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2 Polynomial and Rational Functions
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,
P( x) x 4 6 x 2 5
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,
P( x) x 5
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
P( x) 5
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
P( x) an x n an 1 xn
1
an 2 x n
2
 a1 x a0
where n is a non negative integer and an 0 . The numbers ai , i 0, 1, 2,, n are the
coefficients of the polynomial. The number a0 is called the constant coefficient of the
polynomial, which is the y – intercept when we plot the polynomial. The number an 0
is the leading coefficient of the polynomial and an x n 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 P ( x) x n . Below we have the graphs of
monomials when n is odd or even.
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2 Polynomial and Rational Functions
If a polynomial consists of
just a single term, then it
is called a monomial. The
simplest polynomial
functions are the
monomials P ( x) x n .
Adjacent graphs are
showing monomials for n
is odd or even.
n =1
n =2
n =3
n =4
n =5
Fig 2.1
n =7
Quadratic polynomial
Let us consider a quadratic polynomial of the form P( x) a( x h) 2 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 P( x) ax 2 bx c . We will now produce few examples
on finding standard form of quadratic function from its general form.
1. Transform P( x) x 2 8 x 5 into its standard form and find its vertex. Also
discuss the transformation of the graph with the monomial P ( x) x 2 (See
Section 1.6)
Solution
P( x) x 2 8 x 5
( x 4) 2 42 5
take half of 8 and subtract square of it
2
( x 4) 11
Comparing with standard form the vertex is at ( 4, 11)
To get the graph of P( x) ( x 4) 2 11 we transform the graph of P ( x)
horizontally 4 units to the left and then 11 units downward vertically.
2. Transform P ( x)
x2
2 x 2 8 x 5 into its standard form and find its vertex. Also
discuss the transformation of the graph with the monomial P ( x)
x2
Solution
P( x)
2 x2 8x 5
2( x 2 4 x) 5
2[( x 2) 2 22 ] 5
take half of 4 and subtract square of it
2( x 2) 2 13
Comparing with standard form the vertex is at (2,13)
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2 Polynomial and Rational Functions
2( x 2) 2 13 we transform the graph of P ( x) x 2
To get the graph of P( 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 P ( x) x 2 then the corresponding
2( x 2) 2 13 will be (1 2, 2 1 13) (3,11)
point on P( x)
3. Transform P( x) 3 x 2 23x 5 into its standard form and find its vertex. Also
discuss the transformation of the graph with the monomial P ( x) x 2 . Consider a
point (1, 1) on P ( x) x 2 and discuss its corresponding position on
P( x) 3 x 2 23 x 5 .
Solution
P ( x) 3 x 2 23 x 5
3 x2
23
x
3
3
23
6
x
23
3 x
6
2
2
5
23
6
2
5
take half of 23/3 and subtract square of it
469
12
23 469
. The corresponding
,
6
12
23
469
29
433
position of the point (1, 1) will be 1
,1 3
,
6
12
6
12
Comparing with standard form the vertex is at
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 P( x) a( x h) 2 k , we have h
3, k
2
It passes through (1, 5), so we have 5 a(1 3) 4 a 1/16
1
The standard form of the function is P( x)
( x 3)2 4
16
Maximum and minimum values P( x) a( x h) 2 k
4
The vertex (h, k ) is the maximum point on the graph if a 0, ( shape)
The vertex (h, k ) is the minimum point on the graph if a 0, ( shape)
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2 Polynomial and Rational Functions
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
A( x) x(200 x)
x 2 200 x
( x 2 200 x)
x
y
( x 100) 2 10000
y
x
The area function has vertex at (100, 10000), which the maximum point as
a
1 0, ( shape) . 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
x y 10
P
xy
x(10 x)
x 2 10 x
( x 5)2 25
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
x y 11
P
xy
x( x 11)
x 2 11x
( x 11/ 2) 2 121/ 4
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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2 Polynomial and Rational Functions
Other Higher Order Polynomials
A cubic Polynomial y ax3 bx 2 cx d is the general form a cubic polynomial.
Similarly a fourth order polynomial looks like y ax 4 bx 3 cx 2 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 P( x) ax30 bx 2 cx d . When x approaches to
, the
polynomial P( x) also approach to
for a 0 and approaches to
for a 0 . On the
other hand when x approaches to
, the polynomial P ( x) ax 27 3x11 2 x 10
approaches to
for a 0 and approaches to
for a 0 . But for x approaches to
,
27
11
3x
2 x 10 approaches to
the polynomial P ( x) ax
for a 0 and approaches to
for a 0 .
P( x) an x n an 1 xn 1 an 2 x n 2  a1 x a0
x
x
x
P( x)
When a 0
P( x)
When a 0
P( x)
When a 0
P( x)
When a 0
P( x)
When a 0
P( x)
When a 0
8. Find the end behavior of the graph of P( x)
3x17
n is even
n is odd
20 x10 2 x 2 13
Solution
Look at the leading term 3x17 , which has odd exponent, graph will move in two
different directions.
When x
, 3x17
and x
, 3x17
. The graph moves upward on
the left and downward on the right side.
9. Find the end behavior of the graph of P( x)
3x16
20 x10 2 x 2 13
Solution
Look at the leading term 3x16 , which has even exponent, graph will move in one
direction only.
When x
, 3x16
and x
, 3x16
. The graph moves downward
on both the sides.
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2 Polynomial and Rational Functions
2.2 Rational Functions
P( x )
, Q( x) 0 are called the rational functions, where both
Q( x )
P( x) and Q( x) are polynomial functions. The domain of Q( x) consists of all real values
of x for which Q( x) 0 . For an example the domain of the rational function
Functions of the type f ( x)
f ( x)
3x 2 5
is all real numbers except x
x2 5x 6
2, 3 , the roots of Q ( x)
x2 5x 6
10. Find the domain of the following rational functions.
3x 2 5
3x 2 5 x
a) f ( x)
b)
g
(
x
)
x 2 x 12
x 1
Solution
3x 2 5
3x 2 5
. The domain consists of all real values of x
x 2 x 12 ( x 4)( x 3)
except –3 and 4.
The domain of the function is in interval notation ( , 3) ( 3, 4) (4, ) .
a) f ( x)
3x 2 5 x
. The domain consists of all real values of x except 0.
x
The domain of the function is in interval notation ( ,0) (0, ) . There is a hole on
the graph at x = 0.
b) g ( x)
Vertical asymptote, Horizontal asymptote, Slant asymptote and a Hole
P( x )
, Q( x) 0
Q( x )
Vertical asymptote If at x a the function f ( x) is not defined, Q(a) 0 but P(a) 0 .
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
We consider the rational function f ( x)
Note that when we right x
a we mean x a and x
As an example we see that f ( x)
x
2 and at x
3x 2 5
,x
x2 5x 6
a
x a
2, 3 has two vertical asymptotes at
3.
Horizontal asymptote The horizontal line y a is a horizontal asymptote to the graph of
or f ( x) a as x
.
f ( x) if f ( x) a as x
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2 Polynomial and Rational Functions
Note that a rational function cannot have more than one horizontal asymptote. We
P( x )
following three different cases for the rational function f ( x)
, Q( x) 0
Q( x )
The horizontal asymptote is y 0 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)
leading coefficient of P( x)
The horizontal asymptote is y
when both P( x) and
leading coeffiient of Q( x)
Q( x) are of the same degrees.
For example the polynomial f ( x)
asymptote y
3x 2 5
,x
x2 5x 6
2, 3 has a horizontal
3
3
1
The polynomial f ( x)
The polynomial f ( x)
3x 5
,x
x 5x 6
3x5 5
,x
x2 5x 6
2
2, 3 has horizontal asymptote y 0
2, 3 has no horizontal asymptote.
P( x )
, Q( x) 0 has a slant asymptote only if
Q( x )
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
.
3x3 5
Look at the example f ( x)
. Use long division to get the form
x2 5x 6
3x3 5
57 x 85
.
y
3x 15 2
2
x 5x 6
x 5x 6
57 x 85
When x
, the last term 2
0.
x 5x 6
And we have the slant asymptote y 3x 15
Slant asymptote The polynomial f ( x)
Note that the slant asymptote and horizontal asymptote cannot occur for a certain
polynomial.
Hole
P( x )
, Q( x) 0 has a common factor x a on both the
Q( x )
polynomial P( x) and Q( x) then the graph of f ( x) has a hole at x a rather than a
If the polynomial function f ( x)
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2 Polynomial and Rational Functions
vertical asymptote. In the graph of the polynomial f ( x)
3x 6
x 5x 6
2
3( x 2)
has
( x 2)( x 3)
a hole at x 2 and a vertical asymptote at x 3 .
x3 2 x 2 7 x 4
11. Graph the rational function g ( x)
x2 1
Solution
Observe that the graph has slant asymptote. Find it using long division as follows
x 2
2
3
2
x 1
x 2x 7 x 4
Subtract to get next line
x3
x
2 x2 6 x 4
2 x2
2
6x 6
3
2
x 2x 7x 4
6( x 1)
x 2
We find g ( x)
2
x 1
( x 1)( x 1)
The graph has a slant asymptote y x 2 a vertical asymptote at x
at x 1 .
y
y x 2
-2
-1
1
1 and a hole
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
1
, where we say that 7 is dividend, 3 is the divisor, 2 is the quotient and 1 is the
2
3
3
remainder
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2 Polynomial and Rational Functions
12. Divide a given polynomial P ( x) 3 x 3 5 x 7 by another polynomial
Q( x) 2x 3
Find a factor and multiply
2
3 / 2x
9 / 4 x 7 / 8 with 2x to get 3x3
2 x 3 3x3 0x2 5x 7
Subtract to get next line
3x3 9 / 2 x 2
9 / 2 x2 5x 7
9 / 2 x 2 27 / 4 x
7 / 4x 7
7 / 4 x 21/ 8
35/ 8
Thus we have
Dividend
Remainder
P( x)
Q( x)
3x
3
5x 7
2x 3
Divisor
Verify that 3x3 5 x 7
3 2
x
2
9
7
x
4
8
35 / 8
2x 3
Divisor
Quotient
3 2
x
2
9
7
x
(2 x 3) 35 / 8
4
8
Division Algorithm for Polynomials
For two polynomial functions P( x) and Q( x) 0 there are unique polynomial functions
q( x) and r ( x) such that
P( x)
r ( x)
q( x)
Q( x)
Q( x)
Or 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
iff P(a) 0 .
Let P( x) be a polynomial function. Then x a is a factor of P( x)
Note that from factor theorem we get following two statements because of iff (if and only
if)
1. If P(a) 0 , then x a is a factor of P( x)
2. If x a is a factor of P( x) , then P(a) 0
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2 Polynomial and Rational Functions
Proof
1. Let us divide P( x) by x a and we write the following by division algorithm
q( x) ( x a) r ,
P( x)
When x
a we find P(a)
q(a) (a a) r
Now if P(a) 0 then P( x)
x a and it is a factor of P( x)
r is a constant
r and we have
P( x) q( x) ( x a) P(a)
q( x) ( x a) , which means P( x) is divisible evenly by
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 P(a) q(a) (a a) 0
13. Show that x 3 is a factor of the polynomial P( x)
x 3 27
Solution
Using factor theorem we need to show that P(3) 0
Given P( x) x 3 27 , then P(3) 33 27 0
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 5x2 3x 7 by x 2 using synthetic division
Solution: we only write coefficients as follows and perform the operations
a =2:
2
5
5
-3
10
7
7
14
21
Rule: Pull down 5, then
multiply 5 by 2 and put below
–3, add and continue this way
The remainder is 21 and quotient is Q( x) 5x 7 and we have
5 x 2 3x 7
x 2
5x 7
21
x 2
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2 Polynomial and Rational Functions
15. Divide 5x2 3x 7 by x 2 using synthetic division
Solution
a =-2:
-2
5
5
-3
-10
-13
7
26
33
Rule: Pull down 5, then
multiply 5 by -2 and put
below –3, add and continue
this way.
The remainder is 33 and quotient is Q( x) 5x 13 and we have
5 x 2 3x 7
x 2
5x 13
33
x 2
16. Divide 5x2 3x 7 by 2 x 4 using synthetic division
Solution: We write 2x 4 2( x 2) then
a =2:
2
5
5
-3
10
7
5 x 2 3x 7
2( x 2)
7
14
21
Rule: Pull down 5, then
multiply 5 by 2 and put below
–3, add and continue this way
The remainder is 21/2 and quotient is Q( x) (5x 7) / 2 and we have
5 x 2 3x 7
2x 4
5x 7
2
21
2( x 2)
The polynomial of degree n can have at most n real zeros. For example the polynomial
Q( x) 5 x 2 15 x 10 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 Q( x) 5( x 1)( x 2) . The
polynomial Q ( x) x 2 x 2 does not have any real zero and also no real factors. Look
at the graphs of these two functions below:
a) Q( x) 5x2 15x 10
b) Q( x) x2 x 2
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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2 Polynomial and Rational Functions
17. Find all real factors and real zeros of the polynomial f ( x)
Solution: Remember the formula x 2 a 2 ( x a )( x a )
f ( x)
x 4 81
x 4 81 ( x 2 ) 2 92
( x 2 9)( x 2 9)
( x 3)( x 3)( x
The given polynomial has two real zeros at x
intercepts.
2
-3
0
3
9)
3, 3 , which are also the x –
18. Find all real factors and real zeros of the polynomial f ( x)
Solution:
f ( x) 2 x3 x 2 15 x
2 x3
x 2 15 x
x( x 2 2 x 15)
x(2 x 5)( x 3)
5
2
Number of zeros or roots: A polynomial of degree n has at most n roots or zeros.
p
Rational zero test: If the rational number , (where p and q are in lowest terms) is a
q
root of the polynomial
P( x) an x n an 1 x n 1  a1 x a0 .
Where the coefficients an , an 1 , a1 , a0 are integers with an 0, a0 0 , then
the term p is a factor of a0
and the term q is a factor of an
The zeros are at x
0, 3,
19. List all possible rational zeros of f ( x) 12 x 7 2 x 6 3x 3 10 x 3
Solution: q = 1, 2, 3, 4, 6, 12 are the factors of an 12 and p = 1,
3 are the factors of a0 3 .
1
2
4
p
The possible rational zeros are : 1, 2, 3, 4, 6, 12,
,
, and
3
3
3
q
3
2
20. List all possible zeros of f ( x) 2 x 3x 4 . Use synthetic division to find all
zeros of the given function.
1
Solution: We have the following possible zeros 1, 2, 4,
. Using synthetic
2
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
2
Other zeros can be found from the quadratic from 2x x 2 0 , but we do not have
real solution.
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2 Polynomial and Rational Functions
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 n 1, then the equation f ( x) 0 has at least one
complex zero.
Note that real zeros are also referred to as complex zeros.
The Linear Factorization Theorem
If f ( x) an xn an 1x n 1 an 2 x n 2  a1x a0 is a polynomial of degree n 1, an 0 ,
and if ci , i 1, 2,3, , n are the complex zeros (not necessarily distinct) then we have
f ( x) an ( x c1 )( x c2 )( x c3 )  ( x cn )
The terms ( x c1 ), ( x c2 ), ( x c3 ),  , ( x cn ) 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 ) = a2 b2 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
( x 2), ( x 4), ( x 7) and we can write the polynomial as
f ( x) 5( x 2)( x 4)( x 7) 5 x3 65 x 2 250 x 280
22. If 2, 4, and 3i are the zeros of the fourth degree polynomial with f (1) 3 , find
the polynomial.
Solution: The polynomial is
f ( x) a( x 2)( x 4)( x 3i)( x 3i )
a( x 2)( x 4)( x 2 9)
Now
f (1) 30 a(1 2)(1 4)(1 9)
a 1
Thus the polynomial is f ( x) 1( x 2)( x 4)( x 2 9) x 4 6 x 3 17 x 2 54 x 72
23. Find all zeros of f ( x)
2 x3 3x 2 4 .
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2 Polynomial and Rational Functions
Solution: In example 20, we have found that x = 2 is a zero and other zeros are in
2 x2
x 2 0 , where x
1
12 4(2)(2)
2(2)
1
15
4
1 i 15
are the
4
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 f ( x)
2 x3 3x 2 4
Solution: f ( x) 2 x 3 3x 2 4 has one variation of sing, it has one positive zero.
2 x3 3x 2 4 has no variation in signs, it does not have negative zero.
And f ( x)
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
f ( x) 5 x3 65 x 2 250 x 280
Solution: f ( x) 5 x3 65 x 2 250 x 280 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 f ( x) 5 x 2 125
Solution: f ( x) 5 x 2 125 has no variation in signs, it has no real positive zeros.
Also f ( x) 5 x 2 125 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 x 5i, 5i .
48