MATHEMATICS EXTENSION 2
4 UNIT MATHEMATICS
TOPIC 7: POLYNOMIALS
7.1
INTRODUCTION TO FUNCTIONS and POLYNOMIALS
A polynomial is a function of the form
y  f ( x )  a0  a1 x  a2 x 
2
n
 an x   ai x i
n
i 0
The degree of the polynomial is n (n integer n  0,1,2,
). Such a function is defined for all
values of x and x is finite. A polynomial is a single valued, continuous and differentiable
function of x.
A linear function (n = 1) is a polynomial of degree 1.
A polynomial of degree 2 (n = 2) is called a quadratic function
y  a0  a1 x  a2 x 2
The quadratic function is mostly expressed as
y  a x2  b x  c
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The graph of a quadratic function is a parabola. If there are real values of x for which y = 0,
the parabola will intersect the X-axis at
real roots
 b  b 2  4a c
x
2a
b 2  4a c  0
Polynomial functions are called single-valued functions because there is only one value of y
for each value of x. The function y 2  x is a multi-valued function since there are two
values of y for each value of x:  x1 and  x1
Functions can depend upon a number of variables. For example, the pressure p of a gas in a
container depends upon the volume V of the container and the temperature T of the gas.
p
n RT
V
variabels  p, T ,V 
constants  n, R 
This is an example of an explicit function, since the equation can be rearranged to make the
variables V or T the subject of the equation
p
n RT
V
V
n RT
p
T
pV
nR
explicit function
This is not the case for the equation below in regard to the variable V. This is an example of
an implicit function

n2 a 
 p  2  V  n b   n RT
V 

POLYNOMIALS
implicit function
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A useful classification of functions is into even and odd functions.
An even function of x is one that remains unchanged when the sign of x is reversed
f ( x)  f ( x)
even function
whereas an odd function changes sign
f ( x)   f ( x)
odd function
Online activity: Graphing Polynomials
Long Division of polynomials
Let P( x) , A( x) , Q( x) and R( x) be polynomial functions in x. Then we can divide the P( x) by
A( x) such that
P( x )  A( x ) Q( x )  R( x )
where
A( x) is the divisor
Q( x) is the quotient
R( x) is the remainder
The degree of R( x) must be less than that of A( x) . The functions Q( x) and R( x) are unique
when this condition is satisfied.
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Example P( x )  6 x 3  7 x 2  4 x  3 A( x )  3x  1
find Q( x) and R( x) .
Solution Make a table as shown below to do the long division to find Q( x) and R( x)
(1)
(2)
(3)
(4)
(5)
(6)
(7)
A(x)
3x+1
Q(x)
2x2
P(x)
(1)-(2)
3x+1
-3x
(3)-(4)
3x+1
7/3
(5)-(6) --> R(x)
x3
6x3
6x3
x2
-7x2
2x2
x1
4x
0
x0
-3
0
0
0
0
0
0
-9x2
-9x2
0
0
0
4x
-3x
7x
7x
0
-3
0
-3
7/3
-16/3
Q( x)  2 x 2  3x  7 / 3 R( x)  16/ 3
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Example P( x )  6 x 3  7 x 2  4 x  3 A( x )  3x  1
find Q( x) and R( x) .
Solution Make a table as shown below to do the long division to find Q( x) and R( x) \
(1)
(2)
(3)
(4)
(5)
A(x)
x2-x+1
P(x)
(1)-(2)
x2-x+1
-7
(3)-(4) --> R(x)
Q( x )  2 x  7
POLYNOMIALS
Q(x)
2x
x3
2x3
2x3
x2
-9x2
-2x2
x1
0
2x
x0
15
0
0
0
0
-7x2
-7x2
0
-2x
7x
-9x
15
-7
22
R( x )  9 x  22
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Roots of polynomials
Consider the polynomial equation of degree n
y  f ( x )  P( x )  a0  a1 x  a2 x 
2
n
 an x   ai x i  0
n
i 0
then real numbers x which satisfy this equation are called the real roots of the equation.
 For large values of x then y  f ( x )  an x n
 A polynomial where n is odd (odd degree) always has at least one real root
 At least one maximum or minimum value of the polynomial f ( x) occurs between any
two distinct real roots.
y  f ( x )  P( x )  a0  a1 x  a2 x 
2
n
 an x   ai x i  0
n
i 0
This equation can be expressed as
P( x )  an  x  1   x   2 
n
 x  n 1  x  n    an  x  i   0
i 1
The polynomial of degree n has n roots. Some of the roots maybe real and others may be
imagery, also there may be multiple roots (eg  2   3   6 ) .
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If  i tis a real, then the term  x  i  is a factor of P( x)
Quadratic Equation
Consider the quadratic equation
P( x )  a x 2  b x  c  0
Since the degree of the quadratic is n = 2, there are two roots which we designate as  and  .
Hence we can express the polynomial as
P( x )  a  x    x     0
We can find the relationships between the coefficients a, b and c and the two roots  and 
.
b
c
x2    x     0
a
a
x 2      x     0
 
POLYNOMIALS
b
a
 
c
a
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Cubic Equation
Consider the cubic equation
P( x )  a x 3  b x 2  c x  d  0
Since the degree of the cubic is n = 3, there are three roots which we designate as  ,  and  .
Hence we can express the polynomial as
P( x )  a  x    x    x     0
We can find the relationships between the coefficients a, b, c and d and the three roots  , 
and  .
b
c
d 
x3    x2    x     0
a
a
a
x 3        x 2           x      0
     
POLYNOMIALS
b
a
       
c
d
 
a
a
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Quartic Equation
Consider the quartic equation
P( x )  a x 4  b x 3  c x 2  dx  e  0
Since the degree of the cubic is n = 4, there are four roots which we designate as  ,   and 
Hence we can express the polynomial as
.
P( x )  a  x    x    x    x     0
We can find the relationships between the coefficients a, b, c and d and the three roots  , 
and  .
b
c
d 
e
   x3    x2    x     0
a
a
a
a
x 4          x 3                    x 2
x4
                 x      0
      
b
a
              
              
POLYNOMIALS
d
a
 
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c
a
e
a
9
Finding the real roots of a polynomial
Halving the interval
Suppose we have two values x1 and x2 such that P ( x1 ) P ( x2 )  0 and since P( x) is a continuous
function, there is a root of P( x) in the interval x1  x  x2 . Now calculate the value of x3 the midpoint in the interval x1 to x2.
x3  12  x1  x2 
If P ( x3 )  0 then x3 is the desired root.
If P ( x3 ) P ( x2 )  0 then replace x1 by x3
If P( x3 ) P( x1 )  0 then replace x2 by x3
and repeat the process until you get a reasonable of the value for the root.
This process is very tedious because you have to calculate P( x) many times. It is best to use a
spreadsheet and not a calculator. The example below shows how to use a spreadsheet.
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Example Find the roots of the equation P( x )  x 3  2 x 2  x  2  0
Enter values for x1 and x2 only
x1
p(x1)
start values
0.0000
2.0000
x3 --> x1
0.6500
0.7796
x3 --> x1
0.9750
0.0506
x3 --> x2
0.9750
0.0506
x3 --> x2
0.9750
0.0506
x3 --> x2
0.9750
0.0506
x3 --> x1
0.9953
0.0094
x3 --> x2
0.9953
0.0094
x3 --> x2
0.9953
0.0094
x3 --> x1
0.9979
0.0043
x3 --> x1
0.9991
0.0018
x2
1.3000
1.3000
1.3000
1.1375
1.0563
1.0156
1.0156
1.0055
1.0004
1.0004
1.0004
p(x2)
-0.4830
-0.4830
-0.4830
-0.2535
-0.1092
-0.0310
-0.0310
-0.0109
-0.0008
-0.0008
-0.0008
p(x1)*p(x2) < 1
-0.9660
-0.3766
-0.0244
-0.0128
-0.0055
-0.0016
-0.0003
-0.0001
0.0000
0.0000
0.0000
x3
0.6500
0.9750
1.1375
1.0563
1.0156
0.9953
1.0055
1.0004
0.9979
0.9991
0.9998
x2
2.4000
2.0500
2.0500
2.0500
2.0063
2.0063
2.0063
2.0008
p(x2)
1.9040
0.1601
0.1601
0.1601
0.0189
0.0189
0.0189
0.0023
p(x1)*p(x2) < 1
-1.0796
-0.0908
-0.0504
-0.0171
-0.0020
-0.0009
-0.0003
0.0000
x3
2.0500
1.8750
1.9625
2.0063
1.9844
1.9953
2.0008
1.9980
First root is x = +1.
Enter values for x1 and x2 only
x1
p(x1)
start values
1.7000
-0.5670
x3 --> x2
1.7000
-0.5670
x3 --> x1
1.8750
-0.3145
x3 --> x1
1.9625
-0.1069
x3 --> x2
1.9625
-0.1069
x3 --> x1
1.9844
-0.0459
x3 --> x1
1.9953
-0.0140
x3 --> x2
1.9953
-0.0140
Second root is x = 2
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Enter values for x1 and x2 only
x1
p(x1)
start values -2.0000
-12.0000
x3 --> x1
-1.2500
-1.8281
x3 --> x2
-1.2500
-1.8281
x3 --> x1
-1.0625
-1.0625
x3 --> x2
-1.0625
-0.3948
x3 --> x1
-1.0156
-0.0950
x3 --> x2
-1.0156
-0.0950
x3 --> x1
-1.0039
-0.0235
x3 --> x2
-1.0039
-0.0235
x3 --> x1
-1.0010
-0.0059
x3 --> x2
-1.0010
-0.0059
x2
-0.5000
-0.5000
-0.8750
-0.8750
-0.9688
-0.9688
-0.9922
-0.9922
-0.9980
-0.9980
-0.9995
p(x2)
1.8750
1.8750
0.6738
0.6738
0.1826
0.1826
0.0466
0.0466
0.0117
0.0117
0.0029
p(x1)*p(x2) < 1
-22.5000
-3.4277
-1.2318
-0.7159
-0.0721
-0.0173
-0.0044
-0.0011
-0.0003
-0.0001
0.0000
x3
-1.2500
-0.8750
-1.0625
-0.9688
-1.0156
-0.9922
-1.0039
-0.9980
-1.0010
-0.9995
-1.0002
Third root is x = -1
The three roots are (-1, 1, 2)
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Go to the online active:
POLYNOMIALS
http://www.physics.usyd.edu.au/teach_res/hsp/sim/sim_poly.htm
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Newton’s Method
Suppose that xn is close to a root of f ( x)  0 . We can make an improved estimate of the root
f ( x)  0 by Newton’s Method which involves f ( x) and f '( x ) as shown in the figure
The tangent to the curve intersects the X-axis at xn 1 at a
point which should be closer to the root than xn . The
gradient f '( xn ) of the tangent to the curve is
approximated by
f '( xn ) 
0  f ( xn )
xn 1  x0
Rearranging this equation, we can the new estimate xn 1
xn 1  xn 
f ( xn )
f '( xn )
This method may fail if the function has a point of inflection, or other bad behaviour near the
root of the function.
POLYNOMIALS
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Example Find the roots of the equation by Newton’s Method P( x )  x 3  2 x 2  x  2  0
Solution This method is much easier to estimate the roots than the half-interval method. Again,
it is a simple matter to perform the repeated calculations in a spreadsheet.
Starting value x = -2
Root x = -1
Starting value x = 0.5
Root x = 1
Starting value x = 3
Root x = 2
POLYNOMIALS
Newton's Method
n
xn
f(xn)
f'(xn)
xn+1
1 -2.0000 -12.0000
19 -1.36842
2 -1.36842 -2.9392 10.09141 -1.07716
3 -1.07716 -0.4932 6.789494 -1.00452
4 -1.00452 -0.0272 6.045263 -1.00002
5 -1.00002 -0.0001 6.000169
-1
6
-1
0.0000
6
-1
Newton's Method
n
xn
1
0.5000
2
1
3
1
4
1
5
1
6
1
f(xn)
1.1250
0.0000
0.0000
0.0000
0.0000
0.0000
Newton's Method
n
xn
1
3.0000
2 2.428571
3 2.12782
4 2.017076
5 2.000375
6
2
f(xn)
f'(xn)
xn+1
8.0000
14 2.428571
2.0991 6.979592 2.12782
0.4509 4.07157 2.017076
0.0524 3.137486 2.000375
0.0011
3.003
2
0.0000 3.000001
2
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f'(xn)
-2.25
-2
-2
-2
-2
-2
xn+1
1
1
1
1
1
1
15