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Mathematics LibreTexts

1.6: Polynomials and Their Operations

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  • LibreTexts

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Learning Objectives

  • Identify a polynomial and determine its degree.
  • Add and subtract polynomials.
  • Multiply and divide polynomials.

Definitions

A polynomial112 is a special algebraic expression with terms that consist of real number coefficients and variable factors with whole number exponents. Some examples of polynomials follow:

3x2 7xy+5 32x3+3x212x+1 6x2y4xy3+7
Table 1.6.1

The degree of a term113 in a polynomial is defined to be the exponent of the variable, or if there is more than one variable in the term, the degree is the sum of their exponents. Recall that x0=1; any constant term can be written as a product of x0 and itself. Hence the degree of a constant term is 0.

Term Degree
3x2 2
6x2y 2+1=3
7a2b3 2+3=5
8 0, since 8=8x0
2x 1, since 2x=2x1
Table 1.6.2

The degree of a polynomial114 is the largest degree of all of its terms.

Polynomial Degree
4x53x3+2x1 5
6x2y5xy3+7 4, because 5xy3 has degree 4.
12x+54 1, because 12x=12x1
Table 1.6.3

Of particular interest are polynomials with one variable115, where each term is of the form anxn. Here an is any real number and n is any whole number. Such polynomials have the standard form:

anxn+an1xn1++a1x+a0

Typically, we arrange terms of polynomials in descending order based on the degree of each term. The leading coefficient116 is the coefficient of the variable with the highest power, in this case, an.

Example 1.6.1:

Write in standard form: 3x4x2+5x3+72x4.

Solution

Since terms are defined to be separated by addition, we write the following:

3x4x2+5x3+72x4=3x+(4)x2+5x3+7+(2)x4

In this form, we can see that the subtraction in the original corresponds to negative coefficients. Because addition is commutative, we can write the terms in descending order based on the degree as follows:

=(2)x4+5x3+(4)x2+3x+7=2x4+5x34x2+3x+7

Answer:

2x4+5x34x2+3x+7

We classify polynomials by the number of terms and the degree:

Expression Classification Degree
5x7 Monomial117 (one term) 7
8x61 Binomial118 (two terms) 6
3x2+x1 Trinomial119 (three terms) 2
5x32x2+3x6 Polynomial (many terms) 3
Table 1.6.4

We can further classify polynomials with one variable by their degree:

Polynomial Name
5 Constant120 (degree 0)
2x+1 Linear121 (degree 1)
3x2+5x3 Quadratic122 (degree 2)
x3+x2+x+1 Cubic123 (degree 3)
7x4+3x37x+8 Fourth-degree polynomial
Table 1.6.5

In this text, we call any polynomial of degree n4 an nth-degree polynomial. In other words, if the degree is 4, we call the polynomial a fourth-degree polynomial. If the degree is 5, we call it a fifth-degree polynomial, and so on.

Example 1.6.2

State whether the following polynomial is linear or quadratic and give the leading coefficient: 25+4xx2.

Solution

The highest power is 2; therefore, it is a quadratic polynomial. Rewriting in standard form we have

x2+4x+25

Here x2=1x2 and thus the leading coefficient is 1.

Answer:

Quadratic; leading coefficient: 1

Adding and Subtracting Polynomials

We begin by simplifying algebraic expressions that look like +(a+b) or (a+b). Here, the coefficients are actually implied to be +1 and 1 respectively and therefore the distributive property applies. Multiply each term within the parentheses by these factors as follows:

+(a+b)=+1(a+b)=(+1)a+(+1)b=a+b(a+b)=1(a+b)=(1)a+(1)b=ab

Use this idea as a means to eliminate parentheses when adding and subtracting polynomials.

Example 1.6.3:

Add: 9x2+(x25).

Solution

The property +(a+b)=a+b allows us to eliminate the parentheses, after which we can then combine like terms.

9x2+(x25)=9x2+x25=10x25

Answer:

10x25

Example 1.6.4:

Add: (3x2y24xy+9)+(2x2y26xy7).

Solution

Remember that the variable parts have to be exactly the same before we can add the coefficients.

(3x2y24xy+9)+(2x2y26xy7)=3x2y2_4xy__+9___+2x2y2_6xy__7___=5x2y210xy+2

Answer:

5x2y210xy+2

When subtracting polynomials, the parentheses become very important.

Example 1.6.5:

Subtract: 4x2(3x2+5x).

Solution

The property (a+b)=ab allows us to remove the parentheses after subtracting each term.

4x2(3x2+5x)=4x23x25x=x25x

Answer:

x25x

Subtracting a quantity is equivalent to multiplying it by 1.

Example 1.6.6:

Subtract: (3x22xy+y2)(2x2xy+3y2).

Solution

Distribute the 1, remove the parentheses, and then combine like terms. Multiplying the terms of a polynomial by 1 changes all the signs.

19157971f267dfed73125fabab9d74ae.png
Figure 1.6.1

=3x22xy+y22x2+xy3y2=x2xy2y2

Answer:

x2xy2y2

Exercise 1.6.1

Subtract: (7a22ab+b2)(a22ab+5b2).

Answer

6a24b2

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Multiplying Polynomials

Use the product rule for exponents, xmxn=xm+n, to multiply a monomial times a polynomial. In other words, when multiplying two expressions with the same base, add the exponents. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,

7x48x3=78x4x3Commutativeproperty=56x4+3Productruleforexponents=56x7

To multiply a polynomial by a monomial, apply the distributive property, and then simplify each term.

Example 1.6.7:

Multiply: 5xy2(2x2y2xy+1).

Solution

Apply the distributive property and then simplify.

2603f8e94a90164b1620a009c51f7eaf.png
Figure 1.6.2

=5xy22x2y25xy2xy+5xy21=10x3y45x2y3+5xy2

Answer:

10x3y45x2y3+5xy2

To summarize, multiplying a polynomial by a monomial involves the distributive property and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add exponents of variables where the bases are the same.

In the same manner that we used the distributive property to distribute a monomial, we use it to distribute a binomial.

(a+b)(c+d)=(a+b)c+(a+b)d=ac+bc+ad+bd=ac+ad+bc+bd

Here we apply the distributive property multiple times to produce the final result. This same result is obtained in one step if we apply the distributive property to a and b separately as follows:

7be734ec5c654af428357e9bd13b2a08.png
Figure 1.6.3

This is often called the FOIL method. Multiply the first, outer, inner, and then last terms.

Example 1.6.8:

Multiply: (6x1)(3x5).

Solution

Distribute 6x and 1 and then combine like terms.

(6x1)(3x5)=6x3x6x5+(1)3x(1)5=18x230x3x+5=18x233x+5

Answer:

18x233x+5

Consider the following two calculations:

(a+b)2=(a+b)(a+b)=a2+ab+ba+b2=a2+ab+ab+b2=a2+2ab+b2 (ab)2=(ab)(ab)=a2abba+b2=a2abab+b2=a22ab+b2
Table 1.6.6

This leads us to two formulas that describe perfect square trinomials124:

(a+b)2=a2+2ab+b2(ab)2=a22ab+b2

We can use these formulas to quickly square a binomial.

Example 1.6.9:

Multiply: (3x+5)2

Solution

Here a=3x and b=5. Apply the formula:

6f9efdeb4dd8ce181d4304e3f8ada495.png
Figure 1.6.4

Answer:

9x2+30x+25

This process should become routine enough to be performed mentally. Our third special product follows:

(a+b)(ab)=a2ab+bab2=a2ab+abb2=a2b2

This product is called difference of squares125:

(a+b)(ab)=a2b2

The binomials (a+b) and (ab) are called conjugate binomials126. When multiplying conjugate binomials the middle terms are opposites and their sum is zero; the product is itself a binomial.

Example 1.6.10:

Multiply: (3xy+1)(3xy1).

Solution

(3xy+1)(3xy1)=(3xy)23xy+3xy12=9x2y21

Answer:

9x2y21

Exercise 1.6.2

Multiply: (x2+5y2)(x25y2).

Answer

(x425y4)

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Example 1.6.11:

Multiply: (5x2)3.

Solution

Here we perform one product at a time.

94bf5b5472fe2ef54e090df103f39d85.png
Figure 1.6.5

Answer:

125x2150x2+60x8

Dividing Polynomials

Use the quotient rule for exponents, xmxn=xmn, to divide a polynomial by a monomial. In other words, when dividing two expressions with the same base, subtract the exponents. In this section, we will assume that all variables in the denominator are nonzero.

Example 1.6.12:

Divide: 24x7y58x3y2.

Solution

Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.

24x7y58x3y2=248x73y52=3x4y3

Answer:

3x4y3

When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and break up the fraction using the following property:

a+bc=ac+bc

Applying this property will result in terms that can be treated as quotients of monomials.

Example 1.6.13:

Divide: 5x4+25x315x25x2.

Solution

Break up the fraction by dividing each term in the numerator by the monomial in the denominator, and then simplify each term.

5x4+25x315x25x2=5x45x2+25x35x215x25x2=55x42+255x32155x22=1x2+5x13x0=x2+5x31

Answer:

x2+5x3

We can check our division by multiplying our answer, the quotient, by the monomial in the denominator, the divisor, to see if we obtain the original numerator, the dividend.

DividendDivisor=Quotient 5x4+25x315x25x2=x2+5x3
or or
Dividend=DivisorQuotient 5x4+25x315x2=5x2(x2+5x3)
Table 1.6.7

The same technique outlined for dividing by a monomial does not work for polynomials with two or more terms in the denominator. In this section, we will outline a process called polynomial long division127, which is based on the division algorithm for real numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.

Example 1.6.14:

Divide x3+3x28x4x2:

Solution

Here x2 is the divisor and x3+3x28x4 is the dividend. To determine the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor.

597c1846212bb42a41ee60d1ea564951.png
Figure 1.6.6

Multiply the first term of the quotient by the divisor, remembering to distribute, and line up like terms with the dividend.

5d4e3953ce01091edb8660ea2cbbcd05.png
Figure 1.6.7

Subtract the resulting quantity from the dividend. Take care to subtract both terms.

59834d557ea8a06825dbda67dbb2956b.png
Figure 1.6.8

Bring down the remaining terms and repeat the process.

8e8f8648f001e0f2ebe3d06303378e50.png Figure 1.6.9

Notice that the leading term is eliminated and that the result has a degree that is one less. The complete process is illustrated below:

12ac7ce3d68c9c0bc218027f83ce92f0.png
Figure 1.6.10

Polynomial long division ends when the degree of the remainder is less than the degree of the divisor. Here, the remainder is 0. Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown above the division bar.

x3+3x28x4x2=x2+5x+2

To check the answer, multiply the divisor by the quotient to see if you obtain the dividend as illustrated below:

x3+3x28x4=(x2)(x2+5x+2)

This is left to the reader as an exercise.

Answer:

x2+5x+2

Next, we demonstrate the case where there is a nonzero remainder.

e7aecc5db5f3a690ad915a7012bd1474.png
Figure 1.6.11

Just as with real numbers, the final answer adds to the quotient the fraction where the remainder is the numerator and the divisor is the denominator. In general, when dividing we have:

DividendDivisor=Quotient+RemainderDivisor

If we multiply both sides by the divisor we obtain,

Dividend=Quotient×Divisor+Remainder

Example 1.6.15:

Divide: 6x25x+32x1.

Solution

Since the denominator is a binomial, begin by setting up polynomial long division.

907e3b78b0543ab03b8e48c0db9fc92a.png
Figure 1.6.12

To start, determine what monomial times 2x1 results in a leading term 6x2.This is the quotient of the given leading terms: (6x2)÷(2x)=3x. Multiply 3x times the divisor 2x1, and line up the result with like terms of the dividend.

0bf0959dce4e14a83cd7c030a7415b9d.png
Figure 1.6.13

Subtract the result from the dividend and bring down the constant term +3.

6637d797c48b15341303c31af6093ade.png
Figure 1.6.14

Subtracting eliminates the leading term. Multiply 2x1 by 1 and line up the result.

e197703db2295db06f984ba307bcf5c2.png
Figure 1.6.15

Subtract again and notice that we are left with a remainder.

da6de19c485bfb98d24aee7dbea7dae9.png
Figure 1.6.16

The constant term 2 has degree 0 and thus the division ends. Therefore,

6x25x+32x1=3x1+22x1

To check that this result is correct, we multiply as follows:

quotient×divisor+remainder=(3x1)(2x1)+2=6x23x2x+1+2=6x25x+2=dividend

Answer:

3x1+22x1

Occasionally, some of the powers of the variables appear to be missing within a polynomial. This can lead to errors when lining up like terms. Therefore, when first learning how to divide polynomials using long division, fill in the missing terms with zero coefficients, called placeholders128.

Example 1.6.16:

Divide: 27x3+643x+4.

Solution

Notice that the binomial in the numerator does not have terms with degree 2 or 1. The division is simplified if we rewrite the expression with placeholders:

27x3+64=27x3+0x2+0x+64

Set up polynomial long division:

3580ae29e84dc2142822da2ab8c2e345.png
Figure 1.6.17

We begin with 27x3÷3x=9x2 and work the rest of the division algorithm.

c1b11a0e625faf8dfcc54428eb1b21e5.png
Figure 1.6.18

Answer:

9x212x+16

Example 1.6.17:

Divide: 3x42x3+6x2+23x7x22x+5.

Solution

179294b59c7365cd517ffacf1e4e7086.png
Figure 1.6.19

Begin the process by dividing the leading terms to determine the leading term of the quotient 3x4÷x2=3x2. Take care to distribute and line up the like terms. Continue the process until the remainder has a degree less than 2.

6712e898302af732f5e899c4fc6abc3b.png
Figure 1.6.20

The remainder is x2. Write the answer with the remainder:

3x42x3+6x2+23x7x22x+5=3x2+4x1+x2x22x+5

Answer:

3x2+4x1+x2x22x+5

Polynomial long division takes time and practice to master. Work lots of problems and remember that you may check your answers by multiplying the quotient by the divisor (and adding the remainder if present) to obtain the dividend.

Exercise 1.6.3

Divide: 6x413x3+9x214x+63x2.

Answer

2x33x2+x423x2

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Key Takeaways

  • Polynomials are special algebraic expressions where the terms are the products of real numbers and variables with whole number exponents.
  • The degree of a polynomial with one variable is the largest exponent of the variable found in any term. In addition, the terms of a polynomial are typically arranged in descending order based on the degree of each term.
  • When adding polynomials, remove the associated parentheses and then combine like terms. When subtracting polynomials, distribute the 1, remove the parentheses, and then combine like terms.
  • To multiply polynomials apply the distributive property; multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
  • When dividing by a monomial, divide all terms in the numerator by the monomial and then simplify each term.
  • When dividing a polynomial by another polynomial, apply the division algorithm.

Exercise 1.6.4

Write the given polynomials in standard form.

  1. 1xx2
  2. y5+y2
  3. y3y2+5y3
  4. 812a2+a3a
  5. 2x2+6x5x3+x4
  6. a35+a2+2a4a5+6a
Answer

1. x2x+1

3. y33y2+y+5

5. x45x3x2+6x+2

Exercise 1.6.5

Classify the given polynomial as a monomial, binomial, or trinomial and state the degree.

  1. x2x+2
  2. 510x3
  3. x2y2+5xy6
  4. 2x3y2
  5. x41
  6. 5
Answer

1. 7. Trinomial; degree 2

3. Trinomial; degree 4

5. Binomial; degree 4

Exercise 1.6.6

State whether the polynomial is linear or quadratic and give the leading coefficient.

  1. 19x2
  2. 10x2
  3. 2x3
  4. 100x
  5. 5x2+3x1
  6. x1
  7. x62x2
  8. 15x
Answer

1. Quadratic, 9

3. Linear, 2

5. Quadratic, 5

7. Quadratic, 2

Exercise 1.6.7

  1. (5x23x2)+(2x26x+7)
  2. (x2+7x12)+(2x2x+3)
  3. (x2+5x+10)+(x210)
  4. (x21)+(4x+2)
  5. (10x2+3x2)(x26x+1)
  6. (x23x8)(2x23x8)
  7. (23x2+34x1)(16x2+52x12)
  8. (45x258x+106)(310x223x+35)
  9. (x2y2+7xy5)(2x2y2+5xy4)
  10. (x2y2)(x2+6xy+y2)
  11. (a2b2+5ab2)+(7ab2)(4a2b2)
  12. (a2+9ab6b2)(a2b2)+7ab
  13. (10x2y8xy+5xy2)(x2y4xy)+(xy2+4xy)
  14. (2m2n6mn+9mn2)(m2n+10mn)m2n
  15. (8x2y25xy+2)(x2y2+5)+(2xy3)
  16. (x2y2)(5x22xyy2)(x27xy)
  17. (16a22ab+34b2)(53a2+45b2)+118ab
  18. (52x22y2)(75x212xy+73y2)12xy
  19. (x2n+5xn2)+(2x2n3xn1)
  20. (7x2nxn+5)(6x2nxn8)
  21. Subtract 4y3 from y2+7y10.
  22. Subtract x2+3x2 from 2x2+4x1.
  23. A right circular cylinder has a height that is equal to the radius of the base, h=r. Find a formula for the surface area in terms of h.
  24. A rectangular solid has a width that is twice the height and a length that is 3 times that of the height. Find a formula for the surface area in terms of the height.
Answer

1. 7x29x+5

3. 2x2+5x

5. 9x2+9x3

7. 12x274x12

9. x2y2+2xy1

11. 2a2b2+12ab8

13. 9x2y+6xy2

15. 7x2y23xy6

17. 32a258ab120b2

19. 32a258ab120b2

21. y2+3y7

23. SA=4πh2

Exercise 1.6.8

Multiply

  1. 8x22x
  2. 10x2y5x3y2
  3. 2x(5x1)
  4. 4x(3x5)
  5. 7x2(2x6)
  6. 3x2(x2x+3)
  7. 5y4(y22y+3)
  8. 52a3(24a26a+4)
  9. 2xy(x27xy+y2)
  10. 2a2b(a23ab+5b2)
  11. xn(x2+x+1)
  12. xn(x2nxn1)
  13. (x+4)(x5)
  14. (x7)(x6)
  15. (2x3)(3x1)
  16. (9x+1)(3x+2)
  17. (3x2y2)(x25y2)
  18. (5y2x2)(2y23x2)
  19. (3x+5)(3x5)
  20. (x+6)(x6)
  21. (a2b2)(a2+b2)
  22. (ab+7)(ab7)
  23. (4x5y2)(3x2y)
  24. (xy+5)(xy)
  25. (x5)(x23x+8)
  26. (2x7)(3x2x+1)
  27. (x2+7x1)(2x23x1)
  28. (4x2x+6)(5x24x3)
  29. (x+8)2
  30. (x3)2
  31. (2x5)2
  32. (3x+1)2
  33. (a3b)2
  34. (7ab)2
  35. (x2+2y2)2
  36. (x26y)2
  37. (a2a+5)2
  38. (x23x1)2
  39. (x3)3
  40. (x+2)3
  41. (3x+1)3
  42. (2x3)3
  43. (x+2)4
  44. (x3)4
  45. (2x1)4
  46. (3x1)4
  47. (x2n+5)(x2n5)
  48. (xn1)(x2n+4xn3)
  49. (x2n1)2
  50. (x3n+1)2
  51. Find the product of 3x2 and x25x2.
  52. Find the product of x2+4 and x31.
  53. Each side of a square measures 3x3 units. Determine the area in terms of x.
  54. Each edge of a cube measures 2x2 units. Determine the volume in terms of x.
Answer

1. 16x3

3. 10x22x

5. 14x342x2

7. 5y6+10y515y4

9. 2x3y14x2y2+2xy3

11. xn+2+xn+1+xn

13. x2x20

15. 6x211x+3

17. 3x416x2y2+5y4

19. 9x225

21. a4b4

23. 12x315x2y24xy+5y3

25. x38x2+23x40

27. 2x4+11x324x24x+1

29. x2+16x+64

31. 4x220x+25

33. a26ab+9b2

35. x4+4x2y2+4y4

37. a42a3+11a10a+25

39. x39x2+27x27

41. 27x3+27x2+9x+1

43. x4+8x3+24x2+32x+16

45. 16x432x3+24x28x+1

47. x4n25

49. x4n2x2n+1

51. 3x317x2+4x+4

53. 9x6 square units

Exercise 1.6.9

Divide.

  1. 125x5y225x4y2
  2. 256x2y3z564x2yz2
  3. 20x312x2+4x4x
  4. 15x475x3+18x23x2
  5. 12a2b+28ab24ab4ab
  6. 2a4b3+16a2b2+8ab32ab2
  7. x3+x23x+9x+3
  8. x34x29x+20x5
  9. 6x311x2+7x62x3
  10. 9x39x2x+13x1
  11. 16x3+8x239x+174x3
  12. 12x356x2+55x+302x5
  13. 6x4+13x39x2x+63x+2
  14. 25x410x3+11x27x+15x1
  15. 20x4+12x3+9x2+10x+52x+1
  16. 25x445x326x2+36x115x2
  17. 3x4+x21x2
  18. x4+x3x+3
  19. x310x2
  20. x3+15x+3
  21. y5+1y+1
  22. y6+1y+1
  23. x44x3+6x27x1x2x+2
  24. 6x4+x32x2+2x+43x2x+1
  25. 2x37x2+8x3x22x+1
  26. 2x4+3x36x24x+3x2+x3
  27. x4+4x32x24x+1x21
  28. x4+x1x2+1
  29. x3+6x2y+4xy2y3x+y
  30. 2x33x2y+4xy23y3xy
  31. 8a3b32ab
  32. a3+27b3a+3b
  33. Find the quotient of 10x211x+3 and 2x1.
  34. Find the quotient of 12x2+x11 and 3x2.
Answer

1. 5x

3. 5x23x+1

5. 3a+7b1

7. x22x+3

9. 3x2x+2

11. 4x2+5x614x3

13. 2x3+3x25x+3

15. 10x3+x2+4x+3+22x+1

17. 3x3+6x2+13x+26+51x2

19. x2+2x+42x2

21. y4y3+y2y+1

23. x23x+13x2x+2

25. 2x3

27. x2+4x1

29. x2+5xyy2

31. 4a2+2ab+b2

33. 5x3

Footnotes

112An algebraic expression consisting of terms with real number coefficients and variables with whole number exponents.

113The exponent of the variable. If there is more than one variable in the term, the degree of the term is the sum their exponents.

114The largest degree of all of its terms.

115A polynomial where each term has the form anxn, where an is any real number and n is any whole number.

116The coefficient of the term with the largest degree.

117Polynomial with one term.

118Polynomial with two terms.

119Polynomial with three terms.

120A polynomial with degree 0.

121A polynomial with degree 1.

122A polynomial with degree 2.

123A polynomial with degree 3.

124The trinomials obtained by squaring the binomials (a+b)2=a2+2ab+b2 and (ab)2=a22ab+b2.

125The special product obtained by multiplying conjugate binomials (a+b)(ab)=a2b2.

126The binomials (a+b) and (ab).

127The process of dividing two polynomials using the division algorithm.

128Terms with zero coefficients used to fill in all missing exponents within a polynomial.


This page titled 1.6: Polynomials and Their Operations is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform.

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