Beaver Mastersreport 2010
Beaver Mastersreport 2010
by
2010
The Report Committee for Donald Wayne Beaver
Certifies that this is the approved version of the following report:
APPROVED BY
SUPERVISING COMMITTEE:
Supervisor:
Efraim Armendariz
Mark Daniels
Exploring Methods For Finding Solutions to Polynomial Equations
by
Report
Presented to the Faculty of the Graduate School of
in Partial Fulfillment
of the Requirements
Master of Arts
I dedicate this report to my wife of twenty eight years, Gayla Beaver, who has
supported and encouraged me to be the person I am today. Without her help, I would
never have attempted the Masters program at the University of Texas and for that I am
grateful. Throughout our married life Gayla has insisted that I persist in my endeavors
and studies to improve my knowledge and teaching. I thank her for that in this dedication.
Acknowledgements
I would like to acknowledge Drs. Efraim Armendariz and Mark Daniels for their
support and guidance throughout this Masters program. Their hard work and dedication
to the teachers of today has been an inspiration for all of us in the past three years.
August 2010
v
Abstract
There are many methods for solving polynomial equations. Dating back to the
Greek and Babylonian mathematicians, these methods have been explored throughout the
centuries. The introduction of the Cartesian Coordinate Plane by Rene Descartes greatly
The invention of the graphing calculator has been a tremendous aid in the
teaching of solutions of polynomial equations. Students are able to visualize what these
solutions represent graphically. This report explores these methods and their uses.
vi
Table of Contents
List of Tables ...................................................................................................... viii
Chapter 5: Conclusion............................................................................................31
References .............................................................................................................32
Vita .......................................................................................................................33
vii
List of Tables
zeros. .................................................................................................20
viii
List of Figures
Figure 6: y1 2 x 3; y 2 x 2 ........................................................................12
ix
Chapter 1: Introduction
Typical secondary curricula introduce the basic concept of functions with first
order linear equations. Teachers invest great time, effort, and creativity as they convey
the ideas of obtaining solutions and the meanings behind those solutions. In solving
linear equations, the concept of balancing an equation can be made concrete and tangible
by allowing students to experience the phenomenon with a balance scale. There exists a
seemingly endless supply of manipulatives which students can use to visualize most
fundamental mathematics concepts. However, making the transition from solving these
basic first order equations to higher order polynomials can prove quite challenging for
both student and teacher. It is at this point in mathematics when one must move from the
concrete to the abstract. Whereas the balance scale provides a core foundational
understanding of solving linear equations, the zero product theorem is not so easily
demonstrated. While the basic idea of the theorem itself seems attainable, using this as a
Additionally, because so much emphasis is placed on finding the one solution to a first
order equation, students often have great difficulty grasping the concept that more than
The introduction of the graphing calculator has been a tremendous aid in allowing
the student to actually see the points where the function crosses the x-axis.
Unfortunately, making the connection from the viewing window to the written
polynomial equation requires students make the abstract connection between the zero
product theorem and x-intercepts. While solving a linear equation can be done without
much mathematical depth, solving quadratics demands students have a fairly in depth
1
and its domain and range, and how the function itself produces the points associated with
the graph. The students are now able to really connect the solutions obtained by the
quadratic formula, completing the square, and factoring to what they see in the viewing
window. Fortunately, once a student reaches this level of understanding, the transition to
cubic, quartic, and higher degree equations proves far less taxing.
Once certain concepts are understood by the student, the x-intercepts are easily
seen as the inputs into a polynomial function that generate the number 0 as the output, or
Similarly, cubic, quartic, and higher degree equations are seen in the same
manner. The methods used to find the solution to quadratic equations, such as completing
the square, factoring, and the quadratic formula, are taught in first year Algebra classes,
but a connection (or meaning) comes to fruition when the actual graphs are seen.
Another problem arises when solutions to quadratics are complex numbers that
are not real numbers. It is hard for the student to visualize what is taking place since the
complex solutions are not often seen or are not easy to graph. The quadratic formula
yields these complex solutions, but what is their meaning? And when discussing the
solutions to a cubic equation, how do these solutions appear in the graph of the
to polynomial functions and their graphs is “local behavior,” finding the zeros of a
Even though each of these methods is of very little value in themselves, when used in
combination with each other, a function’s behavior can be discovered. Further use of
methods learned in calculus such as the derivative of a function or relative extrema, are
has its own purpose. The three forms and their use are:
preferred. Once the student has manipulated the equation into this form, a simple chart
with the vertex in the center entry and two values of x to the left and right (or
above/below the y in the case of horizontal parabolas) is sufficient to see its shape.
If the graph of the parabola intersects the x-axis, these are referred to as the real
zeros of the function. In the case of parabolas that do not cross the x-axis, the numbers
that can be substituted for x that generate the number 0 for y are still complex solutions.
f ( x) c( x a ) 2 b .
3
As mentioned earlier, the vertex is (a, b). Letting f(x) = 0 and solving for x,
0 = c(x – a)2 + b
b
(x – a)2
c
b
=x–a
c
b
x= a .
c
Depending on the algebraic signs of b and c, the parabola will have real or
complex roots, but the spacing about the axis of symmetry is the same. In Figure 1, the
roots are i , where the x-coordinate, or abscissa of the vertex, and is half the
length of the chord determined by the horizontal line y = 2b and where b is the y-
coordinate, or ordinate, of the vertex. The chord is referred to as the latus rectum and
passes through the focus of the parabola. As the parabola is shifted down, these become
real roots and the horizontal line is simply the x-axis, as illustrated in the transition from
4
Figure 1. Graph of f ( x) c( x a ) 2 b [6, p. 248]
A simple example would be to consider the parabola with vertex (0, -1) and c = 1
f ( x) x 2 1
5
and with real roots at ( 1, 0) . If the parabola was shifted up two units to have a new
vertex at (1, 0), but c remains the same, the new complex roots are ( i, 0) .
6
Chapter 2: A Geometrical Approach
Long before Rene Descartes introduced the Cartesian Coordinate Plane, the Greek
that in a right triangle the altitude drawn to the hypotenuse is the geometric mean
a b
between the segments into which it is divided. In the proportion , b is the
b c
Simple geometric constructions can be made using a straight edge and compass,
as shown in Figure 3. Then some use of simple algebra and geometry provides visual
ii. x 2 bx c 0
iii. x 2 bx c 0
iv. x 2 bx c 0
7
Figure 3. Euclidean constructions for i. and ii. [4, p. 363]
8
The first equation (i.) can be verified by setting x1 AC and x2 BC . The previously
AC c
c BC
or
c AC BC .
But since
x1 AC and BC b x1 ,
it follows that
x1 (b x1 ) c
bx1 x 2 c
x 2 bx1 c 0.
9
Similarly the second equation ii. can be solved. However, equations iii. and iv.
make use of another well known theorem stating that the square of a tangent segment
drawn from a point in the exterior of a circle is equal to the product of the secant segment
and the outer secant segment.
Carlyle’s Method.
In the early 1800’s another method was suggested using the intersection of a
circle with the x-axis to show solutions to the equation x 2 bx c 0 . For if the
equation of a particular circle with a diameter having endpoints at (0, 1) and (-b,c) given
by x 2 y 2 bx (1 c) y c 0 is evaluated for y = 0, the equation simplifies to the
desired quadratic.
To illustrate, consider the quadratic equation
x 2 2x 3 0 .
If the circle whose diameter has endpoints at (0, 1) and (-2, -3) is graphed, the roots of
the quadratic equation are also the abscissas of the points of intersection of the circle with
the x-axis as shown in Figure 5. In this case x1 3 and x 2 1 .
10
y
(0,1)
(-3,0) (1,0)
x
(-1,-1)
(-2,-3)
Figure 5. Graph of ( x 1) 2 ( y 1) 2 5
x 2 bx c 0
phenomenon is easily seen because if the left side of the equation is solved for x2, the
equation becomes
x 2 bx c.
11
In the example used earlier, y = x2 + 2x – 3, the points of interest are the intersection of
y1 = -2x + 3 with y2 = x2 , i.e. (-3, 9) and (1, 1) as shown in Figure 6 below:
(-3,9)
(1,1)
Figure 6. y1 2 x 3; y 2 x 2
The abscissas of these points are the solutions to the original quadratic equation.
12
Chapter 3: Iteration as a Technique
P( x) x 2 2 x 2 .
The goal is to find roots of the polynomial, that is the x-values that make P(x) = 0 or, the
x-values that satisfy the equation
x 2 2x 2 0 .
Instead of proceeding with the Rational Root Theorem, Butts suggests putting the
equation in an alternate form and applying iteration to this equation to find a fixed point.
A fixed point, x0 , is a particular value of x such that f ( x 0 ) x 0 .
x2 2
x .
2
Iteration is the process by which a seed (or initial) value x0 is substituted into the right
side of this equation, then the result f ( x 0 ) is then substituted for x until the two values
are identical, the aforementioned fixed point. For if the output (functional value) is equal
to the input (x-value) accurate to any desired decimal approximation, that particular value
13
satisfies the alternate form and therefore will satisfy the original form, yielding the
desired result. If a seed value converges to a fixed point, the root has been found.
However, sometimes the seed value will diverge to in which case iteration fails. To
ensure convergence, the following definition and ensuing theorem guarantee intervals of
convergence.
f ( xo ) f ( xo )
MF ( f ( xo )) = lim .
0 2
This definition leads us to the theorem: The Fixed Point Iteration Algorithm converges if
MF ( f ( x)) 1 for values of x near the seed value x0 " [2, p. 5]. In this example,
f ( xo ) f ( xo )
| MF ( f ( xo )) | lim
0 2
( xo ) 2 2 ( xo ) 2 2
lim 2 2
0 2
x o 2xo 2 2 x o 2xo 2 2
2 2
lim
0 2
4xo
lim
0 4
xo ,
14
so that convergence is guaranteed if xo 1, i.e., 1 xo 1.
15
Chapter 4: Finding Roots Of Higher Degree Polynomials
Finding the roots of higher degree polynomials is much more difficult than
finding the roots of linear or quadratic functions. A few theorems and properties make the
process easier.
i. If r is a root of a polynomial function, then ( x r ) is a factor of the
polynomial,
ii. Any polynomial function with real coefficients can be written as the
product of linear factors ( x r ) and quadratic factors (ax 2 bx c)
A quadratic factor that is irreducible over the reals is a quadratic function with no
real zeros; equivalently those that have a negative discriminant.
Another helpful theorem is Descartes’ Rule of Signs, which states that the number
of variations in (algebraic) signs throughout the polynomial determines the number of
positive roots that the function will have. The Rule will not tell where the polynomial’s
roots are, but will tell how many to expect when finding them. Consider the following
polynomial in its original form:
f ( x) 2 x 5 x 4 2 x 3 4 x 2 x 3.
Without concern for the actual values of the coefficients themselves, notice that the
algebraic signs change four times. Thus there will be, at most, four positive roots.
Conveniently, the roots come in “pairs,” so if one positive root is found, finding another
positive root is in order.
16
To find the number of negative roots, f(-x) is generated:
f ( x ) 2 x 5 x 4 2 x 3 4 x 2 x 3 .
Since there is only one sign change, this polynomial has exactly one negative root. So
once that root is found, looking for another negative root becomes moot. Therefore, there
are 4, 2, or 0 positive roots and exactly 1 negative root. To illustrate the “nature of the
roots,” that is, the number and type (real or complex), a chart is generated.
4 1 0
2 1 2
0 1 4
The Rational Root Theorem also can be used as an aid to finding roots of a
polynomial function of the form:
f ( x ) a n x n a n 1 x n 1 ... a1 x a 0 .
17
Assuming the coefficients are all integers and a root of the polynomial is rational, the
numerator of the root is always a factor of a0 and the denominator is a factor of an. In the
example above, if any rational roots exist, they must come from the “bank” of
1 3
1, 3, , or .
2 2
As stated earlier, each of these theorems is not all that helpful alone, but when
used in combination with one another, finding roots becomes easier. The idea is to use
these tools in combination in order to find a rational root. This is usually done by
synthetic division, then compressing the equation and repeating the process until the
polynomial is expressed as the product of linear factors or irreducible quadratic factors
that can be solved using the quadratic formula. Students sometimes find this process
laborious and tedious. If, however, a student realizes some important facts about the
process, there comes a realization that even though a root is not found by synthetic
division, it is not a waste of time. The remainder from synthetic division is synonymous
with the functional value, so it still yields a point on the graph. Also, if f (0) is positive
1
but f (1) is negative, a logical choice to try next would be ; since the polynomial is a
2
function its graph must cross the x-axis between 0 and 1. That is, the graph cannot “go
around” that portion because then it would not be a function. Of course, there is always
the possibility that the root may be irrational.
Taking the Rational Root Theorem to a higher level, Redmond [7] proves a
theorem that states “If P(x) is a polynomial in the form
f ( x ) a n x n a n 1 x n 1 ... a1 x a 0 n2
18
where a0 , an, and f (1) are all three odd numbers, then f (x) has no rational roots.” [7]
While this does not apply to the previous function (a n 2) , let’s consider the example
Redmond gives:
f ( x) x 5 7 x 4 28 x 3 125 x 2 x 3275
in which a5 = 1, a 0 = -3275, and f (1) = -3169, all odd numbers. Redmond assures that
this particular polynomial has no rational roots. An easier example, one which will be
discussed later, is
f ( x) x 2 x 1
1 i 3
a2 = 1, a0 = 1, and f(1) = 3 (all odd numbers) which has as its roots { },
2
P ( x ) a n x n a n 1 x n 1 ... a1 x a 0
19
p q
is an nth degree polynomial function with integer coefficients. If x i i
r r
are rational imaginary zeros of P(x), where and 0 are rational, p, q, and r are
P( x) x 4 2 x 3 6 x 2 2 x 5 . (1.1)
The possible rational roots are 1, -1, 5, and -5. Synthetically dividing with these possible
roots yields remainders of 16, 8, 1040, and 520, respectively, therefore there are no
rational roots. An application of the imaginary rational root theorem indicates that since
1 = 02 + 12 and 5 = 12 + 22, the possible complex rational roots would be given by
1 2i, 1 2i, and i with corresponding factors x 2 2 x 5, x 2 2 x 5, and x2 + 1.
The chart below gives the results when P(x) is divided by each of these potential factors.
Table 2. Results of dividing (1.1) by quadratic factors with rational complex zeros.
1 2i x 2 2x 5 x 2 4x 9 -40
1 2i x 2 2x 5 x2 1 0
i x2 1 x 2 2x 5 0
Therefore,
x 4 2 x 3 6 x 2 2 x 5 ( x 2 1)( x 2 2 x 5)
and
20
x 4 2 x 3 6 x 2 2 x 5 = 0 if x i or x 1 2i .
Another interesting (but perhaps not so useful) theorem is afforded by Luthar. [5]
According to Luthar, Luddhar’s Theorem states “If P ( x) ax 3 bx 2 cx d , with a, b,
c, and d integers, a 0, b 0, the function has a rational root if there exist non-zero
lm pq
integers l, m, p, and q such that c = l + m, b = p + q, p, and l . That rational
d a
l
root is given by - [5, p. 107]. The following example illustrates Luddhar’s Theorem.
p
Let
P ( x) 6 x 3 2 x 2 15 x 5 .
6 15
Since , the function can be written as
2 5
P( x) 2 x 2 (3x 1) 5(3x 1)
= (3x 1)(2 x 2 5)
1
which yields a rational root of - . The theorem and its process essentially amount to
3
“factoring by grouping” for this example, but the next problem illustrates its worth.
Consider a polynomial that is not factorable by grouping:
P( x) x 3 6 x 2 11x 6 .
21
Select two integers whose sum is 11, the coefficient of the term involving x. For example,
let l = 9 and m = 2. Now p and q must be determined. So
lm 9 2
p= 3
d 6
so that
q = b – p = 6 – 3 = 3.
la 9 1
q= 3;
p 3
thus Luddhars’s conditions are satisfied. Consequently, the original function can now be
written as
P ( x) x 3 3x 2 3x 2 9 x 2 x 6
x 2 ( x 3) 3x( x 3) 2( x 3)
( x 3)( x 2 3x 2)
( x 3)( x 1)( x 2) .
22
Consider a quadratic function with complex roots given by a bi . Using a well
known fact that the quadratic function with roots r1 and r2 is given by
f ( x) x 2 (r1 r2 ) x (r1 r2 )
x 2 2ax a 2 b 2 .
y b 2 ( x a) 2
from which it is evident that, if OA and AP are measured as shown in Figure 7. The
complex roots will be given by a ib OA i AP .
23
To illustrate, consider the following quadratic function:
f ( x) x 2 2 x 4
or
f ( x ) x 2 2 1 x 12 ( 3 ) 2 ,
where a 1 , b 3. It is easily shown by completing the square that the function can be
f ( x) ( x 1) 2 3 .
The vertex of this parabola is (1, 3) and OA = 1 while AP = 3. According to Yanosik [8],
the complex roots would be given by
OA i AP 1 i 3 .
Using the quadratic formula to generate the roots, this fact is seen to be true. Similarly,
the cubic function given by
f ( x) ( x a )( x 2 2bx b 2 c 2 )
will yield the real root a and the complex roots b ic by the following method. As
24
b ic BP i m .
f ( x) x 3 8
( x 2)( x 2 2 x 4)
The graphs of this function and the corresponding tangent are shown in Figure 9.
Figure 9. Graphs of f ( x) x 3 8; g ( x) 3x 6
From the graph (and the factored form of the function), it can be seen that the real root is
–2. If Yanosik is correct, the complex roots are given by
b ic BP i m ,
indicating that BP = 1 and the slope of the tangent is 3. The line shown tangent in Figure
9 indeed has a slope of 3 and is tangent at the point (1, 9) yielding BP = 1.
25
In a previously mentioned polynomial function, the real roots of the polynomial
seem to “disappear” as the graph is shifted up. For example, Figure 10 shows a graph of
the following function:
f ( x) x 2 x 1
1 5
Notice that the function has real roots at x = . However when the function with
2
equation
26
f ( x) x 2 x 1
in a clock. That is, a time of 20 o’clock is considered to be 8 o’clock (mod 12). It’s the
residue, or remainder, when 20 is divided by 12. In using the Rational Root Theorem
mentioned earlier, when a possible root is “checked” by synthetic division, if the
remainder is zero, the number is a root. However if there is a remainder other than zero,
that remainder is the same as the functional value at that value of x. Therefore, it is a
point on the graph of the polynomial.
Similarly, this phenomenon occurs with polynomial residues to the modulus
y 2 1 . If any polynomial in y (with real coefficients) is divided by y 2 1 the remainder
is of the form a by with a and b being real numbers. The polynomial mentioned before
P ( x) x 2 x 1
x 2 x 1 0 (mod y 2 1)
1 y 3 1 y 3
has solutions x because if x then
2 2
2
1 y 3 1 y 3
x x 1
2
1
2 2
27
1 2 y 3 3y 2 2 2 y 3 4
4 4 4
1 2 y 3 3y 2 2 2 y 3 4
4
3 3y 2
4
3
(1 y 2 ) 0 (mod y 2 1).
4
Long and Hern use modulus surfaces to explain this phenomenon. In Figure 10,
the graph shows the real roots mentioned earlier. [4] Figure 12 shows taking the absolute
value of the function, gives somewhat of an insight as to the appearance of the modulus
surface associated with this function as seen in Figure 13. The “low points” of the surface
signify the real roots. If the function is shifted up two units, the surface is that shown in
Figure 14. The appearance of the two are similar, but seem to be rotated horizontally
2
radians.
Figure 12. g ( x) x 2 x 1
28
Figure 13. f ( x ) x 2 x 1 Figure 14. f ( x ) x 2 x 1
29
with the beginning and end positions as shown in Figure 15. Since the zeros are given by
1 1 4a 0 1
, they will remain real until a 0 at which point they become complex.
2 4
1 1
When a 0 , the polynomial z 2 z is a perfect square and has a real zero at
4 4
1
x . Considering cubic functions of the nature f ( x ) x 3 a 0 , as a 0 varies over an
2
interval [- , ] , it is seen that the roots appear to be at the endpoints of the “spokes of a
wheel,” with a rotation of radians, depending on whether is positive or negative as
3
shown in Figure 16. This phenomenon is consistent with the concepts illustrated by
DeMoivre’s Theorem.
30
Chapter 5: Conclusion
Several methods for solving polynomial equations have been introduced in this
report. Depending on the nature of the polynomial and what is known, some methods are
more appropriate than others. The underlying fact is that the more methods that are
known, the more proficient one becomes at solving these equations.
Most secondary mathematics teachers agree that the introduction of the graphing
calculator has greatly aided in student learning. While some might argue that the
understanding of algorithms and processes is essential to advancement in mathematics,
the introduction of the graphing calculator into lessons can assist in enhancing the
understanding of these concepts.
31
References
1. Sharon Barrs, James Braselton, Lorraine Braselton, A Rational Root Theorem for
Imaginary Roots, The College Mathematics Journal 34, No. 5, (2003) 280-382.
2. Thomas Butts, Fixed Point Iteration – An Interesting Way to Begin a Calculus Course,
The Two-Year College Mathematics Journal 12 (1991) 2-7.
3. E John Hornsby, Jr. Geometrical and Graphical Solutions of Quadratic Equations, The
College Mathematics Journal 21 (1990) 362-369.
4. Cliff Long and Thomas Hern, Graphing the Complex Zeros of Polynomials Using
Modulus Surfaces, The College Mathematics Journal 20 (1989) 98-105.
5. R.S. Luthar, Luddhar’s Method of Solving a Cubic Equation with a Rational Root, The
Two-Year College Mathematics Journal 11 (1980) 107-110.
6. Alec Norton and Benjamin Lotto, Complex Roots Made Visible¸The College
Mathematics Journal (1984) 248-249.
8. George Yanosik, Graphical Solutions for Complex Roots of Quadratics, Cubics, and
Quartics, National Mathematics Magazine 17 (1943) 147-150.
32
Vita
33