chapter 1: polynomial

By: Dr Norfaradilla Wahid

Prof. Dr. Abd Samad Hasan Basari

Chapter 1:

Polynomial

BIC 10303

CHAPTER OVERVIEW

POLYNOMIAL

1.1 Basic concept (W1)

1.2 Algebra operation for polynomial (W1)

1.3 Remainder theorem and factor theorem (W2)

1.4 Synthetic divider (W2)

1.5 Solving Quadratic equation (W3)

1.6 Solving Inequality (linear, quadratic, absolute)

(W3)

Basic Concept1.1

1.1 Basic Concept

Monomials are terms that contain variables with only

whole numbers as exponent

Polynomial is made up from a monomial or finite sum

of monomials.

-3y25x 14ab3 Term

Literal

factor

Numerical

coefficiant

1.1 Basic Concept (W1)

A polynomial is an algebraic expression in which the exponents of the variables are nonnegative integers and there are no variables in the denominator

• An expression that can be written in the form

• a + bx + cx2 + dx3 +ex4 + ….

• Things with Surds (e.g. x + 4x +1 ) and reciprocals (e.g. 1/x + x) are not polynomials

• The degree is the highest index

• e.g 4x5 + 13x3 + 27x is of degree 5

Set-up a multiplication table3x2 -5x -3

2x4

+4x2

6x6 -10x5 -6x4

+12x4 -20x3 -12x2

Gather like terms = 6x6 - 10x5 + 6x4 - 20x3 -12x2

They can be added: 2x2 - 5x - 3 + 2x4 + 4x2 + 2x = 2x4 + 6x2 -3x -3

(2x4 + 4x2)(3x2 - 5x - 3)Or multiplied:

6x6 -10x5 -6x4 +12x4 -20x3 -12x2

Polynomials can be combined to give

new polynomials

Function Notation

• An polynomials function can be written as

• f(x) = a + bx + cx2 + dx3 +ex4 + ….

• f(x) means ‘function of x’

• instead of y = ….

• e.g f(x) = 4x5 + 13x3 + 27x

• f(3) means ….

– “the value of the function when x=3”

– e.g. for f(x) = 4x5 + 13x3 + 27x

– f(3) = 4 x 35 + 13 x 33 + 27 x 3

– f(3) = 972 + 351 + 81 = 1404

x

f(x)

Example

Identify which expressions are polynomials. If an expression is

not a polynomial, explain why.

(a)

(b)

(c)

(d)

235 2 ++ xx

xx

35 −

9

232

1 −+− xx

Some polynomials have special names depending on

the number of terms contained in the polynomial.

A monomial is a polynomial containing one term. A

term may have more than one factor.

A binomial is a polynomial containing two terms.

A trinomial is a polynomial containing three terms.

4

3,7,5,2,3

22 a

xyabx−

4,52,3 2 yxxxx +−+

57

2,43, 2 −++−++

axxxcba

14 +x

83 3 −x

1425 2 −+ xx

2

Polynomials in one variable

The degree of a polynomial in one variable is the largest exponent of that variable.

A constant has no variable. It is a 0 degreepolynomial.

This is a 1st degree polynomial. 1st degree polynomials are linear.

This is a 2nd degree polynomial. 2nd degree polynomials are quadratic.

This is a 3rd degree polynomial. 3rd degree polynomials are cubic.

Classify the polynomials by degree and number of terms.

Polynomial

a.

b.

c.

d.

5

42 −x

xx +23

14 23 +− xx

Degree

Classify by

degreeClassify by

number of terms

Zero Constant Monomial

First Linear Binomial

Second Quadratic Binomial

Third Cubic Trinomial

Example

• Try this:

• Identify the degree of each term in the polynomial

Identify the Degree of Terms

Exponents of variable factors must be integers greater than 0.

For a term that is constant, the degree is 0.

For a term that has only one variable factor, the degree is the same as the exponent of the variable.

For a term that has more than one variable factor, the degree is the sum of the exponents of the variables.

3325 23 +−+ xxx

• A linear term has degree 1

• A quadratic term has

degree 2

• A cubic term has degree 3

Identify the Degree of Polynomials

Identify the degree of each term of the polynomial.

Compare the degrees of each term of the

polynomial and select the greatest degree as the

degree of the polynomial.

• Try this:

• Identify the degree of the polynomial

• (a) has a degree of 2

• (b) has a degree of 3 and is a cubic poly

• (c) has a degree of 0 and is a constant

• (d) has a degree of 1 and is a linear polynomial

• (e)

2

1

7

543

125

23

2

−

−+−

−+

x

xxx

xx

𝑥2 +2

𝑦

Algebra operation for polynomial

(W1)1.2

1.2 Basic Operations with Polynomials

(1) Add and Subtract Polynomials

Combine the coefficients of the like terms using the rules for adding or subtracting signed numbers

(2) Multiply Polynomials

Multiply the coefficients using the rules for signed numbers.

Multiply the letter factors using the laws of exponents for factors with like base

Distribute if multiplying a polynomial by a monomial

(3) Divide Polynomials

Divide the coefficients using the rules for signed numbers.

Divide the variable factors using the laws of exponents for factors with like base

24

4

x

x−

( )22 332 yyyy +−+

( )xxx 42 2 −

Take a break

• What does simplify mean?

• - instruction to ‘simplify the expression’ are vague but

often used.

• To simplify an expression means to write the expression

using fewer terms or reduced or with lower coefficients

or exponents.

• Examine the expression and see which laws or operations

allow you to rewrite the expression in a simpler form.

Exercises

Simplify the algebraic expressions by combining

like terms.

(a) (e)

(b) (f)

(c) (g)

(d) mnm

aaaa

xx

xx

−+

−++

−

+

53

634

4

25

2323

55

33

( )

( ) ( )2524453

532

5

2323

22

32

+−−−+−−

+−+

+−+

mmmmmm

yyyy

yyyy

Exercises (multiply polynomials)

( )( )

( )( )

( )( )

( )( )

( )( )yaya

xx

aa

yy

xx

−+

++

−−

−

22

24

3

26

34

32

2

( )

( )

( )

( )( )653232

23

52

2

22

2

2

2

+−−+

−

−

+

xxxx

ba

x

x

Exercise (Divide polynomials)

Express the answers with positive exponents

x

xe

x

xd

x

xc

x

xb

y

ya

4

3

2

4

2

3

5

2.

12

3.

15

5.

4

4.

2

6.

−

−

−

2

23

3

23

234

2

26.

3.

3

1291518.

x

xxh

x

xxg

a

aaaaf

+

−

−−−

Divide polynomials (cont..)

There are two procedures for dividing polynomials:

1. Long division

2. Synthetic division

These procedures are especially valuable in

factoring and finding zeros of polynomial functions.

Break-even points in financial applications.

Intervals where polynomial functions are positive,

negative.

Divide P(x) = x3 – 7x – 6 by x – 4

P(x) = x3 – 7x – 6 is a dividend; Q(x) = x – 4 is a divisor

1. The divisor goes on the outside of the

box. The dividend goes on the inside of the

box.

2. Divide 1st term of dividend by first term

of divisor to get first term of the quotient.

Line up the first term of the quotient with

the term of the dividend that has the same

degree.

3. Take the term found in step 2 and

multiply it times the divisor.

4. Subtract this from the line above.

5. Repeat until done (until the degree of

the "new" dividend is less than the degree

of the divisor).

x3 –0x2 - 7x – 6x – 4

x3 –0x2 - 7x – 6x – 4

x2

x3 –0x2 - 7x – 6x – 4

x2

x3 –4x2

4x2 - 7x

1. Long division

23

Divide P(x) = x3 – 7x – 6 by x – 4

1. Long division (cont..)

P(x) = D(x)Q(x) + R

dividend =

divisor x quotient + remainder

P(x) = D(x)Q(x) + R

x3 – 7x – 6 =

(x - 4) (x2 + 4x + 9) + 30

x3 –0x2 - 7x – 6x – 4

x2 + 4x + 9

x3 –4x2

4x2 - 7x

4x2 – 16x

9x – 6

9x –36

30

Take a break

• Divide using long division and use the result to

factor the polynomial completely.

( )( )4

12175 2

−

−−

x

xx

Synthetic Divider (W2)1.3

Synthetic Division is a special case of dividing by a linear

factor binomial (x - k) or (jx - k) if x is variable, and j = const,

k = const.

Divider is a constant , which turns a binomial divisor into zero:

for a divisor (x – k) use the divider k;

for a divisor (jx – k) use the divider k/j.

Divide x4 - 3x + 5 by (x - 4),

(x- 4) is a divisor, then divider is 4.

2. Synthetic Division

2. Synthetic Division (cont..)

Divide x4 - 3x + 5 by (x - 4)

1. Write dividend in descending powers and insert 0's for any missing

terms x4 + 0x3 + 0x2 - 3x + 5

2. Find the number which turns the divisor into 0: x- 4 = 0, if x = 4

3. Write the coefficients of the dividend to the right:

4| 1 0 0 -3 5

4. Bring down the leading coefficient to the bottom row

4| 1 0 0 -3 5

1

5. Multiply c by the value just written on the bottom row

4| 1 0 0 -3 5

4 16 64 244

1 4 16 61 249

Note; Vertical

pattern : add

terms Diagonal

pattern :

Multiply by k

4| 1 0 0 -3 5

4 16 64 244

1 4 16 61 249

The numbers in the last row make up your coefficients of the

quotient as well as the remainder:

quotient = x3 + 4x2 + 16x + 61 remainder = 249

x4 - 3x + 5 = (x - 4)*(x3 + 4x2 + 16x + 61 ) + 249

2. Synthetic Division (cont..)

P(x) = D(x)Q(x) + R

dividend = divisor x quotient + remainder

Exercise

• Use synthetic division to divide ( )( )4

865 23

−

+−

x

xx

Remainder Theorem and Factor

Theorem (W2)1.4

3.4 Remainder Theorem and Factor

Theorem

1. Remainder Theorem

The remainder obtained in the synthetic division process has an

important interpretation, as described below

The Remainder Theorem tells you that synthetic division can be

used to evaluate a polynomial function.

That is, to evaluate a polynomial function f(x) when x = k,

divide f(x) by x – k. The remainder will be f(k)

Remainder Theorem

If a polynomial f(x) is divided by x – k , the remainder

is

r = f(k)

5 3611

3315

Example:

Use the Remainder Theorem to evaluate the following function at

Using synthetic division, you obtain the following

Because the remainder is r = 36, you can conclude that f(3) = 36.

This means that (3,36) is a point on the graph of f(x). You can check

this by substituting x=3 into original solution.

3 5 –4 3

Example:

f(x) = 5x2 – 4x + 3

f(3) = 5(3)2 – 4(3) + 3

f(3) = 5 ∙ 9 – 12 + 3

f(3) = 45 – 12 + 3

f(3) = 36

Check!

Notice that the value obtained when evaluating the

function at f(3) and the value of the remainder when

dividing the polynomial by x – 3 are the same.

Example:

Use Remainder Theorem and synthetic division to find the

function value

2. Factor Theorem

Lets look at this example again

2. Factor Theorem

• Another important theorem is the Factor Theorem. This theorem

states that you can test to see whether a polynomial has (x – k)

as a factor by evaluating the polynomial at x = k. If the result

is 0, (x – k) is a factor.

If the remainder is zero, that means f(k) = 0. This also means

that the divisor resulting in a remainder of zero is a factor of

the polynomial.

Factor Theorem

A polynomial f (x) has a factor (x – k)

if and only if f (k) = 0

Show that (x – 3) is factors of f(x) = x3 + 4x2 – 15x – 18. Then find

the remaining factors of f(x)

3 1 4 –15 –18

3 21 18

1 7 6 0

Using synthetic division, you obtain the following

Then function f(x) can be written as

f(x) = (x – 3)( x2 + 7x + 6 )

f(x) = (x – 3)(x + 6)(x + 1)

Therefore, (x – 3) is factors of f(x) = x3 + 4x2 – 15x – 18.

Remaining factors of f(x) are (x + 6)(x + 1)

Since the remainder is zero, f(3) =

0, show that (x – 3) is a factor of

x3 + 4x2 – 15x – 18.

Factoring x2 + 7x +6

x2 + 7x +6 = (x + 6)(x + 1)

Example:

The Factor Theorem

(x – 3)(x + 6)(x + 1).

Compare the factors

of the polynomials

to the zeros as seen

on the graph of

x3 + 4x2 – 15x – 18.

Example:

Given a polynomial and one of its factors, find the remaining

factors of the polynomial.

x3 – 11x2 + 14x + 80

x – 8

Ans:

(x – 8)(x – 5)(x + 2)

Factoring

Factoring is the breaking apart of a polynomial into

a product of other smaller polynomials. If you

choose, you could then multiply these factors

together, and you should get the original

polynomial.

For example, one set of factors for 24 is 6 and 4

because 6 times 4 = 24.

Root/Zero of a polynomial

A root or zero of a function is

a number that, when plugged in for the variable, makes

the function equal to zero. Thus, the roots of a

polynomial P(x) are values of x such that P(x) = 0.

3. The Rational Zero Test

The Rational Zero Theorem:

Suppose P(x) = anxn + an-1x

n-1 + an-2xn-2+ …+ a2x

2 + a1x + a0

is a polynomial with integer coefficients, and

Rational zero = p/q is a rational zero of P(x).

Then p is a factor of constant term, q is a factor of leading

coefficient.

Find the rational zeros of f(x) =1x3 - 3x2 - 2x + 6

The constant term, p = 6 and leading

coefficient, q = 1

p = {1, 2, 3, 6}

q = {1}

Possible rational-zeros:

p/q = {1, 2, 3, 6}

f(-1) = (-1)3 - 3(-1)2 -2(-1) + 6 = 4

f(1) = (1)3 - 3(1)2 -2(1) + 6 = 2

f(-2) = (-2)3 - 3(-2)2 -2(-2) + 6 = -10

f(2) = (2)3 - 3(2)2 -2(2) + 6 = -2

f(-3) = (-3)3 - 3(-3)2 -2(-3) + 6 = -42

f(3) = (3)3 - 3(3)2 -2(3) + 6 = 0

List all rational numbers whose numerators

are factors of the constant terms and whose

denominators are factors of the leading

coefficient.

List all the possible rational zeros

Use trial-and-error method to determine which ,

if any, are actual zeros of the polynomial.

(x - 3) is a factor, gives that x = 3 is a

rational zero

f(x) =x3 - 3x2 - 2x + 6 = (x - 3)(…………….…)

Cont. Find the rational zeros of f(x) =x3 - 3x2 - 2x + 6

Now find the remaining factor

Applying synthetic division

Rewrite the polynomial by including the factor

Factor of x2 – 2

produce x = - 2 and x = 2

f(x) = x3 - 3x2 - 2x + 6 = (x - 3)(…………….…)?

x3 - 3x2 - 2x + 6

1 -3 -2 6

3 0 -6

1 0 -2 0

x2 - 2

3

f(x) = x3 - 3x2 - 2x + 6 = (x - 3)(x2 - 2)

f(x) = (x - 3)( x - 2)(x + 2)

We conclude that the rational zeros of f are x = 3 , x = - 2 and x = 2

If a leading coefficient is not 1, the list of possible

rational zeros can increase dramatically.

E.g.

p = {1, 3}

q = {1, 2}

Possible rational zeros : p/q = {1, 3, 1

2,

3

2}

Rational Zero Test/Theorem can be used to find all

the rational zeros of a polynomial

Rational Zero Test often useful when we need to

factorize a polynomial with degree higher than 2!

Example:

Find the rational zeros of the following

f(x) = x4 - 6x2 - 27

Solving Quadratic Equations (W3)1.5

1.5 Solving Quadratic Equations

In this section, we will learn 4 methods for solving

quadratic equations:

i) Factoring

ii) Extracting square roots

iii) Completing square

iv) The Quadratic formula.

49

(i) Factoring

Some quadratic equations can be solved by factoring and

using basic property of real numbers called Zero-Factor

Property.

Be sure you see that Zero-Factor Property works only for

equations written in general form (in which the right side of

the equation is zero). So, all terms must be collected on one

side before factoring.

To use this property, write the left hand side of the general

form of quadratic equation as the product of two linear

factors.

Then find the solutions of the quadratic equation by setting

each linear factor equal to zero.

50

Factoring51

Exercise52

(ii) Extracting Square Roots53

Exercise

Solve each equation by extracting square roots.

54

(iii) Completing the Square

… is to write a quadratic function as a perfect square. Here are

some examples of perfect squares!

x2 + 6x + 9

x2 - 10x + 25

x2 + 12x + 36

Try to factor these (they’re easy).

Perfect Square Trinomials

x2 + 6x + 9

x2 - 10x + 25

x2 + 12x + 36

Can you see a numerical connection between …

6 and 9 using 3

-10 and 25 using -5

12 and 36 using 6

=(x+3)2

=(x-5)2

=(x+6)2

For a perfect square, the following relationships will always be

true …

x2 + 6x + 9

x2 - 10x + 25

The Perfect Square Connection

Half of these values squared … are these values

The Perfect Square Connection

In the following perfect square trinomial, the constant term is missing. Can you predict what it might be?

X2 + 14x + ____

Find the constant term by squaring half the coefficient of the linear term.

(14/2)2

X2 + 14x + 49

Perfect Square Trinomials

Create perfect square

trinomials.

x2 + 20x + ___

x2 - 4x + ___

x2 + 5x + ___

100

4

25/4

Solve the following equation by completing the square:

Step 1: Move the constant term (i.e. the number) to right side of the equation

2 8 20 0x x+ − =

2 8 20x x+ =

Solving Quadratic Equations by

Completing the Square

Step 2: Find the term that

completes the square on the left

side of the equation. Add that

term to both sides.

2 8 =20 + x x+ +

21( ) 4 then square it, 4 16

28• = =

2 8 2016 16x x+ + = +



Step 3: Factor the

perfect square

trinomial on the left

side of the

equation. Simplify

the right side of the

equation.

2 8 2016 16x x+ + = +

2

( 4)( 4) 36

( 4) 36

x x

x

− − =

− =



Solve this by

extracting the square root𝑥 + 4 = ± 36

𝑥 = −4 ± 36



Step 4: Set up

the two

possibilities

and solve

4 6

4 6 an

d 4 6

10 and 2 x=

x

x x

x

= −

=

= −

− − = − +

Completing the Square-Example #2

Solve the following equation by

completing the square:

Step 1: Move the constant to

the right side of the equation.

22 7 12 0x x− + =

22 7 12x x− = −



Step 2: Find the term that

completes the square on the left

side of the equation. Add that

term to both sides.

The quadratic coefficient must be

equal to 1 before you complete

the square, so you must divide

all terms by the quadratic

coefficient first.

2

2

2

2 7

2

2 2 2

7 12

7

2

=-12 +

6

x x

x x

xx

− +

− +

− +

= − +

= − +

21 7 7 49

( ) then square it, 2 62 4 4 1

7 • = =

2 49 49

16 1

76

2 6x x− + = − +



Step 3: Factor the

perfect square

trinomial on the left

side of the

equation. Simplify

the right side of the

equation.

2

2

2

76

2

7 96 49

4 16 16

7 47

4

49 49

16 1

16

6x x

x

x

− + = − +

− = − +

− = −

Use calculator to do this!Use calculator to do this!Use calculator to do this!Use calculator to do this!Use calculator to do this!

(iv) Quadratic Formula67

Quadratic Formula68

Example69

Exercises:

a) b)

70

Applications (a) Find the dimension of the room

71

Solving Inequality (linear, quadratic, absolute)

(W3)1.6

Properties of Inequalities73 Chapter 0

Linear Inequality74

An inequality in one variable is a linear inequality if

it can be written in one of the following forms.

The solution set can be written as set notation, (or

represented as interval notation or graph)

𝑎𝑥 + 𝑏 ≤ 0 𝑎𝑥 + 𝑏 < 0 𝑎𝑥 + 𝑏 ≥ 0 𝑎𝑥 + 𝑏 > 0

Example: Solving a Linear Inequality75

Quadratic Inequality76

Use test intervals to solve quadratic inequalities

Solving Quadratic Inequality77

Example: Solve the inequality 𝑥2 − 5x < 0.

First, find the critical numbers of 𝑥2 − 5x < 0 by adding the solutions of the

equation 𝑥2 − 5x = 0𝑥2 − 5x = 0𝑥 𝑥 − 5 = 0x = 0, x = 5

Test intervals are (−∞, 0), (0,5) 𝑎𝑛𝑑 (5,+∞)

Because the inequality is

satisfied by the middle

test only, we can

conclude that the

solution set in the

interval (0,5)

Critical numbers

Exercise78

Solve the inequality 2𝑥2 + 5𝑥 ≥ 12.

Absolute Value Equation79

Consider the absolute value equation

𝑥 = 3

The only solutions are x=-3 and x=3, because these

are the two real numbers whose distance from zero is

3.

Recap!

For |x|=a,

The solution is given by

x=-a or x=a

Absolute Inequality80

Example:

Solving an absolute value inequality81

Example:


Next Chapter

Chapter 2:

SEQUENCE AND SERIES

2.1 Sequence, series

2.2 Arithmetic sequence and series

2.3 Geometric sequence and series

chapter 1: polynomial

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