Appendix A: The Real Line and the Cartesian Plane
The main purpose of this appendix is to present many properties of the real numbers you hopefully already know. A second goal is to introduce you to specific notation. A third is to give you a sense of just how highly developed the real number system is. Of course the same is true of the Cartesian Plane.
Subsection 1: The Real Line
A natural number is a non-negative whole number. We use the set of natural numbers, IN, for counting. It is useful to know that the sum or product of two natural numbers is again a natural number. We say that IN is closed with respect to addition and multiplication. However
IN is not closed with respect to subtraction, since for example 2 − 4 = − 2 and − 2 is not a non-negative whole number.
An integer is a whole number. The set of integers, ZZ, contains the set of natural numbers.
It is also closed with respect to addition and multiplication, and has the added advantage of being closed with respect to subtraction. It is not closed with respect to division, since for example 12 ÷ 5 = 2 2
5 is not a whole number.
The set of rational numbers, Q (q for quotient), consists of all ratios of integers with non-zero denominators. Q is closed with respect to addition, multiplication, and subtraction.
Also as long as we don’t divide by zero, the ratio of two rational numbers is again a rational number. In fact Q is the smallest set containing IN where we can perform all four basic binary operations, barring only dividing by zero. End of story? NO.
The ancient Greeks knew that certain necessary numbers were not rational.
Fact:
√
2 is not rational.
√
2 ∈ Q. Then
√
2 = p
Proof: Suppose q
, where p, q ∈ ZZ, and q = 0. We may also suppose p that p and q have no common prime divisor. We say that is in lowest terms. By squaring q we obtain 2 = p 2
. Clearing the denominator gives 2 q 2 q 2 even (it is divisible by 2). If p were odd, then p 2
= p 2 . From which we deduce that p 2 is would also be odd. We therefore conclude that p is even. Thus p = 2 k for some whole number k . But this implies 2 q 2 = p 2 = (2 k ) 2 = 4 k 2 .
Which reduces to q 2 = 2 k 2 . As above we deduce that q is even. But this is a contradiction since now 2 is a common prime divisor of p and q . Since we arrived at a contradiction our
√ original hypothesis must be false. Therefore 2 6∈ Q.
60

Appendix A: The Real Line and the Cartesian Plane
To form the real numbers, IR, it is customary to consider a continuous horizontal line of infinite length. One point on the line is designated as 0. Another point to its right is designated as 1. The distance from 1 to 0 forms a unit by which all other distances are measured. For any point x on the line exactly one of the following statements is true: 1) x is to the left of
0 ( x is negative), 2) x is to the right of 0 ( x is positive), or 3) x = 0. This property is called the trichotomy law. We denote the distance from x to 0 by | x | . When x is to the left of zero we have x = −| x | . Otherwise x = | x | . This basically defines the absolute value function
√ f ( x ) = | x | = x 2 .
This model also gives rise to the notion of order on the line. We write a < b , for real numbers a and b , when a − b = −| a − b | . Also a ≤ b is equivalent to b − a = | b − a | . The trichotomy law generalizes in that if a and b are real then exactly one of a < b, b < a and a = b is true. In fact one can prove that the order relations have the following properties.
Theorem: (Inequality Properties) For real numbers a, b and c
1) if a < b , then a + c < b + c
2) if a < b and 0 < c , then ac < bc
3) if a < b and c < 0 , then bc < ac
To begin to understand the importance of this construction we first state and prove the
Pythagorean Theorem. The reader is strongly encouraged to draw the necessary figures to illustrate the proof.
Theorem: For a right triangle with hypotenuse of length c and sides of lengths a and b , we have a 2 + b 2 = c 2 .
Proof: Let A = ( a + b ) 2 be the area of a square of side a + b . Let B denote the area of a right triangle with dimensions as per the theorem statement.
1) Draw a square S with side a + b and two sides horizontal. Let C
1 denote the point on the left hand side which is b units down from the top left corner. Let C
2 be the point on the bottom side which is b units to the right of the lower left corner. Let C
3 be the point on the right hand side which is b units up from the lower right hand corner. Let C
4 denote the point on the top side which is b units to the left of the upper right hand corner. Connect C
1 and C
2 by a line segment. Repeat by connecting C
2 to C
3
, C
3 to C
4 and C
4 to C
1
.
S now contains four non-overlapping right triangles whose sides measure a , b and c as above. The remaining
61

Appendix A: The Real Line and the Cartesian Plane area inside S is the area of a square of side c . Therefore A = 4 B + c 2 .
2) Draw another square S 0 of side a + b and two sides horizontal. Let D
1
= C
1 and D
2
= C
2
.
Let D
3 be the point on the right side of S 0 which lies b units down from the top right hand corner. Let D
4 be the point on the upper side of S 0 which lies b units to the right of the upper left hand corner. With line segments connect D
1 to D
2
, D
1 to D
3
, D
4 to D
2
, and D
4 to D
3
. Now S 0 contains four non-overlapping right triangles with sides of lengths a, b and c .
The remaining area inside S 0 is the area of two squares, one of side a , and the other of side b .
Therefore A = 4 B + a 2 + b 2 .
Since S and S 0 have the same area we deduce a 2 + b 2 = c 2 .
Because
√
2 is the length of the hypotenuse of an isosceles right triangle with sides of unit length, it is a real number. Real numbers which are not rational are called irrational. The set
√ of irrational numbers is not closed with respect to multiplication since for example 2
√
2 = 2.
Consequently we have no standard symbol for this set.
The absolute value function also plays a role in the definition of distance on the real line. The distance from a to b , denoted d ( a, b ) equals | a − b | . Naturally d ( a, b ) = d ( b, a ) and d ( a, b ) ≥ 0 with equality if and only if a = b . Seemingly innocuous is the statement that the midpoint of the line segment joining a and b is ( a + b ) / 2. This fact is used repeatedly in section
7.
Finally, the absolute value function satisfies the following properties.
Theorem: (Absolute Value Properties) For real numbers a, b and c
1) | ab | = | a || b | a
2) If b = 0, =
| a | b | b |
3) | c | < a if and only if − a < c < a
4) | c | ≤ a if and only if − a ≤ c ≤ a
3) | c | > a ≥ 0 if and only if c < − a or c > a
3) | c | ≥ a ≥ 0 if and only if c ≤ − a or c ≥ a
In turn the order relations are used to define intervals. For real numbers a and b , [ a, b ] denotes all real numbers x with a ≤ x ≤ b . Similarly [ a, b ) = { x ∈ IR | a ≤ x < b } , ( a, b ] =
{ x ∈ IR | a < x ≤ b } , and ( a, b ) = { x ∈ IR | a < x < b } . We use the infinity symbol, ∞ when we have an interval of the real line which extends indefinitely in the positive direction. So
62

Appendix A: The Real Line and the Cartesian Plane
( a, ∞ ) = { x ∈ IR | a < x } . Also ( −∞ , b ] = { x ∈ IR | x ≤ b } . It is to be understood that ∞ is a symbol, not a real number.
Intervals are special subsets of the real line. Whenever we have sets A and B we denote their union by A ∪ B . The union of two sets is the set whose elements are in at least one of the two original sets. Also the intersection of two sets A ∩ B is the set whose elements are in both A and in B . We can build more general subsets of IR using unions and intersections of intervals.
The last computational tool we introduce in this subsection is the sign graph. The premise behind this is contained in part 3 of the properties of inequalities. To paraphrase “minus times minus is plus”, “minus times plus is minus”, and “plus times plus is plus”. So for example we can determine all x ∈ IR so that x 3 − 4 x ≥ 0. To proceed we factor x 3 − 4 x = x ( x 2 − 4) = x ( x + 2)( x − 2). For each linear factor the trichotomy law allows us to determine intervals where the factor is positive and negative. We combine these subsolutions according to the rules above.
x – – – – – – – – 0 + + + + + + + + x − 2 – – – – – – – – – – – – 0 + + + + x + 2 – – – – – 0 + + + + + + + + + + + x 3 − 4 x – – – – 0 + + + 0 – – – 0 + + + +
| | | | | | | | |
-4 -3 -2 -1 0 1 2 3 4
So x 3 − 4 x ≥ 0 when x ∈ [ − 2 , 0] ∪ [2 , ∞ ).
Subsection 2: The Cartesian Plane
One of the greatest mathematical ideas of all time is due to Ren´e DesCartes. Given an infinite flat plane choose a point O . Draw two perpendicular lines concurrent at O , one horizontal and the other vertical. Call the horizontal line the x -axis and the vertical line the y -axis. On the x -axis pick a point P to the right of O and let the distance from P to O serve as 1 unit of length in the horizontal direction. Pick another point Q above O on the y -axis and use the distance from O to Q as a unit of vertical distance. Any point in the plane can then be assigned a unique ordered pair of Cartesian coordinates. To label a point
63

Appendix A: The Real Line and the Cartesian Plane
( a, b ) means that it lies | a | units from O horizontally (to the left if a = −| a | , and to the right if a = | a | .) and | b | units from O vertically (down if b = −| b | , and up otherwise). The points with ( a, b ) = ( | a | , | b | ) lie in the first quadrant. Those with ( a, b ) = ( −| a | , | b | ) are in the second quadrant. The third quadrant consists of those points with ( a, b ) = ( −| a | , −| b | ). The points with ( a, b ) = ( | a | , −| b | ) make up the fourth quadrant. In essence each coordinate axis is a copy of the real line. We denote the Cartesian plane by IR
2
= IR × IR.
Starting with ( x
1
, y
1
) and ( x
2
, y
2
) in the Cartesian plane form a third point ( x
2
, y
1
). These points form the vertices of a (possibly degenerate) right triangle. Measured against the x -axis we see that the horizontal side of the triangle has length | x
2
− x
1
| = p ( x
2
− x
1
) 2 . The vertical side has length | y
2
− y
1
| = p
( y
2
− y
1
) 2 . By the Pythagorean Theorem the distance between the two points is d (( x
1
, y
1
) , ( x
2
, y
2
)) = p
( x
2
− x
1
) 2 + ( y
2
− y
1
) 2 .
A circle is defined to be the set of points all of which have a prescribed distance from a given point called the center. The equation for a circle follows from the distance formula. If the center is the point ( h, k ) and the distance from any point on the circle to the center is | r | , then the circle consists of all points ( x, y ) so that ( x − h ) 2 + ( y − k ) 2 = r 2 .
In practice the equation of a circle rarely comes in this standard form. The standard form can be recaptured by completing squares. The basic formula to remember is ( z + a
2
) 2 = z 2 + az + ( a
2
) 2 . Given an equation x 2 + ax + y 2 + by = c , we strategically add zero two times.
The first time we add zero in the form ( a
2
) 2 − ( a
2
) 2 . The second time we add zero in the form ( b
2
) 2 − (
2 b ) 2 . The result via substitution transforms the equation x 2 + ax + y 2 + by = c first to x 2 + ax + ( a
2
) 2 − ( a
2
) 2 + y 2 + by = c . Which after adding ( a
2
) 2 to both sides becomes
( x +( a
2
)) 2 + y 2 + by = c +( a
2
) 2 . At this point we have completed the square in x . After completing the square in y and simplifying we arrive at ( x + ( a
2
)) 2 + ( y + ( b
2
)) 2 = c + ( a
2
) 2 + ( b
2
) 2 .
We close this appendix with another seemingly innocent remark that the midpoint of the line segment joining the points with coordinates ( x
1
, y
1
) and ( x
2
, y
2
) has coordinates
(( x
1
+ x
2
) / 2 , ( y
1
+ y
2
) / 2).
Exercises:
1. Use interval notation to describe the values of x which satisfy x 2 − 3 x − 4 > 0.
2. Solve the inequality
1 x + 2
≤
2 x − 3
. Write your answer using interval notation.
3. a) Find the distance between the points (2 , 1) and ( − 1 , − 3).
64

Appendix A: The Real Line and the Cartesian Plane b) Find the midpoint of the line segment joining (2 , 1) and ( − 1 , − 3).
4. a) Complete the squares to find the center and radius of the circle with equation x 2 − 2 x + y 2 + 4 y = 11.
b) Sketch a graph of the circle in part a). Choose equal scales for both axes.
c) Repeat part b) but choose the y -scale to be twice as long as the x -scale.
5. Find an equation of the circle with center ( − 3 , 4) which is tangent to the y -axis.
65

Appendix B: Lines
A linear equation in x and y is of the form Ax + By = C , where we take A, B, C ∈ IR.
If A = B = 0, the equation is either inconsistent (like 2 = 0) and has empty graph, or is the trivial equation 0 = 0 whose graph is all of IR
2
. So let us suppose that at least one of A or B is non-zero.
Thus if B = 0, we must have A = 0 and our equation is equivalent to x = C/A . The graph of this contains all points ( C/A, y ) as y ∈ IR. So the graph is a vertical line.
Otherwise B = 0 and we have the equivalent equation y = ( − A/B ) x + ( C/B ). In which case we set m = − A/B and b = ( C/B ). When x = 0 we see that y = b is the y -intercept of the graph. Too if ( x
1
, y
1
) is on the graph we have y
1
= mx
1
+ b . Therefore for w ∈ IR, y
1
+ mw = mx
1
+ mw + b = m ( x
1
+ w ) + b which is to say that every point of the form
( x
1
+ w, y
1
+ mw ) as w ∈ IR is on the graph. In fact if ( x
2
, y
2
) is a point on the graph other than ( x
1
, y
1
), then simultaneously we have y
1
= mx
1
+ b and y
2
= mx
2
+ b . Subtracting the first equation from the second gives y
2
− y
1
= mx
2
− mx
1
= m ( x
2
− x
1
). In this case when we set w = x
2
− x
1 so that x
2
= x
1
+ w , we have y
2
− y
1
= mw which simplifies to y
2
= y
1
+ mw .
So the graph of the equation consists exactly of all points of the form ( x
1
+ w, y
1
+ mw ) as w ∈ IR. We say that the equation is determined by one point and the value of m which we call the slope. Moreover from y
2
− y
1
= m ( x
2
− x
1
) with ( x
1
, y
1
) = ( x
2
, y
2
) and B = 0 we have that x
2
= x
1 wherefore m = ( y
2
− y
1
) / ( x
2
− x
1
). So the graph is also determined by any two distinct points. Given two distinct points on the graph their midpoint is
( x
1
+ x
2
2
, y
1
+ y
2
) = ( x
1
+
2 x
2
− x
1
2
, y
1
+ y
2
− y
1
) = (( x
1
+
2 x
2
− x
1
2
, y
1
+ m x
2
− x
1
2
)
And since the shortest distance between points in IR
2 is a straight line, the graph must contain the line through the two points. Of course every point on the graph is also on the line as argued above.
So every line in IR
2 has an equation of the form Ax + By = C , where not both A and B are zero. Every non-vertical line in IR
2 also has an equation of the form y = mx + b . This is the slope-intercept form. Equivalently we can describe the line as all points ( x, y ) satisfying an equation of the form ( y − y
1
) = m ( x − x
1
). This is called the point-slope form.
We use the convention that a vertical line has infinite slope. A horizontal line has slope
0. This allows us to define parallel lines as those whose graphs have the same slope when
66

Appendix B: Lines sketched on the same coordinate axes.
Two lines are perpendicular if they meet at right angles. Since the lines are not parallel, they intersect at a point ( x
0
, y
0
). If one line is horizontal the other is vertical and vice versa.
When neither is horizontal, then both lines have non-zero slope. Suppose that the first line has slope m and the second has slope n . For w = 1 we have that the first line contains the point
( x
1
, y
1
) = ( x
0
+ 1 , y
0
+ m ), while the second line contains the point ( x
2
, y
2
) = ( x
0
+ 1 , y
0
+ n ).
The three points ( x
0
, y
0
) , ( x
1
, y
1
) , and ( x
2
, y
2
) form the vertices of a right triangle. Therefore by the Pythagorean Theorem we have that
[( x
1
− x
0
)
2
+ ( y
1
− y
0
)
2
] + [( x
2
− x
0
)
2
+ ( y
2
− y
0
)
2
] = [( x
2
− x
1
)
2
+ ( y
2
− y
1
)
2
]
But x
1
− x
0
= 1 = x
2
− x
0
, y
1
− y
0
= m, y
2
− y
0
= n, x
2
− x
1
= 0, and y
2
− y
1
= n − m . So after substitution we have 1 2 + m 2 + 1 2 + n 2 = ( n − m ) 2 = n 2 − 2 mn + m 2 . Which implies that mn = − 1. So a line is perpendicular to a line with non-zero slope if and only if its slope is the negative reciprocal of the first line’s slope.
Exercises:
1. Find the equation in slope-intercept form of the line a) through (1 , 1) and (2 , 3) b) through (0 , − 2) and ( − 3 , − 6) c) through (1 , 2) with slope 5 d) through ( − 1 , 2) parallel to y = 3 x − 4 e) through (0 , 1) perpendicular to y = 3 x + 4
2. Convert every equation from exercise 1 to point-slope form.
3. Graph the lines whose equations are given. Scales for your axes should be clearly labelled and chosen so that all interesting features of the graph can be clearly seen.
a) 4 x + 3 y = 12 b) y = − 2 x + 4 c) x = 3 d) ( y − 3) = 2( x − 1) e) 4 x − 3 y = 12
67

Appendix C: Factoring Polynomials and Complex Arithmetic
This appendix deals with finding factors of polynomials. The first major theorem is the division algorithm for polynomials which mirrors the division algorithm for integers.
Theorem: (The Division Algorithm) If a ( x ) and b ( x ) are polynomials with real coefficients, then there exist unique polynomials q ( x ) and r ( x ) with real coefficients, so that a ( x ) = b ( x ) q ( x ) + r ( x ) and either r = 0 or the degree of r is strictly less than the degree of b .
The polynomial q ( x ) is called the quotient and the polynomial r ( x ) is the remainder.
When b ( x ) is a non-zero polynomial and a ( x ) = b ( x ) q ( x ) (with remainder zero) we say that b ( x ) divides a ( x ). We call b ( x ) a factor of a ( x ).
Theorem: (The Factor Theorem) A polynomial p ( x ) = a n x n + a n − 1 x n − 1 + . . .
+ a
1 x + a
0
, where a
0
, a
1
, . . . , a n
∈ IR and a n
= 0 has p ( c ) = 0 if and only if x − c divides p ( x ) .
Proof: By the division algorithm p ( x ) = ( x − c ) q ( x ) + r ( x ), where r ( x ) = 0 or its degree is strictly less that the degree of ( x − c ), which is 1. So r ( x ) is zero, or a non-zero constant.
Now if we set x = c , we get p ( c ) = ( c − c ) q ( c ) + r ( c ) = r ( c ). So r is the constant p ( c ).
Thus r = 0 if and only if p ( c ) = 0 if and only if x − c divides p ( x ).
Of course we are interested in such linear factors of p ( x ), since c is a root of p ( x ). We therefore would like to how many possible values of c there are. For this we need to know
Theorem: (The Fundamental Theorem of Algebra) A polynomial p ( x ) = a n x n + a n −
1 x n − 1 +
. . .
+ a
1 x + a
0
, where a
0
, a
1
, . . . , a n
∈ IR and a n
= 0 can be written uniquely (up to reordering terms) as p ( x ) = a n
( x − α
1
)( x − α
2
) . . .
( x − α n
) , where the roots α
1
, α
2
, . . . , α n are complex numbers.
So a polynomial of degree n has n complex roots counting multiplicity. To access the theorem we therefore need some knowledge of the complex numbers. The complex numbers are denoted by C and consist of all objects of the form z = a + bi , where i 2 = − 1. The real part of z is a and we denote it by Re ( z ). The imaginary part of z is b and we denote it by Im ( z ).
So every complex number corresponds to an ordered pair ( a, b ). We can therefore think of the complex numbers forming a plane where the x -axis is the real axis, and the y -axis is the imaginary axis.
68

Appendix C: Factoring Polynomials and Complex Arithmetic
Complex numbers are operated on by addition, subtraction, or multiplication as if they are linear polynomials. The reduction i 2 = − 1 is used, if necessary, after the operation. So for example (2 + 3 i ) + (1 − 4 i ) = 3 − i, ( − 2 + i ) − (3 − 2 i ) = 1 + 3 i, and (2 + 3 i )( − 1 − i ) =
− 2 + ( − 2 + ( − 3)) i − 3 i 2 = − 2 − 5 i − 3( − 1) = 1 − 5 i . Dividing a complex number by a nonzero complex number can also be done. To do this we first introduce the conjugate of z = a + ib , which is the complex number a − ib , and is denoted by z . Next we observe that zz reduces to a 2 + b 2 , which we call the norm of z , and denote | z | . It is critical that | z | = 0 if and only if z = 0. This follows since | z | is the square of the distance of the point z to the origin in the complex plane. Also z = z if and only if z = Re ( z ). At any rate, if z = 0, we have 1 /z = z/ | z | .
So to divide by z , we multiply by its conjugate and divide by its norm (which is a real number).
Complex conjugation has several other important properties. If z = a + bi and w = c + di , then z + w = ( a + c ) + ( b + d ) i = ( a + c ) − ( b + d ) i = ( a − bi ) + ( c − di ) = z + w . The conjugate of a sum is the sum of the conjugates. Similarly one can show that the conjugate of a product is the product of the conjugates. This applies to sums and products with more than two terms.
Especially we have that the conjugate of a power of z is the power of the conjugate, z n = ( z ) n .
Finally z + z = 2 a ∈ IR for any z ∈ C.
Consequently we arrive at the following fact.
Theorem: If p ( x ) = a n x n + a n −
1 x n − 1 + . . .
+ a
1 x + a
0 is a degree n polynomial with a
0
, a
1
, . . . , a n
∈ IR and z ∈ C is a root of p ( x ) , then so is z .
Proof: We have
0 = 0 = p ( z ) = a n z n + a n −
1 z n − 1 + . . .
+ a
1 z + a
0
= a n z n + a n −
1 z n − 1 + . . .
+ a
1 z + a
0
= a n z n + a n −
1 z n − 1 + . . .
+ a
1 z 1 + a
0
= a n
( z ) n + a n − 1
( z ) n − 1 + . . .
+ a
1
( z ) 1 + a
0
= a n
( z ) n + a n − 1
( z ) n − 1 + . . .
+ a
1
( z ) 1 + a
0
, since a
0
, . . . , a n
∈ IR
= p ( z )
Thus if z ∈ C − IR is a root of a polynomial p ( x ) with real coefficients, then both ( x − z ) and
( x − z ) are factors of p ( x ). Therefore p ( x ) has as factor the quadratic polynomial ( x − z )( x − z ) = x 2 − ( z + z ) x + zz = x 2 − 2 ax + | z | . This quadratic has real coefficients and no real roots. This fact allows us to move from the Fundamental Theorem of Algebra to the theorem we need.
69

Appendix C: Factoring Polynomials and Complex Arithmetic
Theorem: (Real Factors Theorem) If p ( x ) = p n x n + p n − 1 x n − 1 + . . .
+ p
1 x + p
0 is a degree n polynomial with p
0
, p
1
, . . . , p n
∈ IR , then there exist k real numbers r
1
, r
2
, . . . , r k
, not necessarily distinct, and m quadratic polynomials with no real roots x 2 + a l x + b l
, a l
, b l
∈
IR , a l
2 − 4 b l
< 0 , l = 1 , . . . , m , not necessarily distinct, so that n = k + 2 m and p ( x ) = p n
( x − r
1
)( x − r
2
) . . .
( x − r k
)( x 2 + a
1 x + b
1
)( x 2 + a
2 x + b
2
) . . .
( x 2 + a m x + b m
) .
Proof: Since IR ⊂ C, by the Fundamental Theorem of Algebra p ( x ) has n complex roots if we count multiplicities. For each non-real root z ∈ C − IR, its conjugate is also a root. Thus we can combine these two linear factors to form a quadratic factor of the form described in the theorem. After we have done this for all properly complex roots, the remaining roots must be real. This forces n = k + 2 m as in the theorem.
After collecting multiple occurences of linear and quadratic factors together we have p ( x ) = p n
( x − r
1
) e
1 ( x − r
2
) e
2 ( . . .
)( x − r k
) e k ( x
2
+ a
1 x + b
1
) f
1 ( x
2
+ a
2 x + b
2
) f
2 . . .
( x
2
+ a j x + b j
) f j , where n = e
1
+ e
2
+ . . .
+ e k
+ 2( f
1
+ f
2
+ . . .
+ f j
), and a l
2 − 4 b l
< 0, for l = 1 , . . . .j
. In this form the linear and irreducible quadratic factors are distinct.
A proof of the fundamental theorem of algebra is beyond the scope of this course. In fact most undergraduate math majors never see a proof. However it lies behind a common application in calculus.
Fact: A polynomial p ( x ) with real coefficients and odd degree must have at least one real root.
Proof: If the degree of p ( x ) is n = 2 r + 1 for some integer r , then 2 r + 1 = k + 2 m , where k is the number of real roots p ( x ) has. Thus k = 2( r − m ) + 1 is odd. In particular it cannot be zero.
So the graph of any odd degree polynomial crosses the x -axis an odd number of times.
Also the graph of a polynomial with even degree intersects the x -axis an even number of times.
Unfortunately the theorem does not help us find these.
As an aid for picking out likely rational roots we have the rational root test.
Theorem: (Rational Root Test) If the rational number p/q in lowest terms is a root of the polynomial p ( x ) = a n x n + . . .
+ a
1 x + a
0 with real coefficients, and a n
= 0 , then p divides a
0 and q | a n
.
70

Appendix C: Factoring Polynomials and Complex Arithmetic
For fun, you might try to prove this theorem. This test alone does not necessarily solve our problem. For example if p ( x ) = x 2 − 30 the rational root test says that the possible rational roots of p ( x ) are ± 1 , ± 2 , ± 3 , ± 5 , ± 6 , ± 10 , ± 15 , ± 30. However we can easily factor p ( x ) and see that neither of its roots is rational. So testing p ( r ) for each of the above values of r would prove fruitless, except to provide points on the graph of the function. Similarly if we attempted to synthetically divide p ( x ) by x − r for the values of r we would be largely wasting our time. So the rational root test should be employed very carefully, hopefully utilizing additional theorems.
One additional theorem is DesCartes’ Rule of Signs. If we take the form of p ( x ) as given by the real factors theorem and expand it then we see that the coefficients of p ( x ) change sign for each time p has a positive real root, plus possibly two more times for each irreducible quadratic factor. That is, the number of positive real roots of p either equals the number of sign changes in coefficients of p or less this by an even number. Also the number of negative real roots of p either equals the number of sign changes in p ( − x ), or is less than this number by an even number. This rule can therefore help us be a bit more selective as to which and how many values from the rational root test we check.
Exercises:
1. Find the quotient q ( x ) and the remainder r ( x ) when a ( x ) = 3 x 3 − x 2 + 4 x + 2 is divided by b ( x ) = 2 x 2 + x − 1.
2. Given that p ( x ) = x 6 − 8 x 4 + 6 x 3 + 7 x 2 − 6 x has x − 2 and x − 1 as factors, factor p ( x ) completely as per the real factor theorem.
3. List all possible rational roots of p ( x ) = 2 x 3 − 5 x 2 + 7 x − 21.
4. Find all rational zeroes of the polynomial p ( x ) = x 4 − 3 x 3 − 2 x 2 + 5 x + 3. Then determine any irrational zeroes and factor p ( x ) completely.
5. Find a third degree polynomial with zeroes at x = 2 , x = 0 and x = − 4 whose x 2 term has coefficient 7.
6. If z = 1 − 5 i and w = − 2 + 2 i , write z 3 − 2 zw + w 2 in the form a + bi .
7. Given that x = 2 + 3 i is a solution of x 4 − 4 x 3 + 14 x 2 − 4 x + 13 = 0, find all solutions.
8. Find the polynomial with real coefficients and leading coefficient 1 which has roots at − 1 , 2 and (1 + i ) and degree 4.
71

Appendix E: Rotation of Axes
Given a non-degenerate quadratic of the form Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0 we can rotate axes to get a new coordinate system, where the corresponding quadratic equation in these variables has no cross term.
Essentially we want to hold the origin fixed and tilt our heads counter-clockwise at
π an angle θ where 0 < θ < . The Greeks determined that the appropriate angle to use is
2
θ =
1 A − C A − C
2 arccot
B
, which means θ satisfies cot 2 θ =
B
.
To see that this works, with our heads tilted at such an angle we draw in new axes, call them the z and w -axes. So θ is the angle between the positive x -axis and the positive z -axis and also the angle bewteen the positive y -axis and the postive w -axis. Given a point P = ( x, y ) let r = p x 2 + y 2 denote its distance to the origin. Let α denote the angle between the positive z -axis and the line, l , joining the origin and P . Note that θ + α is the angle between the positive x -axis and l . From right triangle trigonometry we know that x = r cos( θ + α ) and y = r sin( θ + α ). With respect to our second set of axes P has coordinates z = r cos α and w = r sin α .
w
...
.......
...
...
...
...
...
...
...
...
....
...
...
...
...
...
...
...
...
...
...
....
...
...
...
...
...
...
...
...
...
...
....
...
...
...
...
...
...
...
...
...
...
....
...
...
...
...
...
...
...
...
...
...
....
...
...
...
...
...
...
...
...
...
...
....
...
...
...
...
...
...
...
y
.......
..
........
........
........
..
..
..
.............
..
........
..
..
...
.....
.............
........
α
........
........
.............
.............
........
(
........
....
.............
x, y
.............
)
..
...
.......
...
...
....
....
....
.....
.....
......
.......
.......
.......
.......
.............
.............
θ
.............
.....
z x
We use the sum formulae for sine and cosine to write x = r cos( θ + α ) = r (cos θ cos α − sin θ sin α ) = z cos θ − w sin θ y = r sin( θ + α ) = r (sin θ cos α + cos θ sin α ) = w cos θ + z sin θ
72

Appendix E: Rotation of Axes
We can temporarily let c = cos θ and s = sin θ , so x = zc − ws and y = wc + zs . Substitution into our general equation yields
0 = Ax 2 + Bxy + Cy 2 + Dx + Ey + F
= A ( zc − ws ) 2 + B ( zc − ws )( wc + zs ) + C ( wc + zs ) 2 + D ( zc − ws ) + E ( wc + zs ) + F which is a general quadratic equation in z and w of the form Gz 2 + Hzw + Iw 2 + Jz + Kw + F .
By expanding the expression and collecting coefficients we find that
G = Ac
2
+ Bsc + Cs
2
I = As
2
− Bsc + Cc
2
J = Dc + Es
K = − Ds + Ec
H = − 2( A − C ) cs + B ( c 2 − s 2 )
But since θ satisfies cot 2 θ =
( A − C )
B we have by the double angle identities cot 2 θ = cos 2 θ sin 2 θ
= c 2 − s 2
2 cs
=
( A − C )
B
Cross-multiplying gives us 2( A − C ) cs = B ( c 2 − s 2 ). Thus H = 0 as desired.
Exercises:
1. Find the angle θ by which rotation of axes will clear the cross term from the equation x 2 − 2 xy + y 2 + 6 = 0.
2. Find the angle θ by which rotation of axes will clear the cross term from the equation x 2 +
√
3 xy + 2 y 2 + 14 = 0.
73