯π¦π¦2. For each polynomial:
a. Write the polynomial in descending order. b. State the degree of the polynomial and the number of its terms. c. Identify the leading term, the leading coefficient, the coefficient of the linear term, the
coefficient of the quadratic term, and the free term of the polynomial. a. After arranging the terms in descending powers of π₯π₯, polynomial ππ becomes
βππππ + ππππππ β ππππππ + ππ + ππ, while polynomial ππ becomes
ππππππ β ππππππππ β ππππππ + ππππ.
Solution
ππππππππππππππ ππππππππππππππππππππππ
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Notice that the first two terms, π₯π₯π¦π¦2 and β3π₯π₯π¦π¦π₯π₯, are both of the same degree. So, to decide which one should be written first, we look at powers of π₯π₯. Since these powers are again the same, we look at powers of π¦π¦. This time, the power of π¦π¦ in π₯π₯π¦π¦2 is higher than the power of π¦π¦ in β3π₯π₯π¦π¦π₯π₯. So, the term π₯π₯π¦π¦2 should be written first.
b. The polynomial ππ has 5 terms. The highest power of π₯π₯ in ππ is 6, so the degree of the
polynomial ππ is 6. The polynomial ππ has 4 terms. The highest degree terms in ππ are π₯π₯π¦π¦2 and β3π₯π₯π¦π¦π₯π₯,
both third degree. So the degree of the polynomial ππ is 3. c. The leading term of the polynomial ππ = βπ₯π₯6 + 4π₯π₯4 β 6π₯π₯3 + π₯π₯ β 2 is βπ₯π₯6, so the
leading coefficient equals βππ. The linear term of ππ is π₯π₯, so the coefficient of the linear term equals ππ. ππ doesnβt have any quadratic term so the coefficient of the quadratic term equals 0. The free term of ππ equals βππ.
The leading term of the polynomial ππ = π₯π₯π¦π¦2 β 3π₯π₯π¦π¦π₯π₯ β 5π₯π₯2 + 2π¦π¦ is π₯π₯π¦π¦2, so the leading coefficient is equal to ππ.
The linear term of ππ is 2π¦π¦, so the coefficient of the linear term equals ππ. The quadratic term of ππ is β5π₯π₯2, so the coefficient of the quadratic term equals βππ. The polynomial ππ does not have a free term, so the free term equals 0.
Classifying Polynomials
Describe each polynomial as a constant, linear, quadratic, or ππ-th degree polynomial. Decide whether it is a monomial, binomial, or trinomial, if applicable.
a. π₯π₯2 β 9 b. β 3π₯π₯7π¦π¦ c. π₯π₯2 + 2π₯π₯ β 15 d. ππ e. 4π₯π₯5 β π₯π₯3 + π₯π₯ β 7 f. π₯π₯4 + 1 a. π₯π₯2 β 9 is a second degree polynomial with two terms, so it is a quadratic binomial.
b. β 3π₯π₯7π¦π¦ is an 8-th degree monomial.
c. π₯π₯2 + 2π₯π₯ β 15 is a second degree polynomial with three terms, so it is a quadratic trinomial.
d. ππ is a 0-degree term, so it is a constant monomial.
e. 4π₯π₯5 β π₯π₯3 + π₯π₯ β 7 is a 5-th degree polynomial.
f. π₯π₯4 + 1 is a 4-th degree binomial.
Polynomials as Functions and Evaluation of Polynomials
Each term of a polynomial in one variable is a product of a number and a power of the variable. The polynomial itself is either one term or a sum of several terms. Since taking a power of a given value, multiplying, and adding given values produce unique answers,
Solution
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polynomials are also functions. While ππ, ππ, or β are the most commonly used letters to represent functions, other letters can also be used. To represent polynomial functions, we customarily use capital letters, such as ππ, ππ, π π , etc.
Any polynomial function ππ of degree ππ, has the form
π·π·(ππ) = ππππππππ + ππππβππππππβππ + β―+ ππππππππ + ππππππ + ππππ
where ππππ,ππππβ1, β¦ ,ππ2,ππ1,ππ0 β β, ππππ β 0, and ππ β ππ.
Since polynomials are functions, they can be evaluated for different π₯π₯-values.
Evaluating Polynomials
Given ππ(π₯π₯) = 3π₯π₯3 β π₯π₯2 + 4, evaluate the following expressions:
a. ππ(0) b. ππ(β1) c. 2 β ππ(1) d. ππ(ππ)
a. ππ(0) = 3 β 03 β 02 + 4 = ππ
b. ππ(β1) = 3 β (β1)3 β (β1)2 + 4 = 3 β (β1) β 1 + 4 = β3 β 1 + 4 = ππ
When evaluating at negative π₯π₯-values, it is essential to use brackets in place of the variable before substituting the desired value. c. 2 β ππ(1) = 2 β (3 β 13 β 12 + 4)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
π‘π‘βππππ ππππ ππ(1)= 2 β (3 β 1 + 4) = 2 β 6 = ππππ
d. To find the value of ππ(ππ), we replace the variable π₯π₯ in ππ(π₯π₯) with ππ. So, this time the final answer,
ππ(ππ) = ππππππ β ππππ + ππ,
is an expression in terms of ππ rather than a specific number.
Since polynomials can be evaluated at any real π₯π₯-value, then the domain (see Section G3, Definition 5.1) of any polynomial is the set β of all real numbers.
Addition and Subtraction of Polynomials
Recall that terms with the same variable part are referred to as like terms (see Section R3, Definition 3.1). Like terms can be combined by adding their coefficients. For example,
2π₯π₯2π¦π¦ β 5π₯π₯2π¦π¦ = (2 β 5)π₯π₯2π¦π¦οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππ πππππππ‘π‘πππππππππ‘π‘ππππππ πππππππππππππ‘π‘ππ
(πππππππ‘π‘ππππππππππ)
= β3π₯π₯2π¦π¦
Unlike terms, such as 2π₯π₯2 and 3π₯π₯, cannot be combined.
In practice, this step is not necessary to
write.
Solution
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Simplifying Polynomial Expressions
Simplify each polynomial expression.
a. 5π₯π₯ β 4π₯π₯2 + 2π₯π₯ + 7π₯π₯2 b. 8ππ β (2 β 3ππ) + (3ππ β 6)
a. To simplify 5π₯π₯ β 4π₯π₯2 + 2π₯π₯ + 7π₯π₯2, we combine like terms, starting from the highest degree terms. It is suggested to underline the groups of like terms, using different type of underlining for each group, so that it is easier to see all the like terms and not to miss any of them. So,
5π₯π₯ β4π₯π₯2 +2π₯π₯ +7π₯π₯2 = ππππππ + ππππ
b. To simplify 8ππ β (2 β 3ππ) + (3ππ β 6), first we remove the brackets using the distributive property of multiplication and then we combine like terms. So, we have
8ππ β (2 β 3ππ) + (3ππ β 6)
= 8ππ β 2 +3ππ +3ππ β 6
= ππππππ β ππ
Adding or Subtracting Polynomials
Perform the indicated operations.
a. (6ππ5 β 4ππ3 + 3ππ β 1) + (2ππ4 + ππ2 β 5ππ + 9) b. (4π¦π¦3 β 3π¦π¦2 + π¦π¦ + 6) β (π¦π¦3 + 3π¦π¦ β 2) c. [9ππ β (3ππ β 2)] β [4ππ β (3 β 7ππ) + ππ]
a. To add polynomials, combine their like terms. So,
(6ππ5 β 4ππ3 + 3ππ β 1) + (2ππ4 + ππ2 β 5ππ + 9)
= 6ππ5β4ππ3 +3ππ β 1 + 2ππ4+ππ3 β5ππ + 9
= ππππππ + ππππππ β ππππππ β ππππ + ππ
b. To subtract a polynomial, add its opposite. In practice, remove any bracket preceeded by a negative sign by reversing the signs of all the terms of the polynomial inside the bracket. So,
(4π¦π¦3 β 3π¦π¦2 + π¦π¦ + 6) β ( π¦π¦3 + 3π¦π¦ β 2)
= 4π¦π¦3 β 3π¦π¦2+π¦π¦ + 6 βπ¦π¦3 β3π¦π¦ + 2
= ππππππ β ππππππ β ππππ + ππ
Solution
Solution
remove any bracket preceeded by a β+β
sign
To remove a bracket
preceeded by a βββ sign, reverse each sign inside
the bracket.
Remember that the
sign in front of a term belongs to this term.
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c. First, perform the operations within the square brackets and then subtract the resulting polynomials. So,
[9ππ β ( 3ππ β 2)] β [4ππ β ( 3 β 7ππ) + ππ]
= [9ππ β 3ππ + 2] β [4ππ β 3 + 7ππ + ππ]
= [6ππ + 2] β [ 12ππ β 3]
= 6ππ + 2 β 12ππ + 3
= βππππ+ ππ
Addition and Subtraction of Polynomial Functions
Similarly as for polynomials, addition and subtraction can also be defined for general functions.
Definition 1.3 Suppose ππ and ππ are functions of π₯π₯ with the corresponding domains π·π·ππ and π·π·ππ.
Then the sum function ππ + ππ is defined as
(ππ + ππ)(ππ) = ππ(ππ) + ππ(ππ)
and the difference function ππ β ππ is defined as
(ππ β ππ)(ππ) = ππ(ππ) β ππ(ππ).
The domain of the sum or difference function is the intersection π«π«ππ β© π«π«ππof the domains of the two functions.
A frequently used application of a sum or difference of polynomial functions comes from the business area. The fact that profit ππ equals revenue π π minus cost πΆπΆ can be recorded using function notation as
ππ(π₯π₯) = (π π β πΆπΆ)(π₯π₯) = π π (π₯π₯) β πΆπΆ(π₯π₯),
where π₯π₯ is the number of items produced and sold. Then, if π π (π₯π₯) = 6.5π₯π₯ and πΆπΆ(π₯π₯) =3.5π₯π₯ + 900, the profit function becomes
π·π·(ππ) = π π (π₯π₯) β πΆπΆ(π₯π₯) = 6.5π₯π₯ β (3.5π₯π₯ + 900) = 6.5π₯π₯ β 3.5π₯π₯ β 900 = ππππ β ππππππ.
Adding or Subtracting Polynomial Functions
Suppose ππ(π₯π₯) = π₯π₯2 β 6π₯π₯ + 4 and ππ(π₯π₯) = 2π₯π₯2 β 1. Find the following:
a. (ππ + ππ)(π₯π₯) and (ππ + ππ)(2) b. (ππ β ππ)(π₯π₯) and (ππ β ππ)(β1) c. (ππ + ππ)(ππ) d. (ππ β ππ)(2ππ)
a. Using the definition of the sum of functions, we have
collect like terms
before removing the next set of brackets
Solution
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(π·π· + πΈπΈ)(ππ) = ππ(π₯π₯) + ππ(π₯π₯) = π₯π₯2 β 6π₯π₯ + 4οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ(π₯π₯)
+ 2π₯π₯2 β 1οΏ½οΏ½οΏ½οΏ½οΏ½ππ(π₯π₯)
= ππππππ β ππππ + ππ
Therefore, (π·π· + πΈπΈ)(ππ) = 3 β 22 β 6 β 2 + 3 = 12 β 12 + 3 = ππ.
Alternatively, (ππ + ππ)(2) can be calculated without refering to the function (ππ + ππ)(π₯π₯), as shown below.
(π·π· +πΈπΈ)(ππ) = ππ(2) + ππ(2) = 22 β 6 β 2 + 4οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ(2)
+ 2 β 22 β 1οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ(2)
= 4 β 12 + 4 + 8 β 1 = ππ.
b. Using the definition of the difference of functions, we have
(π·π· βπΈπΈ)(ππ) = ππ(π₯π₯) β ππ(π₯π₯) = π₯π₯2 β 6π₯π₯ + 4οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ(π₯π₯)
β (2π₯π₯2 β 1)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ(π₯π₯)
= π₯π₯2 β 6π₯π₯ + 4 β 2π₯π₯2 + 1 = βππππ β ππππ + ππ
To evaluate (ππ β ππ)(β1), we will take advantage of the difference function calculated above. So, we have
(π·π·β πΈπΈ)(βππ) = β(β1)2 β 6(β1) + 5 = β1 + 6 + 5 = ππππ.
c. By Definition 1.3,
(π·π· + πΈπΈ)(ππ) = ππ(ππ) + ππ(ππ) = ππ2 β 6ππ + 4 + 2ππ2 β 1 = ππππππ β ππππ + ππ
Alternatively, we could use the sum function already calculated in the solution to Example 6a. Then, the result is instant: (π·π· + πΈπΈ)(ππ) = ππππππ β ππππ + ππ.
d. To find (ππ β ππ)(2ππ), we will use the difference function calculated in the solution to Example 6b. So, we have
(π·π·β πΈπΈ)(ππππ) = β(2ππ)2 β 6(2ππ) + 5 = βππππππ β ππππππ + ππ.
P.1 Exercises
Vocabulary Check Complete each blank with the most appropriate term or phrase from the given list:
monomials, polynomial, binomial, trinomial, leading, constant, like, degree.
1. Terms that are products of only numbers and variables are called _________________.
2. A _______________ is a sum of monomials.
3. A polynomial with two terms is called a ______________.
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4. A polynomial with three terms is called a _____________.
5. The _____________ coefficient is the coefficient of the highest degree term.
6. A free term, also called a ___________ term, is the term of zero degree.
7. Only __________ terms can be combined.
8. The __________ of a monomial is the sum of exponents of all its variables.
Concept Check Determine whether the expression is a monomial.
9. βπππ₯π₯3π¦π¦2 10. 5π₯π₯β4 11. 5βπ₯π₯ 12. β2π₯π₯4 Concept Check Identify the degree and coefficient.
13. π₯π₯π¦π¦3 14. βπ₯π₯2π¦π¦ 15. β2π₯π₯π¦π¦ 16. β3πππ₯π₯2π¦π¦5
Concept Check Arrange each polynomial in descending order of powers of the variable. Then, identify the degree and the leading coefficient of the polynomial.
17. 5 β π₯π₯ + 3π₯π₯2 β 25π₯π₯3 18. 7π₯π₯ + 4π₯π₯4 β 4
3π₯π₯3 19. 8π₯π₯4 + 2π₯π₯3 β 3π₯π₯ + π₯π₯5
20. 4π¦π¦3 β 8π¦π¦5 + π¦π¦7 21. ππ2 + 3ππ4 β 2ππ + 1 22. 3ππ2 βππ4 + 2ππ3
Concept Check State the degree of each polynomial and identify it as a monomial, binomial, trinomial, or n-th degree polynomial if ππ > 2.
23. 7ππ β 5 24. 4π₯π₯2 β 11π₯π₯ + 2 25. 25
26. β6ππ4ππ + 3ππ3ππ2 β 2ππππ3 β ππ4 27. βππππ6 28. 16ππ2 β 9ππ2 Concept Check Let ππ(π₯π₯) = β2π₯π₯2 + π₯π₯ β 5 and ππ(π₯π₯) = 2π₯π₯ β 3. Evaluate each expression.
29. ππ(β1) 30. ππ(0) 31. 2ππ(1) 32. ππ(ππ)
33. ππ(β1) 34. ππ(5) 35. ππ(ππ) 36. ππ(3ππ)
37. 3ππ(β2) 38. 3ππ(ππ) 39. 3ππ(ππ) 40. ππ(ππ + 1) Concept Check Simplify each polynomial expression.
41. 5π₯π₯ + 4π¦π¦ β 6π₯π₯ + 9π¦π¦ 42. 4π₯π₯2 + 2π₯π₯ β 6π₯π₯2 β 6
43. 6π₯π₯π¦π¦ + 4π₯π₯ β 2π₯π₯π¦π¦ β π₯π₯ 44. 3π₯π₯2π¦π¦ + 5π₯π₯π¦π¦2 β 3π₯π₯2π¦π¦ β π₯π₯π¦π¦2
45. 9ππ3 + ππ2 β 3ππ3 + ππ β 4ππ2 + 2 46. ππ4 β 2ππ3 + ππ2 β 3ππ4 + ππ3
47. 4 β (2 + 3ππ) + 6ππ + 9 48. 2ππ β (5a β 3) β (7ππ β 2)
49. 6 + 3π₯π₯ β (2π₯π₯ + 1) β (2π₯π₯ + 9) 50. 4π¦π¦ β 8 β (β3 + π¦π¦) β (11π¦π¦ + 5)
Perform the indicated operations.
51. (π₯π₯2 β 5π¦π¦2 β 9π₯π₯2) + (β6π₯π₯2 + 9π¦π¦2 β 2π₯π₯2) 52. (7π₯π₯2π¦π¦ β 3π₯π₯π¦π¦2 + 4π₯π₯π¦π¦) + (β2π₯π₯2π¦π¦ β π₯π₯π¦π¦2 + π₯π₯π¦π¦)
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53. (β3π₯π₯2 + 2π₯π₯ β 9) β (π₯π₯2 + 5π₯π₯ β 4) 54. (8π¦π¦2 β 4π¦π¦3 β 3π¦π¦) β (3π¦π¦2 β 9π¦π¦ β 7π¦π¦3)
55. (3ππ6 + 5) + (β7ππ2 + 2ππ6 β ππ5) 56. (5π₯π₯2ππ β 3π₯π₯ππ + 2) + (βπ₯π₯2ππ + 2π₯π₯ππ β 6)
57. (β5ππ4 + 8ππ2 β 9) β (6ππ3 β ππ2 + 2) 58. (3π₯π₯3ππ β π₯π₯ππ + 7) β (β2π₯π₯3ππ + 5π₯π₯2ππ β 1)
59. (10π₯π₯π¦π¦ β 4π₯π₯2π¦π¦2 β 3π¦π¦3)β (β9π₯π₯2π¦π¦2 + 4π¦π¦3 β 7π₯π₯π¦π¦)
60. Subtract (β4π₯π₯ + 2π₯π₯2 + 3ππ) from the sum of (2π₯π₯2 β 3π₯π₯ + ππ) and (π₯π₯2 β 2ππ).
61. Subtract the sum of (2π₯π₯2 β 3π₯π₯ + ππ) and (π₯π₯2 β 2ππ) from (β4π₯π₯ + 2π₯π₯2 + 3ππ).
62. [2ππ β (3ππ β 6)] β οΏ½οΏ½5ππ β (8 β 9ππ)οΏ½ + 4πποΏ½
63. β[3π₯π₯2 + 5π₯π₯ β (2π₯π₯2 β 6π₯π₯)] + οΏ½οΏ½8π₯π₯2 β (5π₯π₯ β π₯π₯2)οΏ½ + 2π₯π₯2οΏ½
64. 5ππ β (5ππ β [2ππ β (4ππ β 8ππ)]) + 11ππ β (9ππ β 12ππ)
Discussion Point
65. If ππ(π₯π₯) and ππ(π₯π₯) are polynomials of degree 3, is it possible for the sum of the two polynomials to have degree 2? If so, give an example. If not, explain why.
For each pair of functions, find a) (ππ + ππ)(π₯π₯) and b) (ππ β ππ)(π₯π₯).
66. ππ(π₯π₯) = 5π₯π₯ β 6, ππ(π₯π₯) = β2 + 3π₯π₯ 67. ππ(π₯π₯) = π₯π₯2 + 7π₯π₯ β 2, ππ(π₯π₯) = 6π₯π₯ + 5
68. ππ(π₯π₯) = 3π₯π₯2 β 5π₯π₯, ππ(π₯π₯) = β5π₯π₯2 + 2π₯π₯ + 1 69. ππ(π₯π₯) = 2π₯π₯ππ β 3π₯π₯ β 1, ππ(π₯π₯) = 5π₯π₯ππ + π₯π₯ β 6
70. ππ(π₯π₯) = 2π₯π₯2ππ β 3π₯π₯ππ + 3, ππ(π₯π₯) = β8π₯π₯2ππ + π₯π₯ππ β 4 Let π·π·(ππ) = ππππ β ππ, πΈπΈ(ππ) = ππππ + ππ, and πΉπΉ(ππ) = ππ β ππ. Find each of the following.
71. (ππ + π π )(β1) 72. (ππ β ππ)(β2) 73. (ππ β π π )(3) 74. (π π β ππ)(0)
75. (π π β ππ)(ππ) 76. (ππ + ππ)(ππ) 77. (ππ β π π )(ππ + 1) 78. (ππ + π π )(2ππ) Analytic Skills Solve each problem.
79. From 1985 through 1995, the gross farm income G and farm expenses E (in billions of dollars) in the United States can be modeled by
πΊπΊ(ππ) = β0.246ππ2 + 7.88ππ + 159 and πΈπΈ(ππ) = 0.174ππ2 + 2.54ππ + 131
where t is the number of years since 1985. Write a model for the net farm income ππ(ππ) for these years.
80. The deflection π·π· (in inches) of a beam that is uniformly loaded is given by the polynomial function π·π·(π₯π₯) = 0.005π₯π₯4 β 0.1π₯π₯3 + 0.5π₯π₯2, where π₯π₯ is the distance from one end of the beam. The maximum deflection occurs when π₯π₯ is equal to half of the length of the beam. Determine the maximum deflection for the 10 ft long beam, as shown in the accompanying figure.
10 ππππ π·π·
π₯π₯
Years since 1985
200
150
100
50
4 6 1
8
Farming
expenses
net income
2
Bill
ions
of d
olla
rs
gross income
190
81. Write a polynomial that gives the sum of the areas of a square with sides of length π₯π₯ and a circle with radius π₯π₯. Find the combined area when π₯π₯ = 10 centimeters. Round the answer to the nearest centimeter square.
82. Suppose the cost in dollars for a band to produce π₯π₯ compact discs is given by πΆπΆ(π₯π₯) = 4π₯π₯ + 2000. If the band sells CDs for $16 each, complete the following.
a. Write a function π π (π₯π₯) that gives the revenue for selling π₯π₯ compact discs. b. If profit equals revenue minus cost, write a formula ππ(π₯π₯) for the profit. c. Evaluate ππ(3000) and interpret the answer.
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ππ + ππ
ππ + ππ
P.2 Multiplication of Polynomials
As shown in the previous section, addition and subtraction of polynomials results in another polynomial. This means that the set of polynomials is closed under the operation of addition and subtraction. In this section, we will show that the set of polynomials is also closed under the operation of multiplication, meaning that a product of polynomials is also a polynomial.
Properties of Exponents
Since multiplication of polynomials involves multiplication of powers, let us review properties of exponents first. Recall:
ππππ = ππ β ππ β β¦ β πποΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ ππππππππππ
For example, π₯π₯4 = π₯π₯ β π₯π₯ β π₯π₯ β π₯π₯ and we read it βπ₯π₯ to the fourth powerβ or shorter βπ₯π₯ to the fourthβ. If ππ = 2, the power π₯π₯2 is customarily read βπ₯π₯ squaredβ. If ππ = 3, the power π₯π₯3 is often read βπ₯π₯ cubedβ.
Let ππ β β, and ππ, ππ β ππ. The table below shows basic exponential rules with some examples justifying each rule.
Power Rules for Exponents
General Rule Description Example
ππππ β ππππ = ππππ+ππ To multiply powers of the same bases, keep the base and add the exponents.
π₯π₯2 β π₯π₯3 = (π₯π₯ β π₯π₯) β (π₯π₯ β π₯π₯ β π₯π₯) = π₯π₯2+3 = π₯π₯5
ππππ
ππππ= ππππβππ
To divide powers of the same bases, keep the base and subtract the exponents.
π₯π₯5
π₯π₯2=
(π₯π₯ β π₯π₯ β π₯π₯ β π₯π₯ β π₯π₯)(π₯π₯ β π₯π₯)
= π₯π₯5β2 = π₯π₯3
(ππππ)ππ = ππππππ To raise a power to a power, multiply the exponents.
(π₯π₯2)3 = (π₯π₯ β π₯π₯)(π₯π₯ β π₯π₯)(π₯π₯ β π₯π₯) = π₯π₯2β3 = π₯π₯6
(ππππ)ππ = ππππππππ To raise a product to a power, raise each factor to that power.
(2π₯π₯)3 = 23π₯π₯3
οΏ½πππποΏ½ππ
=ππππ
ππππ
To raise a quotient to a power, raise the numerator and the denominator to that power.
οΏ½π₯π₯3οΏ½2
=π₯π₯2
32
ππππ = ππ for ππ β ππ ππππ is undefined
A nonzero number raised to the power of zero equals one.
π₯π₯0 = π₯π₯ππβππ =π₯π₯ππ
π₯π₯ππ= 1
ππππππππ
ππππππππππ πππ₯π₯ππππππππππππ
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Simplifying Exponential Expressions
Simplify. a. (β3π₯π₯π¦π¦2)4 b. (5ππ3ππ)(β2ππππ2)
c. οΏ½β2π₯π₯5
π₯π₯2πποΏ½3 d. π₯π₯2πππ₯π₯ππ
a. To simplify (β3π₯π₯π¦π¦2)4, we apply the fourth power to each factor in the bracket. So,
(β3π₯π₯π¦π¦2)4 = (β3)4οΏ½οΏ½οΏ½ππππππππ ππππππππππππππ ππ πππππππππ‘π‘ππππππππππ ππ πππππππππ‘π‘ππππππ
β π₯π₯4 β (π¦π¦2)4οΏ½οΏ½οΏ½πππππππ‘π‘πππππππππππ₯π₯πππππππππππ‘π‘ππ
= ππππππππππππ
b. To simplify (5ππ3ππ)(β2ππππ2), we multiply numbers, powers of ππ, and powers of ππ. So,
(5ππ3ππ)(β2ππππ2) = (β2) β 5 β ππ3 β πποΏ½οΏ½οΏ½ππππππ
πππ₯π₯πππππππππππ‘π‘ππ
β ππ β ππ2οΏ½οΏ½οΏ½ππππππ
πππ₯π₯πππππππππππ‘π‘ππ
= βππππππππππππ
c. To simplify οΏ½β2π₯π₯5
π₯π₯2πποΏ½3, first we reduce the common factors and then we raise every
factor of the numerator and denominator to the third power. So, we obtain
οΏ½β2π₯π₯5
π₯π₯2π¦π¦ οΏ½3
= οΏ½β2π₯π₯3
π¦π¦ οΏ½3
=(β3)3(π₯π₯3)3
π¦π¦3=βππππππππ
ππππ
d. When multiplying powers with the same bases, we add exponents, so π₯π₯2πππ₯π₯ππ = ππππππ
Multiplication of Polynomials
Multiplication of polynomials involves finding products of monomials. To multiply monomials, we use the commutative property of multiplication and the product rule of powers.
Multiplying Monomials
Find each product. a. (3π₯π₯4)(5π₯π₯3) b. (5ππ)(β2ππ2ππ3) c. β4π₯π₯2(3π₯π₯π¦π¦)(βπ₯π₯2π¦π¦) a. (3π₯π₯4)(5π₯π₯3) = 3 β π₯π₯4 β 5οΏ½οΏ½οΏ½
β π₯π₯3 = 3 β 5 β π₯π₯4 β π₯π₯3οΏ½οΏ½οΏ½
= ππππππππ
b. (5ππ)(β2ππ2ππ3) = 5(β2)ππ2ππππ3 = βππππππππππππ
c. β4π₯π₯2(3π₯π₯π¦π¦)(βπ₯π₯2π¦π¦) = (β4) β 3 β (β1)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
π₯π₯2π₯π₯π₯π₯2οΏ½οΏ½οΏ½
π¦π¦π¦π¦οΏ½
= ππππππππππππ
Solution
Solution
3
commutative property
product rule of powers
multiply coefficients
apply product rule of powers
To find the product of monomials, find
the following: β’ the final sign, β’ the number, β’ the power.
The intermediate steps are not necessary to write.
The final answer is immediate if we follow the
order: sign, number, power of each variable.
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To multiply polynomials by a monomial, we use the distributive property of multiplication.
Multiplying Polynomials by a Monomial
Find each product. a. β2π₯π₯(3π₯π₯2 β π₯π₯ + 7) b. (5ππ β ππππ3)(3ππππ2) a. To find the product β2π₯π₯(3π₯π₯2 β π₯π₯ + 7), we distribute the monomial β2π₯π₯ to each term
inside the bracket. So, we have
β2π₯π₯(3π₯π₯2 β π₯π₯ + 7) = β2π₯π₯(3π₯π₯2)β 2π₯π₯(βπ₯π₯)β 2π₯π₯(7)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π‘π‘βππππ πππ‘π‘ππππ ππππππ ππππ ππππππππ πππππππ‘π‘ππππππππ
= βππππππ + ππππππ β ππππππ
b. (5ππ β ππππ3)(3ππππ2) = 5ππ(3ππππ2)β ππππ3(3ππππ2)οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
π‘π‘βππππ πππ‘π‘ππππ ππππππ ππππ ππππππππ πππππππ‘π‘ππππππππ
= 15ππππ3 β 3ππ2ππ5 = βππππππππππ + ππππππππππ
When multiplying polynomials by polynomials we multiply each term of the first polynomial by each term of the second polynomial. This process can be illustrated with finding areas of a rectangle whose sides represent each polynomial. For example, we multiply (2π₯π₯ + 3)(π₯π₯2 β 3π₯π₯ + 1) as shown below
So, (2π₯π₯ + 3)(π₯π₯2 β 3π₯π₯ + 1) = 2π₯π₯3β6π₯π₯2 + 2π₯π₯ +3π₯π₯2 β 9π₯π₯ + 3
= ππππππβππππππ β ππππ + ππ
Multiplying Polynomials by Polynomials
Find each product. a. (3π¦π¦2 β 4π¦π¦ β 2)(5π¦π¦ β 7) b. 4ππ2(2ππ β 3)(3ππ2 + ππ β 1) a. To find the product (3π¦π¦2 β 4π¦π¦ β 2)(5π¦π¦ β 7), we can distribute the terms of the
second bracket over the first bracket and then collect the like terms. So, we have
(3π¦π¦2 β 4π¦π¦ β 2)(5π¦π¦ β 7) = 15π¦π¦3 β 20π¦π¦2 β 10π¦π¦
β 21π¦π¦2 + 28π¦π¦ + 14
= ππππππππ β ππππππππ + ππππππ + ππππ b. To find the product 4ππ2(2ππ β 3)(3ππ2 + ππ β 1), we will multiply the two brackets
first, and then multiply the resulting product by 4ππ2. So,
Solution
Solution
arranged in
decreasing order of powers
π₯π₯2 β 3π₯π₯ + 1
2π₯π₯ +3 β9π₯π₯ 3π₯π₯2 3
2π₯π₯3 2π₯π₯ β6π₯π₯2 line up like terms to
combine them
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ππ + ππ
ππ + ππ ππ2 ππππ
ππ2 ππππ
4ππ2(2ππ β 3)(3ππ2 + ππ β 1) = 4ππ2 οΏ½6ππ3 + 2ππ2 β2ππ β 9ππ2 β3ππ + 3οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½πππππππππππππ‘π‘ ππππππππ π‘π‘ππππππππ ππππππππππππ ππππππππππππππππ π‘π‘βππ πππππππππππππ‘π‘
= 4ππ2(6ππ3 β 7ππ2 β 5ππ + 3) = ππππππππ β ππππππππ β ππππππππ + ππππππππ
In multiplication of binomials, it might be convenient to keep track of the multiplying terms by following the FOIL mnemonic, which stands for multiplying the First, Outer, Inner, and Last terms of the binomials. Here is how it works:
(2π₯π₯ β 3)(π₯π₯ + 5) = 2π₯π₯2 +10π₯π₯ β 3π₯π₯οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½π‘π‘βππ ππππππ ππππ π‘π‘βππ πΆπΆππππππππ ππππππ π°π°ππππππππ π‘π‘ππππππππ ππππππππππππππ π‘π‘βππ ππππππππππππ π‘π‘ππππππ
β 15 = ππππππ + ππππ β ππππ
Using the FOIL Method in Binomial Multiplication
Find each product. a. (π₯π₯ + 3)(π₯π₯ β 4) b. (5π₯π₯ β 6)(2π₯π₯ + 3) a. To find the product (π₯π₯ + 3)(π₯π₯ β 4), we may follow the FOIL method
(π₯π₯ + 3)(π₯π₯ β 4) = ππππ β ππ β ππππ
b. Observe that the linear term of the product (5π₯π₯ β 6)(2π₯π₯ + 3) is equal to the sum of β12π₯π₯ and 15π₯π₯, which is 3π₯π₯. So, we have
(5π₯π₯ β 6)(2π₯π₯ + 3) = ππππππππ + ππππ β ππππ
Special Products
Suppose we want to find the product (ππ + ππ)(ππ + ππ). This can be done via the FOIL method
(ππ + ππ)(ππ + ππ) = ππ2 + ππππ + ππππ + ππ2 = ππππ + ππππππ + ππππ,
or via the geometric visualization:
Solution
First Last
Outer Inner
π₯π₯2 β12
β4π₯π₯ 3π₯π₯
To find the linear
(middle) term try to add the inner and outer products mentally.
195
Observe that since the products of the inner and outer terms are both equal to ππππ, we can use a shortcut and write the middle term of the final product as 2ππππ. We encorage the reader to come up with similar observations regarding the product (ππ β ππ)(ππ β ππ). This regularity in multiplication of a binomial by itself leads us to the perfect square formula:
(ππ Β± ππ)ππ = ππππ Β± ππππππ + ππππ
In the above notation, the " Β± " sign is used to record two formulas at once, the perferct square of a sum and the perfect square of a difference. It tells us to either use a " + " in both places, or a " β " in both places. The ππ and ππ can actually represent any expression. For example, to expand (2π₯π₯ β π¦π¦2)2, we can apply the perfect square formula by treating 2π₯π₯ as ππ and π¦π¦2 as ππ. Here is the calculation.
(2π₯π₯ β π¦π¦2)2 = (2π₯π₯)2 β 2(2π₯π₯)π¦π¦2 + (π¦π¦2)2 = ππππππ β ππππππππ + ππππ
Another interesting pattern can be observed when multiplying two conjugate brackets, such as (ππ + ππ) and (ππ β ππ). Using the FOIL method,
(ππ + ππ)(ππ β ππ) = ππ2 + ππππ β ππππ + ππ2 = ππππ β ππππ,
we observe, that the products of the inner and outer terms are opposite. So, they add to zero and the product of conjugate brackets becomes the difference of squares of the two terms. This regularity in multiplication of conjugate brackets leads us to the difference of squares formula.
(ππ + ππ)(ππ β ππ) = ππππ β ππππ
Again, ππ and ππ can represent any expression. For example, to find the product (3π₯π₯ + 0.1π¦π¦2)(3π₯π₯ β 0.1π¦π¦2), we can apply the difference of squares formula by treating 3π₯π₯ as ππ and 0.1π¦π¦2 as ππ. Here is the calculation.
οΏ½ππππ + ππ.πππππποΏ½οΏ½ππππ β ππ.πππππποΏ½ = (3π₯π₯)2 β (0.1π¦π¦3)2 = ππππππ β ππ.ππππππππ
We encourage to use the above formulas whenever applicable, as it allows for more efficient calculations and helps to observe patterns useful in future factoring.
Useing Special Product Formulas in Polynomial Multiplication
Find each product. Apply special products formulas, if applicable. a. (5π₯π₯ + 3π¦π¦)2 b. (π₯π₯ + π¦π¦ β 5)(π₯π₯ + π¦π¦ + 5) a. Applying the perfect square formula, we have
(5π₯π₯ + 3π¦π¦)2 = (5π₯π₯)2 + 2(5π₯π₯)3π¦π¦ + (3π¦π¦)2 = ππππππππ + ππππππππ + ππππππ b. The product (π₯π₯ + π¦π¦ β 5)(π₯π₯ + π¦π¦ + 5) can be found by multiplying each term of the first polynomial by each term of the second polynomial, using the distributive property. However, we can find the product (π₯π₯ + π¦π¦ β 5)(π₯π₯ + π¦π¦ + 5) in a more efficient way by
Solution
Conjugate binomials have the same first terms and opposite
second terms.
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applying the difference of squares formula. Treating the expression π₯π₯ + π¦π¦ as the first term ππ and the 5 as the second term ππ in the formula (ππ + ππ)(ππ β ππ) = ππ2 β ππ2, we obtain
(π₯π₯ + π¦π¦ β 5)(π₯π₯ + π¦π¦ + 5) = (π₯π₯ + π¦π¦)2 β 52
= ππππ + ππππππ + πππποΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βππππππ ππππ ππππππππππ
π‘π‘βππ πππππππππππππ‘π‘ πππ π ππππππππ ππππππππππππππ
β ππππ
Caution: The perfect square formula shows that (ππ + ππ)ππ β ππππ + ππππ. The difference of squares formula shows that (ππ β ππ)ππ β ππππ β ππππ. More generally, (ππ Β± ππ)ππ β ππππ Β± ππππ for any natural ππ β 1.
Product Functions The operation of multiplication can be defined not only for polynomials but also for general functions.
Definition 2.1 Suppose ππ and ππ are functions of π₯π₯ with the corresponding domains π·π·ππ and π·π·ππ.
Then the product function, denoted ππ β ππ or ππππ, is defined as
(ππ β ππ)(ππ) = ππ(ππ) β ππ(ππ).
The domain of the product function is the intersection π«π«ππ β© π«π«ππof the domains of the two functions.
Multiplying Polynomial Functions
Suppose ππ(π₯π₯) = π₯π₯2 β 4π₯π₯ and ππ(π₯π₯) = 3π₯π₯ + 2. Find the following:
a. (ππππ)(π₯π₯), (ππππ)(β2), and ππ(β2)ππ(β2) b. (ππππ)(π₯π₯) and (ππππ)(1) c. 2(ππππ)(ππ) a. Using the definition of the product function, we have
(ππππ)(π₯π₯) = ππ(π₯π₯) β ππ(π₯π₯) = (π₯π₯2 β 4π₯π₯)(3π₯π₯ + 2) = 3π₯π₯3 + 2π₯π₯2 β 12π₯π₯2 β 8π₯π₯
= ππππππ β ππππππππ β ππππ
To find (ππππ)(β2), we substitute π₯π₯ = β2 to the above polynomial function. So,
(ππππ)(β2) = 3(β2)3 β 10(β2)2 β 8(β2) = 3 β (β8) β 10 β 4 + 16
= β24 β 40 + 16 = βππππ
To find ππ(β2)ππ(β2), we calculate ππ(β2)ππ(β2) = οΏ½(β2)2 β 4(β2)οΏ½(3(β2) + 2) = (4 + 8)(β6 + 2) = 12 β (β4)
= βππππ
Solution
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Observe that both expressions result in the same value. This was to expect, as by the definition, (ππππ)(β2) = ππ(β2) β ππ(β2).
b. Using the definition of the product function as well as the perfect square formula, we
have (ππππ)(π₯π₯) = ππ(π₯π₯) β ππ(π₯π₯) = [ππ(π₯π₯)]2 = (3π₯π₯ + 2)2 = ππππππ + ππππππ + ππ
Therefore, (ππππ)(1) = 9 β 12 + 12 β 1 + 4 = 9 + 12 + 4 = ππππ. c. Since (ππππ)(π₯π₯) = 3π₯π₯3 β 10π₯π₯2 β 8π₯π₯, as shown in the solution to Example 7a, then
(ππππ)(ππ) = 3ππ3 β 10ππ2 β 8ππ. Therefore,
2(ππππ)(ππ) = 2[3ππ3 β 10ππ2 β 8ππ] = ππππππ β ππππππππ β ππππππ
P.2 Exercises
Vocabulary Check Complete each blank with the most appropriate term from the given list: add, binomials,
conjugate, divide, intersection, multiply, perfect.
1. To multiply powers of the same bases, keep the base and _________ the exponents.
2. To __________ powers of the same bases, keep the base and subtract the exponents.
3. To calculate a power of a power, ______________ exponents.
4. The domain of a product function ππππ is the ________________ of the domains of each of the functions, ππ and ππ.
5. The FOIL rule applies only to the multiplication of ______________.
6. A difference of squares is a product of two ______________ binomials, such as (π₯π₯ + π¦π¦) and (π₯π₯ β π¦π¦).
7. A _____________ square is a product of two identical binomials.
Concept Check
8. Decide whether each expression has been simplified correctly. If not, correct it.
a. π₯π₯2 β π₯π₯4 = π₯π₯8 b. β2π₯π₯2 = 4π₯π₯2 c. (5π₯π₯)3 = 53π₯π₯3
d. βοΏ½π₯π₯5οΏ½2
= βπ₯π₯2
25 e. (ππ2)3 = ππ5 f. 45 β 42 = 167
g. 65
32= 23 h. π₯π₯π¦π¦0 = 1 i. (βπ₯π₯2π¦π¦)3 = βπ₯π₯6π¦π¦3
Simplify each expression.
9. 3π₯π₯2 β 5π₯π₯3 10. β2π¦π¦3 β 4π¦π¦5 11. 3π₯π₯3(β5π₯π₯4)
12. 2π₯π₯2π¦π¦5(7π₯π₯π¦π¦3) 13. (6ππ4ππ)(β3ππ3ππ5) 14. (β3π₯π₯2π¦π¦)3
198
15. 12π₯π₯3ππ4π₯π₯ππ2
16. 15π₯π₯5ππ2
β3π₯π₯2ππ4 17. (β2π₯π₯5π¦π¦3)2
18. οΏ½4ππ2
πποΏ½3 19. οΏ½β3ππ
4
ππ3οΏ½2 20. οΏ½β5ππ
2π π πππ π 4
οΏ½3
21. 3ππ2(β5ππ5)(β2ππ)0 22. β3ππ3ππ(β4ππ2ππ4)(ππππ)0 23. (β2ππ)2πππ π 3
6ππ2π π 4
24. (β8π₯π₯ππ)2ππ3
4π₯π₯5ππ4 25. οΏ½β3π₯π₯
4ππ6
18π₯π₯6ππ7οΏ½3 26. ((β2π₯π₯3π¦π¦)2)3
27. ((βππ2ππ4)3)5 28. π₯π₯πππ₯π₯ππβ1 29. 3ππ2ππππ1βππ
30. (5ππ)2ππ 31. (β73π₯π₯)4ππ 32. β12π₯π₯ππ+1
6π₯π₯ππβ1
33. 25π₯π₯ππ+ππ
β5π₯π₯ππβππ 34. οΏ½π₯π₯ππ+πποΏ½ππβππ 35. (π₯π₯2π¦π¦)ππ
Concept Check Find each product.
36. 8π₯π₯2π¦π¦3(β2π₯π₯5π¦π¦) 37. 5ππ3ππ5(β3ππ2ππ4) 38. 2π₯π₯(β3π₯π₯ + 5)
39. 4π¦π¦(1 β 6π¦π¦) 40. β3π₯π₯4π¦π¦(4π₯π₯ β 3π¦π¦) 41. β6ππ3ππ(2ππ + 5ππ)
42. 5ππ2(3ππ2 β 2ππ + 4) 43. 6ππ3(2ππ2 + 5ππ β 3) 44. (π₯π₯ + 6)(π₯π₯ β 5)
45. (π₯π₯ β 7)(π₯π₯ + 3) 46. (2π₯π₯ + 3)(3π₯π₯ β 2) 47. 3ππ(5ππ + 1)(3ππ + 2)
48. 2π’π’2(π’π’ β 3)(3π’π’ + 5) 49. (2ππ + 3)(ππ2 β 4ππ β 2) 50. (2π₯π₯ β 3)(3π₯π₯2 + π₯π₯ β 5)
51. (ππ2 β 2ππ2)(ππ2 β 3ππ2) 52. (2ππ2 β ππ2)(3ππ2 β 5ππ2) 53. (π₯π₯ + 5)(π₯π₯ β 5)
54. (ππ + 2ππ)(ππ β 2ππ) 55. (π₯π₯ + 4)(π₯π₯ + 4) 56. (ππ β 2ππ)(ππ β 2ππ)
57. (π₯π₯ β 4)(π₯π₯2 + 4π₯π₯ + 16) 58. (π¦π¦ + 3)(π¦π¦2 β 3π¦π¦ + 9)
59. (π₯π₯2 + π₯π₯ β 2)(π₯π₯2 β 2π₯π₯ + 3) 60. (2π₯π₯2 + π¦π¦2 β 2π₯π₯π¦π¦)(π₯π₯2 β 2π¦π¦2 β π₯π₯π¦π¦) Concept Check True or False? If it is false, show a counterexample by choosing values for a and b that would
not satisfy the equation.
61. (ππ + ππ)2 = ππ2 + ππ2 62. ππ2 β ππ2 = (ππ β ππ)(ππ + ππ) 63. (ππ β ππ)2 = ππ2 + ππ2
64. (ππ + ππ)2 = ππ2 + 2ππππ + ππ2 65. (ππ β ππ)2 = ππ2 + ππππ + ππ2 66. (ππ β ππ)3 = ππ3 β ππ3 Find each product. Use the difference of squares or the perfect square formula, if applicable.
67. (2ππ + 3)(2ππ β 3) 68. (5π₯π₯ β 4)(5π₯π₯ + 4) 69. οΏ½ππ β 13οΏ½ οΏ½ππ + 1
3οΏ½
70. οΏ½12π₯π₯ β 3π¦π¦οΏ½ οΏ½1
2π₯π₯ + 3π¦π¦οΏ½ 71. (2π₯π₯π¦π¦ + 5π¦π¦3)(2π₯π₯π¦π¦ β 5π¦π¦3) 72. (π₯π₯2 + 7π¦π¦3)(π₯π₯2 β 7π¦π¦3)
73. (1.1π₯π₯ + 0.5π¦π¦)(1.1π₯π₯ β 0.5π¦π¦) 74. (0.8ππ + 0.2ππ)(0.8ππ + 0.2ππ) 75. (π₯π₯ + 6)2
76. (π₯π₯ β 3)2 77. (4π₯π₯ + 3π¦π¦)2 78. (5π₯π₯ β 6π¦π¦)2
199
79. οΏ½3ππ + 12οΏ½2 80. οΏ½2ππ β 1
3οΏ½2 81. (ππ3ππ2 β 1)2
82. (π₯π₯4π¦π¦2 + 3)2 83. (3ππ2 + 4ππ3)2 84. (2π₯π₯2 β 3π¦π¦3)2
85. 3π¦π¦(5π₯π₯π¦π¦3 + 2)(5π₯π₯π¦π¦3 β 2) 86. 2ππ(2ππ2 + 5ππππ)(2ππ2 + 5ππππ) 87. 3π₯π₯(π₯π₯2π¦π¦ β π₯π₯π¦π¦3)2
88. (βπ₯π₯π¦π¦ + π₯π₯2)(π₯π₯π¦π¦ + π₯π₯2) 89. (4ππ2 + 3ππππ)(β3ππππ + 4ππ2) 90. (π₯π₯ + 1)(π₯π₯ β 1)(π₯π₯2 + 1)
91. (2π₯π₯ β π¦π¦)(2π₯π₯ + π¦π¦)(4π₯π₯2 + π¦π¦2) 92. (ππ β ππ)(ππ + ππ)(ππ2 β ππ2) 93. (ππ + ππ + 1)(ππ + ππ β 1)
94. (2π₯π₯ + 3π¦π¦ β 5)(2π₯π₯ + 3π¦π¦ + 5) 95. (3ππ + 2ππ)(3ππβ 2ππ)(9ππ2 β 4ππ2)
96. οΏ½(2ππ β 3) + βοΏ½2 97. οΏ½(4π₯π₯ + π¦π¦) β 5οΏ½2
98. οΏ½π₯π₯ππ + π¦π¦πποΏ½οΏ½π₯π₯ππ β π¦π¦πποΏ½οΏ½π₯π₯2ππ + π¦π¦2πποΏ½ 99. οΏ½π₯π₯ππ + π¦π¦πποΏ½οΏ½π₯π₯ππ β π¦π¦πποΏ½οΏ½π₯π₯2ππ β π¦π¦2πποΏ½
Concept Check Use the difference of squares formula, (ππ + ππ)(ππ β ππ) = ππ2 β ππ2, to find each product.
100. 101 β 99 101. 198 β 202 102. 505 β 495 Find the area of each figure. Express it as a polynomial in descending powers of the variable x.
103. 104. 105.
Concept Check For each pair of functions, ππ and ππ, find the product function (ππππ)(ππ).
106. ππ(π₯π₯) = 5π₯π₯ β 6, ππ(π₯π₯) = β2 + 3π₯π₯ 107. ππ(π₯π₯) = π₯π₯2 + 7π₯π₯ β 2, ππ(π₯π₯) = 6π₯π₯ + 5
108. ππ(π₯π₯) = 3π₯π₯2 β 5π₯π₯, ππ(π₯π₯) = 9 + π₯π₯ β π₯π₯2 109. ππ(π₯π₯) = π₯π₯ππ β 4, ππ(π₯π₯) = π₯π₯ππ + 1 Let π·π·(ππ) = ππππ β ππ, πΈπΈ(ππ) = ππππ, and πΉπΉ(ππ) = ππ β ππ. Find each of the following.
110. (πππ π )(π₯π₯) 111. (ππππ)(π₯π₯) 112. (ππππ)(ππ)
113. (πππ π )(β1) 114. (ππππ)(3) 115. (πππ π )(0)
116. (πππ π )(π₯π₯) 117. (πππ π ) οΏ½12οΏ½ 118. (πππ π )(ππ + 1)
119. ππ(ππ β 1) 120. ππ(2ππ + 3) 121. ππ(1 + β)β ππ(1) Analytic Skills Solve each problem.
122. The corners are cut from a rectangular piece of cardboard measuring 8 in. by 12 in. The sides are folded up to make a box. Find the volume of the box in terms of the variable π₯π₯, where π₯π₯ is the length of a side of the square cut from each corner of the rectangle.
123. A rectangular pen has a perimeter of 100 feet. If π₯π₯ represents its width, write a polynomial that gives the area of the pen in terms of π₯π₯.
2π₯π₯ + 3
π₯π₯β
5
2π₯π₯ + 6
π₯π₯ β 4
3π₯π₯ β 1
2π₯π₯ β 2
2π₯π₯
200
P.3 Division of Polynomials
In this section we will discuss dividing polynomials. The result of division of polynomials is not always a polynomial. For example, π₯π₯ + 1 divided by π₯π₯ becomes
π₯π₯ + 1π₯π₯
=π₯π₯π₯π₯
+1π₯π₯
= 1 +1π₯π₯
,
which is not a polynomial. Thus, the set of polynomials is not closed under the operation of division. However, we can perform division with remainders, mirroring the algorithm of division of natural numbers. We begin with dividing a polynomial by a monomial and then by another polynomial.
Division of Polynomials by Monomials
To divide a polynomial by a monomial, we divide each term of the polynomial by the monomial, and then simplify each quotient. In other words, we use the reverse process of addition of fractions, as illustrated below.
ππ + ππππ
=ππππ
+ππππ
Dividing Polynomials by Monomials
Divide and simplify.
a. (6π₯π₯3 + 15π₯π₯2 β 2π₯π₯) Γ· (3π₯π₯) b. π₯π₯ππ2β8π₯π₯2ππ+6π₯π₯3ππ2
β2π₯π₯ππ2
a. (6π₯π₯3 + 15π₯π₯2 β 2π₯π₯) Γ· (3π₯π₯) = 6π₯π₯3+15π₯π₯2β2π₯π₯3π₯π₯
= 6π₯π₯3
3π₯π₯+ 15π₯π₯2
3π₯π₯β 2π₯π₯
3π₯π₯= ππππππ + ππππ β ππ
ππ
b. π₯π₯ππ2β8π₯π₯2ππ+6π₯π₯3ππ2
β2π₯π₯ππ2= β π₯π₯ππ2
2π₯π₯ππ2+ 8π₯π₯2ππ
2π₯π₯ππ2β 6π₯π₯3ππ2
2π₯π₯ππ2= βππ
ππ+ ππππ
ππβ ππππππ
Division of Polynomials by Polynomials
To divide a polynomial by another polynomial, we follow an algorithm similar to the long division algorithm used in arithmetic. For example, observe the steps taken in the long division algorithm when dividing 158 by 13 and the corresponding steps when dividing π₯π₯2 + 5π₯π₯ + 8 by π₯π₯ + 3. Step 1: Place the dividend under the long division symbol and the divisor in front of this
symbol.
13 ) 158 π₯π₯ + 3οΏ½οΏ½οΏ½ππππππππππππππ
) π₯π₯2 + 5π₯π₯ + 8οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππππππππππππππππ
Solution 2 5 2
3 4 2
201
Remember: Both polynomials should be written in decreasing order of powers. Also, any missing terms after the leading term should be displayed with a zero coefficient. This will ensure that the terms in each column are of the same degree.
Step 2: Divide the first term of the dividend by the first term of the divisor and record the
quotient above the division symbol.
1
13οΏ½
) 15οΏ½
8 π₯π₯ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ π π πππππ‘π‘πππππππ‘π‘
π₯π₯β
+ 3 ) π₯π₯2οΏ½
+ 5π₯π₯ + 8
Step 3: Multiply the quotient from Step 2 by the divisor and write the product under the dividend, lining up the columns with the same degree terms.
113 ) 15
8
13
π₯π₯ π₯π₯
+ 3οΏ½οΏ½οΏ½
) π₯π₯2
+ 5π₯π₯ + 8
π₯π₯2 + 3π₯π₯
Step 4: Underline and subtract by adding opposite terms in each column. We suggest to record the new sign in a circle, so that it is clear what is being added.
113 ) 15
8
β 13 2
π₯π₯ π₯π₯ + 3 ) π₯π₯2
+ 5π₯π₯ + 8
β (π₯π₯2 + 3π₯π₯) 2π₯π₯
Step 5: Drop the next term (or digit) and repeat the algorithm until the degree of the remainder is lower than the degree of the divisor.
1213 ) 15
8
β 13
28
β26 2
π₯π₯ + 2 π₯π₯ + 3 ) π₯π₯2
+ 5π₯π₯ + 8
β (π₯π₯2 + 3π₯π₯)
2π₯π₯ + 8
β(2π₯π₯ + 6) 2
In the example of long division of numbers, we have 158 = 13 β 12 + 2. So, the quotient can be written as 158
13= ππππ + ππ
ππππ.
In the example of long division of polynomials, we have
π₯π₯2
+ 5π₯π₯ + 8 = (π₯π₯ + 3) β (π₯π₯ + 2) + 2.
So, the quotient can be written as π₯π₯2
+5π₯π₯+8
π₯π₯+3= ππ + ππ + ππ
ππ+ππ.
Generally, if ππ, π·π·, ππ, and π π are polynomials, such that ππ(π₯π₯) = π·π·(π₯π₯) β ππ(π₯π₯) + π π (π₯π₯), then the ratio of polynomials ππ and π·π· can be written as
π·π·(ππ)π«π«(ππ)
= πΈπΈ(ππ) +πΉπΉ(ππ)π«π«(ππ)
,
β
β
β
ππππππππππππππππππ
202
where ππ(π₯π₯) is the quotient polynomial, and π π (π₯π₯) is the remainder from the division of ππ(π₯π₯) by the divisor π·π·(π₯π₯).
Observe: The degree of the remainder must be lower than the degree of the divisor, as otherwise, we could apply the division algorithm one more time.
Dividing Polynomials by Polynomials
Divide.
a. (3π₯π₯3 β 2π₯π₯2 + 5) Γ· (π₯π₯2 β 3) b. 2ππ3+ππ+5ππ2
1+2ππ
a. When writing the polynomials in the long division format, we use a zero placeholder
term in place of the missing linear terms in both, the dividend and the divisor. So, we have
ππππ β ππ π₯π₯2 + 0π₯π₯ β 3 ) 3π₯π₯3
β 2π₯π₯2 + 0π₯π₯ + 5
β ( 3π₯π₯3 + 0π₯π₯2 + 9π₯π₯)
β2π₯π₯2 β 9π₯π₯ + 5
β (β2π₯π₯2 β 0π₯π₯ + 6) β ππππ β ππ
Thus, (3π₯π₯3 β 2π₯π₯2 + 5) Γ· (π₯π₯2 β 3) = 3π₯π₯ β 2 + β9π₯π₯β1
π₯π₯2β3= ππππ β ππ β ππππ+ππ
ππππβππ.
b. To perform this division, we arrange both polynomials in decreasing order of powers,
and replace the constant term in the dividend with a zero. So, we have
ππππ β ππ + ππππ
2ππ + 5 ) 2ππ3
+ 3ππ2 + 2ππ + 0
β ( 2ππ3 + 5ππ2)
β2ππ2 + 2ππ
β (β2ππ2 β 5ππ)
7ππ + 0 β οΏ½7ππ + 35
2οΏ½
β ππππππ
Thus, 2ππ3+ππ+5ππ2
5+2ππ= ππ2 β ππ + 7
2+
β 3522ππ+5
= ππππ β ππ + ππππβ ππππ
ππππ+ππππ.
Observe in the above answer that β 3522ππ+5
is written in a simpler form, β 354ππ+10
. This is
because β 3522ππ+5
= β352β 12ππ+5
= β 354ππ+10
.
Solution
β
β
β
β
β¨
203
Quotient Functions
Similarly as in the case of polynomials, we can define quotients of functions.
Definition 3.1 Suppose ππ and ππ are functions of π₯π₯ with the corresponding domains π·π·ππ and π·π·ππ.
Then the quotient function, denoted ππππ, is defined as
οΏ½πππποΏ½ (ππ) =
ππ(ππ)ππ(ππ).
The domain of the quotient function is the intersection of the domains of the two functions, π·π·ππ and π·π·ππ, excluding the π₯π₯-values for which ππ(π₯π₯) = 0. So,
π«π«ππππ
= π«π«ππ β© π«π«ππ\{ππ|ππ(ππ) = ππ}
Dividing Polynomial Functions
Suppose ππ(π₯π₯) = 2π₯π₯2 β π₯π₯ β 6 and ππ(π₯π₯) = π₯π₯ β 2. Find the following:
a. οΏ½πππποΏ½ (π₯π₯) and οΏ½ππ
πποΏ½ (2ππ),
b. οΏ½πππποΏ½ (β3) and οΏ½ππ
πποΏ½ (2),
c. domain of ππππ
.
a. By Definition 3.1, οΏ½πππποΏ½ (π₯π₯) = ππ(π₯π₯)
ππ(π₯π₯)= 2π₯π₯2βπ₯π₯β6
π₯π₯β2= (2π₯π₯+3)(π₯π₯β2)
π₯π₯β2= ππππ + ππ
So, οΏ½πππποΏ½ (β3) = 2(β3) + 3 = βππ. One can verify that the same value is found by
evaluating ππ(β3)ππ(β3)
.
b. Since the equation (2π₯π₯+3)(π₯π₯β2)π₯π₯β2
= 2π₯π₯ + 3 is true only for π₯π₯ β 2, the simplified formula
οΏ½πππποΏ½ (π₯π₯) = 2π₯π₯+ 3 cannot be used to evaluate οΏ½ππ
πποΏ½ (2). However, by Definition 3.1, we have
οΏ½πππποΏ½ (2) =
ππ(2)ππ(2)
=2(2)2 β (2) β 6
(2) β 2=
8 β 2 β 60
=00
= ππππππππππππππππππ
To evaluate οΏ½πππποΏ½ (2ππ), we first notice that if ππ β 1, then 2ππ β 2. So, we can use the
simplified formula οΏ½πππποΏ½ (π₯π₯) = 2π₯π₯ + 3 and evaluate οΏ½ππ
πποΏ½ (2ππ) = 2(2ππ) + 3 = ππππ + ππ.
c. The domain of any polynomial is the set of all real numbers. So, the domain of ππππ
is
the set of all real numbers except for the π₯π₯-values for which the denominator ππ(π₯π₯) =
Solution
οΏ½πππποΏ½ (2) is undefined, so 2 is not in the
domain of ππππ
Notice that this equation holds only
for π₯π₯ β 2.
204
π₯π₯ β 2 is equal to zero. Since the solution to the equation π₯π₯ β 2 = 0 is π₯π₯ = 2, then the value 2 must be excluded from the set of all real numbers. Therefore, π«π«π·π·
πΈπΈ= β \ {ππ}.
P.3 Exercises
Vocabulary Check Complete each blank with the most appropriate term from the given list: divisor,
remainder, decreasing, zero, monomial, lower, domain, factor.
1. The algorithm of long division of polynomials requires that both polynomials are written in _____________ order of powers and any missing term is replaced with a _________ term which plays the role of a placeholder.
2. To check a division problem, multiply the ____________ by the quotient and then add the _____________ .
3. When dividing a polynomial by a ______________, there is no need to use the long division algorithm.
4. In polynomial division, when the degree of the remainder is _________ than the degree of the divisor, the division process ends.
5. The _____________ of a quotient function ππππ does not contain the zeros of the devisor function ππ.
6. If the remainder in the division of ππ(π₯π₯) by ππ(π₯π₯) is zero, then ππ(π₯π₯) is a ____________ of ππ(π₯π₯).
Concept Check
7. True or False? When a tenth-degree polynomial is divided by a second-degree polynomial, the quotient is a fifth-degree polynomial.
8. True or False? When a polynomial is divided by a second-degree polynomial, the remainder is a first-degree polynomial.
Divide.
9. 20π₯π₯3β15π₯π₯2+5π₯π₯5π₯π₯
10. 27ππ4+18ππ2β9ππ9ππ
11. 8π₯π₯2ππ2β24π₯π₯ππ4π₯π₯ππ
12. 5ππ3ππ+10ππ2ππ2β15ππππ3
5ππππ 13. 9ππ5β15ππ4+12ππ3
β3ππ2 14. 20π₯π₯3ππ2+44π₯π₯2ππ3β24π₯π₯2ππ
β4π₯π₯2ππ
15. 64π₯π₯3β72π₯π₯2+12π₯π₯8π₯π₯3
16. 4ππ2ππ2β21ππππ3+18ππππ2
14ππ2ππ3 17. 12ππππ2ππ+10ππ2ππππ+18ππππππ2
6ππ2ππππ
Divide.
18. (π₯π₯2 + 3π₯π₯ β 18) Γ· (π₯π₯ + 6) 19. (3π¦π¦2 + 17π¦π¦ + 10) Γ· (3π¦π¦ + 2)
20. (π₯π₯2 β 11π₯π₯ + 16) Γ· (π₯π₯ + 8) 21. (ππ2 β 7ππ β 9) Γ· (ππ β 3)
205
22. 6ππ3βππ2β103ππ+4
23. 4ππ3+6ππ2+142ππ+4
24. 4π₯π₯3+8π₯π₯2β11π₯π₯+34π₯π₯+1
25. 10π§π§3β26π§π§2+17π§π§β135π§π§β3
26. 2π₯π₯3+4π₯π₯2βπ₯π₯+2π₯π₯2+2π₯π₯β1
27. 3π₯π₯3β2π₯π₯2+5π₯π₯β4π₯π₯2βπ₯π₯+3
28. 4ππ4+6ππ3+3ππβ12ππ2+1
29. 9ππ4+12ππ3β4ππβ13ππ2β1
30. 2ππ3+7ππ2+9ππ+32ππ+2
31. 5π‘π‘2+19π‘π‘+74π‘π‘+12
32. π₯π₯4β4π₯π₯3+5π₯π₯2β3π₯π₯+2π₯π₯2+3
33. ππ3β1ππβ1
34. π₯π₯3+1π₯π₯+1
35. ππ4+16ππ+2
36. π₯π₯5β32π₯π₯β2
Concept Check For each pair of polynomials, ππ(π₯π₯) and π·π·(π₯π₯), find such polynomials ππ(π₯π₯) and π π (π₯π₯) that
ππ(π₯π₯) = ππ(π₯π₯) β π·π·(π₯π₯) + π π (π₯π₯).
37. ππ(π₯π₯) = 4π₯π₯3 β 4π₯π₯2 + 13π₯π₯ β 2 and π·π·(π₯π₯) = 2π₯π₯ β 1
38. ππ(π₯π₯) = 3π₯π₯3 β 2π₯π₯2 + 3π₯π₯ β 5 and π·π·(π₯π₯) = 3π₯π₯ β 2
Concept Check For each pair of functions, ππ and ππ, find the quotient function οΏ½πππποΏ½ (ππ) and state its domain.
39. ππ(π₯π₯) = 6π₯π₯2 β 4π₯π₯, ππ(π₯π₯) = 2π₯π₯ 40. ππ(π₯π₯) = 6π₯π₯2 + 9π₯π₯, ππ(π₯π₯) = β3π₯π₯
41. ππ(π₯π₯) = π₯π₯2 β 36, ππ(π₯π₯) = π₯π₯ + 6 42. ππ(π₯π₯) = π₯π₯2 β 25, ππ(π₯π₯) = π₯π₯ β 5
43. ππ(π₯π₯) = 2π₯π₯2 β π₯π₯ β 3, ππ(π₯π₯) = 2π₯π₯ β 3 44. ππ(π₯π₯) = 3π₯π₯2 + π₯π₯ β 4, ππ(π₯π₯) = 3π₯π₯ + 4
45. ππ(π₯π₯) = 8π₯π₯3 + 125, ππ(π₯π₯) = 2π₯π₯ + 5 46. ππ(π₯π₯) = 64π₯π₯3 β 27, ππ(π₯π₯) = 4π₯π₯ β 3
Let π·π·(ππ) = ππππ β ππ, πΈπΈ(ππ) = ππππ, and πΉπΉ(ππ) = ππ β ππ. Find each of the following. If the value canβt be evaluated, say DNE (does not exist).
47. οΏ½π π πποΏ½ (π₯π₯) 48. οΏ½ππ
π π οΏ½ (π₯π₯) 49. οΏ½π π
πποΏ½ (π₯π₯)
50. οΏ½π π πποΏ½ (2) 51. οΏ½π π
πποΏ½ (0) 52. οΏ½ππ
π π οΏ½ (3)
53. οΏ½π π πποΏ½ (β2) 54. οΏ½π π
πποΏ½ (2) 55. οΏ½ππ
π π οΏ½ (ππ), for ππ β 2
56. οΏ½π π πποΏ½ οΏ½3
2οΏ½ 57. 1
2οΏ½πππ π οΏ½ (π₯π₯) 58. οΏ½ππ
π π οΏ½ (ππ β 1)
Analytic Skills Solve each problem.
59. The area π΄π΄ of a rectangle is 5π₯π₯2 + 12π₯π₯ + 4 and its width ππ is π₯π₯ + 2. a. Find the length πΏπΏ of the rectangle. b. Find the length if the width is 8 meters.
60. The area π΄π΄ of a triangle is 6π₯π₯2 β π₯π₯ β 15. Find its height β, if the base of the triangle is 3π₯π₯ β 5. Then, find the height β, if the base is 7 centimeters.
3π₯π₯ β 5
β
π₯π₯ + 2
πΏπΏ
206
P.4 Composition of Functions and Graphs of Basic Polynomial Functions
In the last three sections, we have created new functions with the use of function operations such as addition, subtraction, multiplication, and division. In this section, we will introduce one more function operation, called composition of functions. This operation will allow us
to represent situations in which some quantity depends on a variable that, in turn, depends on another variable. For instance, the number of employees hired by a firm may depend on the firmβs profit, which may in turn depend on the number of items the firm produces. In this situation, we might be interested in the number of employees hired by a firm as a function of the number of items the firm produces. This illustrates a composite function. It is obtained by composing the number of employees with respect to the profit and the profit with respect to the number of items produced.
In the second part of this section, we will examine graphs of basic polynomial functions, such as constant, linear, quadratic, and cubic functions.
Composition of Functions
Consider womenβs shoe size scales in U.S., Italy, and Great Britain.
U.S. Italy Britain 4 32 2 5 ππ 34 f 3 6 36 4 7 38 5 8 40 6
ππ
The reader is encouraged to confirm that the function ππ(π₯π₯) = 2π₯π₯ + 24 gives the womenβs shoe size in Italy for any given U.S. shoe size π₯π₯. For example, a U.S. shoe size 7 corresponds to a shoe size of ππ(7) = 2 β 7 + 24 = 38 in Italy. Similarly, the function ππ(π₯π₯) = 1
2π₯π₯ β 14
gives the womenβs shoe size in Britain for any given Italian shoe size π₯π₯. For example, an Italian shoe size 38 corresponds to a shoe size of ππ(38) = 1
2β 38 β 14 = 5 in Italy. So, by
converting the U.S. shoe size first to the corresponding Italian size and then to the Great Britain size, we create a third function, β, that converts the U.S. shoe size directly to the Great Britain size. We say, that function β is the composition of functions ππ and ππ. Can we find a formula for this composite function? By observing the corresponding data for U.S. and Great Britain shoe sizes, we can conclude that β(π₯π₯) = π₯π₯ β 2. This formula can also be derived with the use of algebra.
Since the U.S. shoe size π₯π₯ corresponds to the Italian shoe size ππ(π₯π₯), which in turn corresponds to the Britain shoe size ππ(ππ(π₯π₯)), the composition function β(π₯π₯) is given by the formula
ππ(ππ) = ππ(ππ(ππ)) = ππ(2π₯π₯ + 24) =12
(2π₯π₯ + 24) β 14 = π₯π₯ + 12 β 14 = ππ β ππ,
which confirms our earlier observation.
ππ β ππ
input ππ
ππ(ππ)
ππ(ππ(ππ)) ππ
ππ
207
Generally, the composition of functions ππ and ππ, denoted ππ β ππ, is the function that acts on the input π₯π₯ by mapping it by the function ππ to the value ππ(π₯π₯), which in turn becomes the input for function ππ, to be mapped to the value ππ(ππ(π₯π₯)). See the diagram below.
Definition 4.1 If ππ and ππ are functions, then the composite function ππ β ππ, or composition of ππ and ππ, is defined by
(ππ β ππ)(ππ) = πποΏ½ππ(ππ)οΏ½,
for all π₯π₯ in the domain of ππ, such that ππ(π₯π₯) is in the domain of ππ.
Note: We read πποΏ½ππ(π₯π₯)οΏ½ as βππ of ππ of π₯π₯β.
Evaluating a Composite Function
Suppose ππ(π₯π₯) = π₯π₯2 and ππ(π₯π₯) = 2π₯π₯ + 3. Find the following:
a. (ππ β ππ)(β2) b. (ππ β ππ)(β2)
a. (ππ β ππ)(βππ) = πποΏ½ππ(β2)οΏ½ = ππ(2(β2) + 3) = ππ(β1) = (β1)2οΏ½οΏ½οΏ½ππ(β1)
= ππ
b. (ππ β ππ)(βππ) = πποΏ½ππ(β2)οΏ½ = ππ((β2)2) = ππ(4) = 2(4) + 3οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
ππ(4)= ππππ
Observation: In the above example, (ππ β ππ)(β2) β (ππ β ππ)(β2). This shows that composition of functions is not commutative.
Finding the Composition of Two Functions
Given functions ππ and ππ, find (ππ β ππ)(π₯π₯).
a. ππ(π₯π₯) = 2 β 3π₯π₯; ππ(π₯π₯) = π₯π₯2 β 2π₯π₯ b. ππ(π₯π₯) = π₯π₯2 β π₯π₯ + 3; ππ(π₯π₯) = π₯π₯ + 4
a. (ππ β ππ)(ππ) = πποΏ½ππ(π₯π₯)οΏ½ = ππ(π₯π₯2 β 2π₯π₯) = 2 β 3(π₯π₯2 β 2π₯π₯) = βππππππ + ππππ + ππ
Solution
ππ
ππ
ππ
ππ β ππ
ππ(ππ) ππ(ππ(ππ))
ππ(β2)
ππ(β2)
Solution ππ(π₯π₯) ππ(π₯π₯)
208
b. (ππ β ππ)(ππ) = πποΏ½ππ(π₯π₯)οΏ½ = ππ(π₯π₯ + 4) = (π₯π₯ + 4)2 β (π₯π₯ + 4) + 3οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ππ(π₯π₯+4)
= π₯π₯2 + 8π₯π₯ + 16 β π₯π₯ β 4 + 3 = ππππ + ππππ + ππππ
Attention! Do not confuse the composition of functions, (ππ β ππ)(ππ) = ππ(ππ(ππ)), with the multiplication of functions, (ππππ)(ππ) = ππ(ππ) β ππ(ππ).
Graphs of Basic Polynomial Functions
Since polynomials are functions, they can be evaluated for different π₯π₯-values and graphed in a system of coordinates. How do polynomial functions look like? Below, we graph several basic polynomial functions up to the third degree, and observe their shape, domain, and range.
Let us start with a constant function, which is defined by a zero degree polynomial, such as ππ(π₯π₯) = 1. In this example, for any real π₯π₯-value, the corresponding π¦π¦-value is constantly equal to 1. So, the graph of this function is a horizontal line with the π¦π¦-intercept at 1.
Domain: β Range: {1} Generally, the graph of a constant function, ππ(ππ) = ππ, is a horizontal line with the π¦π¦-intercept at ππ. The domain of this function is β and the range is {ππ}.
The basic first degree polynomial function is the identity function given by the formula ππ(ππ) = ππ. Since both coordinates of any point satisfying this equation are the same, the graph of the identity function is the diagonal line, as shown below.
Domain: β Range: β Generally, the graph of any first degree polynomial function, ππ(ππ) = ππππ + ππ with ππ β 0, is a slanted line. So, the domain and range of such function is β.
ππ(π₯π₯)
ππ(π₯π₯)
π₯π₯ 1
1
ππ(π₯π₯)
π₯π₯ 1
1
209
The basic second degree polynomial function is the squaring function given by the formula ππ(ππ) = ππππ. The shape of the graph of this function is refered to as the basic parabola. The reader is encouraged to observe the relations between the five points calculated in the table of values below.
Domain: β Range: [0,β) Generally, the graph of any second degree polynomial function, ππ(ππ) = ππππππ + ππππ + ππ with ππ β 0, is a parabola. The domain of such function is β and the range depends on how the parabola is directed, with arms up or down.
The basic third degree polynomial function is the cubic function, given by the formula ππ(ππ) = ππππ. The graph of this function has a shape of a βsnakeβ. The reader is encouraged to observe the relations between the five points calculated in the table of values below.
Domain: β Range: β
Generally, the graph of a third degree polynomial function, ππ(ππ) = ππππππ + ππππππ + ππππ + ππ with ππ β 0, has a shape of a βsnakeβ with different size waves in the middle. The domain and range of such function is β.
ππ ππ(ππ) = ππππ βππ ππ βππ ππ ππ ππ ππ ππ ππ ππ
ππ ππ(ππ) = ππππ βππ βππ βππ βππ ππ ππ ππ ππ ππ ππ
ππ(π₯π₯)
π₯π₯ 1
1
vertex
sym
met
ry
abou
t the
y-
axis
ππ(π₯π₯)
π₯π₯ 1
1
center
sym
met
ry
abou
t the
or
igin
210
Graphing Polynomial Functions
Graph each function using a table of values. Give the domain and range of each function by observing its graph. Then, on the same grid, graph the corresponding basic polynomial function. Observe and name the transformation(s) that can be applied to the basic shape in order to obtain the desired function.
a. ππ(π₯π₯) = β2π₯π₯ b. ππ(π₯π₯) = (π₯π₯ + 2)2 c. ππ(π₯π₯) = π₯π₯3 β 2 a. The graph of ππ(π₯π₯) = β2π₯π₯ is a line passing through the origin and falling from left to
right, as shown below in green.
Domain of f: β Range of f: β
Observe that to obtain the green line, we multiply π¦π¦-coordinates of the orange line by
a factor of β2. Such a transformation is called a dilation in the ππ-axis by a factor of βππ. This dilation can also be achieved by applying a symmetry in the ππ-axis first, and then stretching the resulting graph in the ππ-axis by a factor of ππ.
b. The graph of ππ(π₯π₯) = (π₯π₯ + 2)2 is a parabola with a vertex at (β2, 0), and its arms are
directed upwards as shown below in green.
Domain: β
Range: [0,β)
Observe that to obtain the green shape, it is enough to move the graph of the basic
parabola by two units to the left. This transformation is called a horizontal translation by two units to the left. The translation to the left reflects the fact that the vertex of the parabola ππ(π₯π₯) = (π₯π₯ + 2)2 is located at π₯π₯ + 2 = 0, which is equivalent to π₯π₯ = β2.
ππ ππ(ππ) = βππππ βππ 2 ππ 0 ππ β2
ππ ππ(ππ) = (ππ + ππ)ππ βππ 4 βππ 1 βππ 0 βππ 1 ππ 4
Solution
ππ(π₯π₯)
π₯π₯ 1
1
vertex
sym
met
ry
abou
t the
π₯π₯
=β
2
ππ(π₯π₯)
π₯π₯ β2
4
211
c. The graph of ππ(π₯π₯) = π₯π₯3 β 2 has the shape of a basic cubic function with a center at (0,β2).
Domain: β Range: β
Observe that the green graph can be obtained by shifting the graph of the basic cubic
function by two units down. This transformation is called a vertical translation by two units down.
P.4 Exercises
Vocabulary Check Complete each blank with the most appropriate term from the given list: composition,
product, identity, parabola, translating, up, left, down, right, basic, cubic, graph.
1. For two functions ππ and ππ, the function ππ β ππ is called the _______________ of ππ and ππ.
2. For two functions ππ and ππ, the function ππ β ππ is called the _______________ of ππ and ππ.
3. The function ππ(π₯π₯) = π₯π₯ is called an ______________ function.
4. The graph of ππ(π₯π₯) = π₯π₯2 is referred to as the basic ______________.
5. The graph of ππ(π₯π₯) = π₯π₯2 + 3 can be obtained by _______________ the graph of a basic parabola by 3 units _______.
6. The graph of ππ(π₯π₯) = (π₯π₯ + 3)2 can be obtained by translating the graph of a _________ parabola by 3 units to the _______.
7. The graph of ππ(π₯π₯) = (π₯π₯ β 3)3 can be obtained by translating the graph of a basic ________ function by 3 units to the ________.
8. The ________ of ππ(π₯π₯) = π₯π₯3 β 3 can be obtained by translating the graph of a basic cubic function by 3 units ________.
ππ ππ(ππ) = ππππ β ππ βππ β10 βππ β3 ππ β2 ππ β1 ππ 6
center
sym
met
ry
abou
t (0,β
2)
ππ(π₯π₯)
π₯π₯ 2
1
212
Concept Check Suppose ππ(π₯π₯) = π₯π₯2 + 2 , ππ(π₯π₯) = 5 β π₯π₯, and β(π₯π₯) = 2π₯π₯ β 3. Find the following values or expressions.
9. (ππ β ππ)(1) 10. (ππ β ππ)(1) 11. (ππ β ππ)(π₯π₯) 12. (ππ β ππ)(π₯π₯)
13. (ππ β β)(β1) 14. (β β ππ)(β1) 15. (ππ β β)(π₯π₯) 16. (β β ππ)(π₯π₯)
17. (β β ππ)(β2) 18. (ππ β β)(β2) 19. (β β ππ)(π₯π₯) 20. (ππ β β)(π₯π₯)
21. (ππ β ππ)(2) 22. (ππ β β) οΏ½12οΏ½ 23. (β β β)(π₯π₯) 24. (ππ β ππ)(π₯π₯)
Analytic Skills Solve each problem.
25. The function defined by ππ(π₯π₯) = 12π₯π₯ computes the number of inches in π₯π₯ feet, and the function defined by ππ(π₯π₯) = 2.54π₯π₯ computes the number of centimeters in π₯π₯ inches. What is (ππ β ππ)(π₯π₯) and what does it compute?
26. If hamburgers are $3.50 each, then πΆπΆ(π₯π₯) = 3.5π₯π₯ gives the pre-tax cost in dollars for π₯π₯ hamburgers. If sales tax is 12%, then ππ(π₯π₯) = 1.12π₯π₯ gives the total cost when the pre-tax cost is π₯π₯ dollars. Write the total cost as a function of the number of hamburgers.
27. The perimeter π₯π₯ of a square with sides of length ππ is given by the formula π₯π₯ = 4ππ. a. Solve the above formula for ππ in terms of π₯π₯. b. If π¦π¦ represents the area of this square, write π¦π¦ as a function of the perimeter π₯π₯. c. Use the composite function found in part b. to find the area of a square with perimeter 6.
28. When a thermal inversion layer occurs over a city, pollutants cannot rise vertically because they are trapped below the layer meaning they must disperse horizontally. Assume a factory smokestack begins emitting a pollutant at 8 a.m., and the pollutant disperses horizontally over a circular area. Suppose that ππ represents the time, in hours, since the factory began emitting pollutants (ππ = 0 represents 8 a.m.), and assume that the radius of the circle of pollution is ππ(ππ) = 2ππ miles. If π΄π΄(ππ) = ππππ2 represents the area of a circle of radius ππ, find and interpret (π΄π΄ β ππ)(ππ).
Discussion Point
29. Sears has Super Saturday sales events during which everything is 10% off, even items already on sale. If a coat was already on sale for 25% off, then does this mean that it is 35% off on Super Saturday? What function operation is at work here? Justify.
Concept Check Graph each function and state its domain and range.
30. ππ(π₯π₯) = β2π₯π₯ + 3 31. ππ(π₯π₯) = 3π₯π₯ β 4 32. ππ(π₯π₯) = βπ₯π₯2 + 4
33. ππ(π₯π₯) = π₯π₯2 β 2 34. ππ(π₯π₯) = 12π₯π₯2 35. ππ(π₯π₯) = β2π₯π₯2 + 1
36. ππ(π₯π₯) = (π₯π₯ + 1)2 β 2 37. ππ(π₯π₯) = βπ₯π₯3 + 1 38. ππ(π₯π₯) = (π₯π₯ β 3)3
ππ
ππ
213
Analytic Skills Guess the transformations needed to apply to the graph of a basic parabola ππ(π₯π₯) = π₯π₯2 to obtain the graph of the given function ππ(π₯π₯). Then graph both ππ(π₯π₯) and ππ(π₯π₯) on the same grid and confirm the the original guess.
39. ππ(π₯π₯) = βπ₯π₯2 40. ππ(π₯π₯) = π₯π₯2 β 3 41. ππ(π₯π₯) = π₯π₯2 + 2
42. ππ(π₯π₯) = (π₯π₯ + 2)2 43. ππ(π₯π₯) = (π₯π₯ β 3)2 44. ππ(π₯π₯) = (π₯π₯ + 2)2 β 1