chapter 2 polynomial and rational functionsfiroz/m170/170ch2.pdf2 polynomial and rational functions...

2 Polynomial and Rational Functions

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Chapter 2 Polynomial and Rational Functions

We have discussed about polynomial expressions in Chapter 0. In this section we will

discuss about functions defined by polynomial expressions, also called polynomial

functions. For example, 4 2( ) 6 5P x x x

is a polynomial function. Polynomial functions are always continuous with domain entire

real line and easy to evaluate because they are defined using only addition, subtraction,

and multiplication. The graphs of polynomial functions can increase or decrease several

times. These functions are mostly useful in modeling many real life problems.

2.1 Polynomial Functions and Their Graphs

The graphs of polynomials of degree zero or one are straight lines (Section 1.3). For

example,

( ) 5P x x

is a polynomial of degree one and the graph of this polynomial is a straight line with

slope 1 and y – intercept 5. On the other hand

( ) 5P x

is polynomial of degree zero and the graph of this polynomial is a straight line with slope

0 and y – intercept 5 (a horizontal line).

The graphs of polynomials of degree two are parabolas (Section 1.5).

A polynomial function of general degree n is a function given by

1 2

1 2 1 0( ) n n n

n n nP x a x a x a x a x a

where n is a non negative integer and 0na . The numbers ia , 0, 1, 2, ,i n are the

coefficients of the polynomial. The number 0a is called the constant coefficient of the

polynomial, which is the y – intercept when we plot the polynomial. The number 0na

is the leading coefficient of the polynomial and n

na x is called the leading term.

Monomial

If a polynomial consists of just a single term, then it is called a monomial. The simplest

polynomial functions are the monomials ( ) nP x x . Below we have the graphs of

monomials when n is odd or even.


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If a polynomial consists of

just a single term, then it

is called a monomial. The

simplest polynomial

functions are the

monomials ( ) nP x x .

Adjacent graphs are

showing monomials for n

is odd or even.

n =1 n =2 n =3

n =4 n =5 n =7

Fig 2.1

Quadratic polynomial

Let us consider a quadratic polynomial of the form 2( ) ( )P x a x h k , which is known

as standard form. It is well known that the graph of a quadratic polynomial is a parabola.

The vertex of the parabola defined above is the point ( , )h k . The quadratic polynomial

has also a general form given as 2( )P x ax bx c . We will now produce few examples

on finding standard form of quadratic function from its general form.

1. Transform2( ) 8 5P x x x into its standard form and find its vertex. Also

discuss the transformation of the graph with the monomial 2( )P x x (See

Section 1.6)

Solution 2

2 2

2

( ) 8 5

( 4) 4 5 take half of 8 and subtract square of it

( 4) 11

P x x x

x

x

Comparing with standard form the vertex is at ( 4, 11)

To get the graph of 2( ) ( 4) 11P x x we transform the graph of 2( )P x x

horizontally 4 units to the left and then 11 units downward vertically.

2. Transform2( ) 2 8 5P x x x into its standard form and find its vertex. Also

discuss the transformation of the graph with the monomial 2( )P x x

Solution 2

2

2 2

2

( ) 2 8 5

2( 4 ) 5

2[( 2) 2 ] 5 take half of 4 and subtract square of it

2( 2) 13

P x x x

x x

x

x

Comparing with standard form the vertex is at (2,13)


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To get the graph of 2( ) 2( 2) 13P x x we transform the graph of 2( )P x x

horizontally 2 units to the right, make a reflection along x - axis, move the graph 13

units upward vertically, and then have vertical stretch (multiply each y coordinate by

- 2).

Note: Observe that if we take the point (1, 1) on 2( )P x x then the corresponding

point on 2( ) 2( 2) 13P x x will be (1 2, 2 1 13) (3,11)

3. Transform 2( ) 3 23 5P x x x into its standard form and find its vertex. Also

discuss the transformation of the graph with the monomial 2( )P x x . Consider a

point (1, 1) on 2( )P x x and discuss its corresponding position on 2( ) 3 23 5P x x x .

Solution 2

2

2 2

2

( ) 3 23 5

233 5

3

23 233 5 take half of 23/3 and subtract square of it

6 6

23 4693

6 12

P x x x

x x

x

x

Comparing with standard form the vertex is at 23 469

,6 12

. The corresponding

position of the point (1, 1) will be 23 469 29 433

1 , 1 3 ,6 12 6 12

4. Find the quadratic function whose graph is a parabola with vertex at (-3, 4) and

passes through (1, 5)

Solution

Let us consider the function 2( ) ( )P x a x h k , we have 3, 4h k

It passes through (1, 5), so we have 25 (1 3) 4 1/16a a

The standard form of the function is 21( ) ( 3) 4

16P x x

Maximum and minimum values 2( ) ( )P x a x h k

The vertex ( , )h k is the maximum point on the graph if 0, ( shape)a

The vertex ( , )h k is the minimum point on the graph if 0, ( shape)a


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Applications

5. Find the area and dimensions of the largest rectangular field that can be enclosed

with 400 feet of fence.

Solution

Suppose x denote the length and y denote the width of the rectangular field, a s

shown in the given figure

Perimeter = x + y + x + y = 2x + 2y = 400

Area = xy

From perimeter we have y = 200 – x

And now Area 2

2 2

( ) (200 ) 200

( 200 ) ( 100) 10000

A x x x x x

x x x

x

y y

x

The area function has vertex at (100, 10000), which the maximum point as

1 0, ( shape)a . The maximum area is 10000 square feet and the dimension

of the field is 100 100 .

6. The sum of two real numbers is 10. Find the numbers so that their product is a

maximum.

Solution

Suppose that the real numbers are x and y. Then we have

2

2

10

(10 ) 10

( 5) 25

x y

P xy x x x x

x

The graph of P is an upside down parabola, which has maximum at the vertex

(5, 25) . The real numbers are x = 5 and y = 10 - 5 = 5 and the their product is 25.

7. The difference of two real numbers is 11. Find the numbers so that their product

is a minimum.

Solution

Suppose that the real numbers are x and y. Then we have

2

2

11

( 11) 11

( 11/ 2) 121/ 4

x y

P xy x x x x

x

The graph of P is an upward parabola, which has minimum at the vertex

(11/ 2, 121/ 4) . The real numbers are x = 11/2 = 5.5 and y = 5.5-11 = -5.5 and the

their product is –121/4.


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Other Higher Order Polynomials

A cubic Polynomial 3 2y ax bx cx d is the general form a cubic polynomial.

Similarly a fourth order polynomial looks like 4 3 2y ax bx cx dx e and so on.

End Behavior and Leading Term

The end behavior of a polynomial is a description of what happens to the graph when we

consider x approach to either positive infinity or negative infinity. Basically we will

consider the graph of polynomials with even and odd exponents to the leading term. Let

us consider the polynomial 30 2( )P x ax bx cx d . When x approaches to , the

polynomial ( )P x also approach to for 0a and approaches to for 0a . On the

other hand when x approaches to , the polynomial 27 11( ) 3 2 10P x ax x x

approaches to for 0a and approaches to for 0a . But for x approaches to ,

the polynomial 27 11( ) 3 2 10P x ax x x approaches to for 0a and approaches to

for 0a . 1 2

1 2 1 0( ) n n n

n n nP x a x a x a x a x a

x ( )P x When 0a

n is even ( )P x When 0a

x ( )P x When 0a

n is odd ( )P x When 0a

x ( )P x When 0a

( )P x When 0a

8. Find the end behavior of the graph of 17 10 2( ) 3 20 2 13P x x x x

Solution

Look at the leading term 173x , which has odd exponent, graph will move in two

different directions.

When x , 173x and x , 173x . The graph moves upward on

the left and downward on the right side.

9. Find the end behavior of the graph of 16 10 2( ) 3 20 2 13P x x x x

Solution

Look at the leading term 163x , which has even exponent, graph will move in one

direction only.

When x , 163x and x , 163x . The graph moves downward

on both the sides.


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2.2 Rational Functions

Functions of the type ( )

( ) , ( ) 0( )

P xf x Q x

Q x are called the rational functions, where both

( )P x and ( )Q x are polynomial functions. The domain of ( )Q x consists of all real values

of x for which ( ) 0Q x . For an example the domain of the rational function 2

2

3 5( )

5 6

xf x

x x is all real numbers except 2, 3x , the roots of 2( ) 5 6Q x x x

10. Find the domain of the following rational functions.

a) 2

2

3 5( )

12

xf x

x x b)

23 5( )

1

x xg x

x

Solution

a) 2 2

2

3 5 3 5( )

12 ( 4)( 3)

x xf x

x x x x. The domain consists of all real values of x

except –3 and 4.

The domain of the function is in interval notation ( , 3) ( 3,4) (4, ) .

b) 23 5

( )x x

g xx

. The domain consists of all real values of x except 0.

The domain of the function is in interval notation ( ,0) (0, ) . There is a hole on

the graph at x = 0.

Vertical asymptote, Horizontal asymptote, Slant asymptote and a Hole

We consider the rational function ( )

( ) , ( ) 0( )

P xf x Q x

Q x

Vertical asymptote If at x a the function ( )f x is not defined, ( ) 0Q a but ( ) 0P a .

The vertical line x a is called the vertical asymptote for the graph of ( )f x . Observe

that

( )f x as x a , or ( )f x as x a

Note that when we right x a we mean x a and x a x a

As an example we see that 2

2

3 5( ) , 2, 3

5 6

xf x x

x x has two vertical asymptotes at

2x and at 3x .

Horizontal asymptote The horizontal line y a is a horizontal asymptote to the graph of

( )f x if ( )f x a as x or ( )f x a as x .


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Note that a rational function cannot have more than one horizontal asymptote. We

following three different cases for the rational function ( )

( ) , ( ) 0( )

P xf x Q x

Q x

The horizontal asymptote is 0y when the polynomial ( )P x is of lower order

than the polynomial ( )Q x

There is no horizontal asymptote when the polynomial ( )P x is of higher order

than the polynomial ( )Q x

The horizontal asymptote is leading coefficient of ( )

leading coeffiient of ( )

P xy

Q x when both ( )P x and

( )Q x are of the same degrees.

For example the polynomial2

2

3 5( ) , 2, 3

5 6

xf x x

x x has a horizontal

asymptote 3

31

y

The polynomial 2

3 5( ) , 2, 3

5 6

xf x x

x x has horizontal asymptote 0y

The polynomial 5

2

3 5( ) , 2, 3

5 6

xf x x

x x has no horizontal asymptote.

Slant asymptote The polynomial ( )

( ) , ( ) 0( )

P xf x Q x

Q x has a slant asymptote only if

the polynomial ( )P x is just one degree higher than the polynomial ( )Q x . The straight of

the form y mx b is the slant asymptote found by dividing ( )P x with ( )Q x and letting

x .

Look at the example 3

2

3 5( )

5 6

xf x

x x. Use long division to get the form

3

2 2

3 5 57 853 15

5 6 5 6

x xy x

x x x x.

When x , the last term 2

57 850

5 6

x

x x.

And we have the slant asymptote 3 15y x

Note that the slant asymptote and horizontal asymptote cannot occur for a certain

polynomial.

Hole

If the polynomial function ( )

( ) , ( ) 0( )

P xf x Q x

Q x has a common factor x a on both the

polynomial ( )P x and ( )Q x then the graph of ( )f x has a hole at x a rather than a


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vertical asymptote. In the graph of the polynomial 2

3 6 3( 2)( )

5 6 ( 2)( 3)

x xf x

x x x x has

a hole at 2x and a vertical asymptote at 3x .

11. Graph the rational function 3 2

2

2 7 4( )

1

x x xg x

x

Solution

Observe that the graph has slant asymptote. Find it using long division as follows

2x 2 1x 3 22 7 4x x x

3x x Subtract to get next line

22 6 4x x

22 2x

6 6x

We find 3 2

2

2 7 4 6( 1)( ) 2

1 ( 1)( 1)

x x x xg x x

x x x

The graph has a slant asymptote 2y x a vertical asymptote at 1x and a hole

at 1x .

y 2y x

-2 -1 1 x

Fig 2.2

2.3 Finding Factors and Zeros of polynomials

In this section we need to have the following concepts:

Long division

Synthetic division

Factoring polynomial

Quadratic formula

Long division

Remember that when we divide 7 by 3 we write the following

7 12

3 3, where we say that 7 is dividend, 3 is the divisor, 2 is the quotient and 1 is the

remainder


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12. Divide a given polynomial 3( ) 3 5 7P x x x by another polynomial

( ) 2 3Q x x

23/ 2 9 / 4 7 / 8x x

Find a factor and multiply

with 2x to get 33x

2 3x 3 23 0 5 7x x x

3 23 9 / 2x x Subtract to get next line

29/ 2 5 7x x

29 / 2 27 / 4x x

7 / 4 7x

7 / 4 21/8x

35/8

Thus we have

Dividend

3

2( ) 3 5 7 3 9 7 35 / 8

( ) 2 3 2 4 8 2 3

P x x xx x

Q x x x

Remainder

Divisor

Divisor Quotient

Verify that 3 23 9 73 5 7 (2 3) 35 /8

2 4 8x x x x x

Division Algorithm for Polynomials

For two polynomial functions ( )P x and ( ) 0Q x there are unique polynomial functions

( )q x and ( )r x such that

( ) ( )( )

( ) ( )

Or ( ) ( ) ( ) ( )

P x r xq x

Q x Q x

P x q x Q x r x

dividend quotient divisor remainder

Remainder Theorem Let ( )P x be a polynomial function. If ( )P x is divided by

x a , then the remainder is ( )P a .

Factor Theorem Let ( )P x be a polynomial function. Then x a is a factor of ( )P x

iff ( ) 0P a .

Note that from factor theorem we get following two statements because of iff (if and only

if)

1. If ( ) 0P a , then x a is a factor of ( )P x

2. If x a is a factor of ( )P x , then ( ) 0P a


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Proof

1. Let us divide ( )P x by x a and we write the following by division algorithm

( ) ( ) ( )P x q x x a r , r is a constant

When x a we find ( ) ( ) ( )P a q a a a r r and we have

( ) ( ) ( ) ( )P x q x x a P a

Now if ( ) 0P a then ( ) ( ) ( )P x q x x a , which means ( )P x is divisible evenly by

x a and it is a factor of ( )P x

2. Suppose that x a is a factor of ( )P x . Then there is a polynomial ( )q x such that

( ) ( ) ( )P x q x x a

When we replace x by a, we get ( ) ( ) ( ) 0P a q a a a

13. Show that 3x is a factor of the polynomial 3( ) 27P x x

Solution

Using factor theorem we need to show that (3) 0P

Given 3( ) 27P x x , then 3(3) 3 27 0P

Synthetic division

Synthetic division is shorter method of dividing a polynomial by a binomial of the form

x a only. This method of dividing uses only the coefficients of the terms. We produce

an example of this method.

14. Divide 25 3 7x x by 2x using synthetic division

Solution: we only write coefficients as follows and perform the operations

a =2: 2 5 -3 7 Rule: Pull down 5, then

multiply 5 by 2 and put below

–3, add and continue this way 10 14

5 7 21

The remainder is 21 and quotient is ( ) 5 7Q x x and we have

25 3 7 21

5 72 2

x xx

x x


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15. Divide 25 3 7x x by 2x using synthetic division

Solution

a =-2: -2 5 -3 7 Rule: Pull down 5, then

multiply 5 by -2 and put

below –3, add and continue

this way.

-10 26

5 -13 33

The remainder is 33 and quotient is ( ) 5 13Q x x and we have

25 3 7 33

5 132 2

x xx

x x

16. Divide 25 3 7x x by 2 4x using synthetic division

Solution: We write 2 4 2( 2)x x then 25 3 7

2( 2)

x x

x

a =2: 2 5 -3 7 Rule: Pull down 5, then

multiply 5 by 2 and put below

–3, add and continue this way 10 14

5 7 21

The remainder is 21/2 and quotient is ( ) (5 7) / 2Q x x and we have

25 3 7 5 7 21

2 4 2 2( 2)

x x x

x x

The polynomial of degree n can have at most n real zeros. For example the polynomial 2( ) 5 15 10Q x x x of degree two has exactly two zeros namely x = 1 and x = 2. The

factors of the polynomial are x –1 and x – 2. Notice that ( ) 5( 1)( 2)Q x x x . The

polynomial 2( ) 2Q x x x does not have any real zero and also no real factors. Look

at the graphs of these two functions below:

a) 2( ) 5 15 10Q x x x b) 2( ) 2Q x x x

Fig 2.3

Note that the x coordinates of x – intercepts are the real zeros of the function. If there is

no x – intercept of a graph there is no real zero.


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17. Find all real factors and real zeros of the polynomial 4( ) 81f x x

Solution: Remember the formula 2 2 ( )( )x a x a x a

4 2 2 2

2 2

2

( ) 81 ( ) 9

( 9)( 9)

( 3)( 3)( 9)

f x x x

x x

x x x

-3 0 3

The given polynomial has two real zeros at 3, 3x , which are also the x –

intercepts.

18. Find all real factors and real zeros of the polynomial 3 2( ) 2 15f x x x x

Solution: 3 2

2

( ) 2 15

( 2 15)

(2 5)( 3)

f x x x x

x x x

x x x

The zeros are at 5

0, 3,2

x

Number of zeros or roots: A polynomial of degree n has at most n roots or zeros.

Rational zero test: If the rational number p

q, (where p and q are in lowest terms) is a

root of the polynomial 1

1 1 0( ) n n

n nP x a x a x a x a .

Where the coefficients 1 1 0, , ,n na a a a are integers with 00, 0na a , then

the term p is a factor of 0a

and the term q is a factor of na

19. List all possible rational zeros of 7 6 3( ) 12 2 3 10 3f x x x x x

Solution: q = 1, 2, 3, 4, 6, 12 are the factors of 12na and p = 1,

3 are the factors of 0 3a .

The possible rational zeros are :p

q1, 2, 3, 4, 6, 12,

1

3,

2

3, and

4

3

20. List all possible zeros of 3 2( ) 2 3 4f x x x . Use synthetic division to find all

zeros of the given function.

Solution: We have the following possible zeros 1

1, 2, 4,2

. Using synthetic

division you can see that only x = 2 is a real zero of the function.

2 2 -3 0 -4

4 2 4

2 1 2 0

Other zeros can be found from the quadratic from 22 2 0x x , but we do not have

real solution.


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2.4 Complex Zeros of Polynomials

We have discussed about complex numbers of the form a ib in chapter 1. We also have

discussed about real zeros of a polynomial of degree n. In this section we will discuss

about complex zeros of a polynomial. The fundamental theorem of algebra (by Carl

Friedrich Gauss a German mathematician) is presented below.

The Fundamental Theorem of Algebra

If ( )f x is a polynomial of degree 1n , then the equation ( ) 0f x has at least one

complex zero.

Note that real zeros are also referred to as complex zeros.

The Linear Factorization Theorem

If 1 2

1 2 1 0( ) n n n

n n nf x a x a x a x a x a is a polynomial of degree 1, 0nn a ,

and if , 1, 2,3, ,ic i n are the complex zeros (not necessarily distinct) then we have

1 2 3( ) ( )( )( ) ( )n nf x a x c x c x c x c

The terms 1 2 3( ), ( ), ( ), , ( )nx c x c x c x c are called the factors of ( )f x .

Complex zeros: If a ib is a zero of a polynomial then a ib is also a zero of the

polynomial. Remember that ( a ib )( a ib ) = 2 2a b a pure real number.

21. If 2, 4, and 7 are the zeros of the third degree polynomial with leading coefficient

5, find the polynomial.

Solution: From the linear factorization theorem we have the following factors

( 2), ( 4), ( 7)x x x and we can write the polynomial as 3 2( ) 5( 2)( 4)( 7) 5 65 250 280f x x x x x x x

22. If 2, 4, and 3i are the zeros of the fourth degree polynomial with (1) 3f , find

the polynomial.

Solution: The polynomial is

2

( ) ( 2)( 4)( 3 )( 3 )

( 2)( 4)( 9)

f x a x x x i x i

a x x x

Now

(1) 30 (1 2)(1 4)(1 9)

1

f a

a

Thus the polynomial is 2 4 3 2( ) 1( 2)( 4)( 9) 6 17 54 72f x x x x x x x x

23. Find all zeros of 3 2( ) 2 3 4f x x x .


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Solution: In example 20, we have found that x = 2 is a zero and other zeros are in

22 2 0x x , where 21 1 4(2)(2) 1 15 1 15

2(2) 4 4

ix are the

complex zeros.

Finding number of positive and negative real zeros: Descartes rule of signs

Descartes rule of signs provides the number of positive and negative real zeros. The rule

is based on considering variation in sign between consecutive coefficients. For the

polynomial ( )f x has exactly r real positive zeros if there are r variations in signs and

there are m negative real zeros if ( )f x has exactly m variation in signs.

24. Find the number of positive and negative real zeros of 3 2( ) 2 3 4f x x x

Solution: 3 2( ) 2 3 4f x x x has one variation of sing, it has one positive zero.

And 3 2( ) 2 3 4f x x x has no variation in signs, it does not have negative zero.

Since the polynomial is of degree 3, other two zeros are imaginary. See example 23.

25. Find the number of positive and negative real zeros of 3 2( ) 5 65 250 280f x x x x

Solution: 3 2( ) 5 65 250 280f x x x x has 3 variations in signs; it has 3 positive

real zeros. As the polynomial is of degree 3, it does not have any negative or

complex (imaginary) zeros. See example 21.

26. Find the number of positive and negative real zeros of 2( ) 5 125f x x

Solution: 2( ) 5 125f x x has no variation in signs, it has no real positive zeros.

Also 2( ) 5 125f x x has nor variation in signs, it has no negative zeros. The

polynomial has exactly two complex zeros; one is the conjugate of the other. The

zeros are 5 , 5x i i .

chapter 2 polynomial and rational functionsfiroz/m170/170ch2.pdf2 polynomial and rational functions...

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