Download Chapter 3: Polynomial and Rational Functions

Lecture Notes
Algebra II Honors
Chapter 3: Polynomial and Rational Functions
3.1 Polynomial Functions
A polynomial on degree n is a function of the form P(x) = anxn + an–1xn–1 + ··· + a1x1 + a0,
where n is a nonnegative integer and an ≠ 0.
Graphs of polynomials are smooth, unbroken curves with no sharp points.
The end behavior is determined by the highest power term.
If the highest power is even and an > 0, then y → ∞ as x → ±∞.
If the highest power is odd and an > 0, then y → ∞ as x → ∞ and y → –∞ as x → –∞.
In both cases, if an < 0 then reflect the end behavior across the x-axis.
If y = P(x) is a polynomial and c is a number such that P(c) = 0, then c is a zero of the poly.
Thus x = c is a root of P(x) = 0, x – c is a factor of P(x), and c is an x-intercept of the graph of
y = P(x). The only place y = P(x) can change signs is at an x-intercept. This is a consequence of
the Intermediate value theorem for polynomials. If the poly is written in factored form, only
factors to odd powers give rise to x-intercepts where the graph crosses the x-axis.
If P(x) is a poly of degree n, then the graph of y = P(x) has at most n – 1 local extrema.
To graph a polynomial equation, write it in factored form. Find the x-intercepts and y-intercept.
Determine using test points where the poly is + or –. Determine the end behavior and graph.
• Example: Graph y = x3 + x2 – x – 1.
We note there are at most 2 local maxima or minima. The end behavior is like that of y = x3, with
y → ∞ as x → ∞ and y → –∞ as x → –∞. In factored form y = x2(x + 1) –1(x + 1) =
(x2 – 1)(x + 1) = (x + 1)2(x – 1). This shows there are x-intercepts at –1 and 1; the graph does not
cross the axis at x = –1 since that factor is raised to an even power. Letting x = 0 we find y = –1
for the y-intercept. The graph is shown below.
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-1
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-1
-2
-3
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Lecture Notes
Algebra II Honors
3.2 Dividing Polynomials
Study long division and synthetic division of polynomials to prepare for factoring.
Example of synthetic division:
3
1
1
0
3
3
0
9
9
x 3 − 27
x −3
–27
+27
0 → Quotient is: 1x2 + 3x + 9 with remainder 0
Thus we see x – 3 is a factor of x3 – 27.
•Remainder Theorem: If poly P(x) is divided by x – c, then the remainder is P(c).
Note above P(3) = 33 – 27 = 0, which was the remainder.
Thus if c is a zero of the poly, so that P(c) = 0, x – c is a factor of P(x).
•Factor Theorem: c is a zero of P if and only if x – c is a factor of P(x).
Example: Find the remainder when 6x1000 – 17x562 + 12x + 26 is divided by x + 1 = x – (–1).
Use the remainder theorem! Evaluate P(–1) and this must be the remainder.
6(–1)1000 – 17(–1)562 + 12(–1) + 26 = 6 – 17 – 12 + 26 = 3. Thus remainder = 3.
Example: Is x – 1 a factor of x567 –3x400 + x9 + 2? Use the Factor Theorem!
If P(1) = 0, then x – 1 is a factor of the poly. (1)567 –3(1)400 + (1)9 + 2 = 1 – 3 + 1 + 2 = 1.
Since P(1) ≠ 0, x – 1 is not a factor of the poly.
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Algebra II Honors
3.3 Real Zeros of Polynomials
•Rational Zeros Theorem: If a poly P(x) = anxn + an–1xn–1 + ··· + a1x1 + a0 has integer coefficients,
then every rational zero of P, and every rational root of P(x) = 0, has the form p / q.
Here p is a factor of the constant coefficient a0 and q is a factor of the leading coefficient an.
•Descartes' Rule of Signs: If P is a poly with real coefficients:
1.
The number of positive real zeros of P(x) is either equal to the number of variations in
sign of P or less than that by an even number.
2.
The number of negative real zeros of P(–x) is either equal to the number of variations in
sign of P or less than that by an even number.
•Upper and Lower Bounds' Theorem: If P is a poly with real coefficients:
1.
Divide P by x – b (b > 0) using synthetic division. If the quotient row has no negative
entry, then b is an upper bound for the real zeros of P.
2.
Divide P by x – a (a < 0) using synthetic division. If the quotient row has entries
alternately nonpositive and nonnegative), then a is an lower bound for the real zeros of P.
Example: Find all the real zeros of P(x) = 2x3 + 5x2 + x – 2.
Since P has integer coefficients we can use the rational zeros theorem. This assures us that the
only possible rational roots are of the form p / q where p is a factor of the constant term –2, and q
is a factor of the leading coefficient 2. Thus the only possible rational roots are ±2, ±1, and ±1/2.
P(x) has 1 variation in sign so there is only 1 positive real zero.
P(–x) = 2(–x)3 + 5(–x)2 + (–x) – 2 = –2x3 + 5x2 – x – 2 has 2 variations in sign and so either 2 or 0
negative real zeros.
We will soon see that every poly of degree n has exactly n zeros, counting complex zeros
discussed in the next two sections. Thus P either has 1 positive real and 2 negative real roots, or 1
positive real root and two complex roots.
Test the possible rational roots using synthetic division. If one is found, the remaining roots can
be found from the reduced quadratic equation which is the quotient.
2
2
1
2
5
1
–2
4
18
38
2
9
19
36 → since remainder ≠ 0, 2 is not a zero. But this bottom row
has no negative entry, and so by the upper bounds theorem there is no real root greater than 2.
1
2
1
5
1
6
1
6
7
–2
7
5 → not a zero, but another upper bound (the lub so far).
2
5
1
–2
1
3
+2
2
6
4
0 → thus 1/2 is a zero. Now find the zeros of 2x2 + 6x + 4.
We can factor 2x2 + 6x + 4 = 2(x2 + 3x + 2) = 2(x + 2)(x + 1) and P(x) = 2(x – 1/2)(x + 2)(x + 1)
Thus the zeros are 1/2, –1, and –2. As Descartes predicted, 1 positive and 2 negative zeros.
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3.4 Complex Numbers
Consider the quadratic equation x2 – 6x + 25 = 0 and the associated P(x) = x2 – 6x + 25.
The discriminant of the equation b2 – 4ac = –64 < 0, indicating no real solutions. No real number
squared can ever be negative. But Italians trying to solve more complicated equations found they
needed to use solutions to equations like our example, in passing, on their way to real number
solutions of the equations that interested them. So imagine, they said, that quadratics like the
above have solutions, and we will use them for our own purposes.
Define an imaginary unit number i such that i2 = –1 (or equivalently, define i =
−1 ).
Then let −4 = 4 ⋅ −1 = 2i and so on, numbers like this being called imaginary.
Further, consider numbers of the form a + bi, where a and b are real. These are complex numbers.
Solve x2 – 6x + 25 = 0 using the quadratic formula.
x=
−(−6) ± (−6) 2 − 4(1)(25) 6 ± −64 6 ± 8i
=
=
= 3 ± 4i
2(1)
2
2
Thus P(x) = x2 – 6x + 25 has complex zeros and factors as P(x) = [x – (3 + 4i)][x – (3 – 4i)]
The two roots, differing only in the sign of the imaginary term, are called complex conjugates.
Any quadratic with real coefficients and discriminant < 0 will always have a pair of complex
conjugate solutions.
Arithmetic is defined for these numbers as if they were binomials, with of course i2 = –1.
Addition:
(3 + 4i) + (–5 + 12i) = (3 + –5) + (4i + 12i) = –2 + 16i
Subtraction:
(3 + 4i) – (–5 + 12i) = (3 – –5) + (4i – 12i) = 8 – 8i
Multiplication: (3 + 4i)(–5 + 12i) = (3)(–5) + (3)(12i) + (4i)(–5) + (4i)(12i) = –63 + 16i
Division:
3 + 4i
3 + 4i −5 − 12i 33 − 56i 33 56
i
=
⋅
=
=
−
169
169 169
−5 + 12i −5 + 12i −5 − 12i
Note in division we multiplied numerator and denominator by the complex conjugate of the
denominator. This is just like rationalizing a denominator (which it is, because i is
−1 ).
Note when we write −4 = 2i we are finding the principal square root. (2i)(2i) = –4 but also
(–2i)(–2i) = –4, so there are really two square roots here just as ±2 are the square roots of 4.
Powers of i form a repeating sequence.
i1 = i
i5 = i4 · i1 = i
2
i = –1
i6 = i4 · i2 = –1
3
2
i = i · i = –i
i7 = i4 · i3 = –i
4
2
2
i =i ·i =1
i8 = i4 · i4 = 1
To find i78 for example, factor out i raised to a power that is the greatest multiple of 4 less than
the power of interest. Thus i78 = i76 · i2 = (i 4 )19 ⋅ i 2 = −1
If z = 3 + 4i, Re(z), the real part of z, is 3. Im(z), the imaginary part of z, is 4.
The complex conjugate z = 3 − 4i (Sometimes this is written z*).
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Algebra II Honors
Complex numbers can be represented geometrically as points in the complex plane.
(Diagram from Wikipedia)
The horizontal axis is the real number line, and the vertical axis is the imaginary number line.
Numbers like 3 + 4i can be considered as representing a coordinate pair (3, 4).
An arrow, or vector, from the origin to (3, 4) has a certain length and makes a certain angle with
the positive "x-axis".
The modulus or magnitude of z, written |z| = 32 + 42 = 5 . This is the length of the vector.
The argument of z is the angle whose tangent is 4/3 or about 53°.
Thus we can write 3 + 4i or 5 ∠ 53° or (3, 4). The TI-86 calculator allows either of the last two
forms and can perform all the above arithmetic operations.
Caution! There is one operation in which complex numbers are unlike radicals.
a ⋅ b = ab , but this is not true if both a and b are
negative. Always convert the square root of a negative quantity to the form i r before
Positive numbers a and b obey the relation
multiplying to avoid this difficulty.
Example:
−4 ⋅ −9 = 2i ⋅ 3i = 6i 2 = −6
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Algebra II Honors
3.5 Complex Zeros
The theorems below help us to factor any poly and tell us that the poly equation y = P(x) has n
roots if P is of degree n.
•Fundamental Theorem of Algebra: Every poly with complex coefficients has at least one
complex zero. This is very hard to prove - it was first proven by Gauss. The following theorems
have much easier proofs, given in the text.
•Complete Factorization Theorem: Every poly of nth degree can be factored into n complex linear
factors or n complex zeros.
•Zeros Theorem: Every poly of degree n ≥ 1 has n zeros, provided that a zero of multiplicity k is
counted k times.
•Linear and Quadratic Factors Theorem: Every poly with real coefficients can be factored into a
product of linear and irreducible quadratic factors.
Solve x3 + 2x2 + 4x + 8 = 0. This factors (grouping) into x2(x + 2) + 4(x + 2) = (x2 + 4)(x + 2) = 0.
Thus x = –2 or x = ± −4 = ±2i. The poly factors into (x + 2)(x + 2i)(x – 2i).
Solutions may be checked with the TI-86. Use the Poly function, give the order (degree) of the
poly, enter the coefficients, and solve. Mathematica also solves algebraic equations.
Solve x3 – 1 = 0. By inspection, 1 is a root. What are the other two?
Factor: x3 – 1 = (x – 1)(x2 + x + 1) = 0. Now solve x2 + x + 1 = 0 with the quadratic formula.
x=
−1 ± 12 − 4 ⋅1 ⋅1 −1 ± i 3
1
3
=
=− ±
i.
2 ⋅1
2
2 2
Since x3 – 1 = 0 has solution x = 3 1 , there are apparently three cube roots of unity if we allow
complex roots. The reader should show that the roots above, when cubed, do equal one.
There are formulas that solve the general cubic and quartic as the quadratic formula solves any
quadratic. These were historically important but are complicated and rarely used. This reference
(http://mathworld.wolfram.com/CubicFormula.html) discusses the formulas.
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Algebra II Honors
3.6 Rational Functions
A rational function is the ratio of two polynomials, assumed to have no factor in common.
We illustrate their properties with an example.
Graph y =
(x − 1)(x + 2)
(x + 1)(x − 3)
•x-intercepts: Look at numerator for values of x that make y = 0. (1, 0) and (–2, 0) are the two
odd x-intercepts. They are called odd because the factors giving rise to them are raised to odd
powers. Recall this means the graph crosses the x-axis at these points.
•y - intercept: Set x = 0 everywhere to find y = 2/3. Thus (0, 2/3) is the y-intercept.
Rational functions need not have x or y intercepts!
•Vertical asymptotes: An asymptote is a line that a graph approaches ever more closely.
Since division by zero is not allowed, there is no point of the graph of the given rational function
at x = –1 and x = 3. These values are found in general by seeing where the denominator is zero.
Near those values y → ∞ or y → – ∞. Indicate a vertical asymptote with a dashed line. If the
factor giving rise to the asymptote is raised to an odd power, the graph skips across the x-axis
(from + to – ∞) at that x value. If raised to an even power, the graph stays on the same side of the
x-axis. Both of the vertical asymptotes for our example are odd: x = –1 and x = 3 are the odd
vertical asymptotes of this example.
•Horizontal asymptotes: Consider the leading (highest power) terms of numerator and
denominator. If the power to which x is raised is larger in the numerator, the end behavior is
increasing y as x increases. But if the power is equal or less than the highest power of the
denominator, as x increases y will approach a constant value, called a horizontal asymptote.
x 2 x −2
1 −2
1+ + 2
+ 2+ 2
2
x2 + x − 2
x
x =
x x = 1 as x → ∞ .
y= 2
= x
x − 2x − 3 x 2 −2x −3 1 + −2 + −3
+ 2 + 2
x x2
x2
x
x
More easily remembered: if y =
Ax n
, then the following can occur:
Bx m
1. if n > m, then there is no horizontal asymptote.
2. if n = m, then the horizontal asymptote is y = A / B.
3. if n < m, then the horizontal asymptote is y = 0.
The graph may cross a horizontal asymptote in the middle of the graph. It is only as x gets large
that the graph gets ever closer to this line. The graph never touched a vertical asymptote.
•Slant asymptotes and other end behavior: If n > m above, there is no horizontal asymptote, but if
n = m + 1 there is a slanted line that is approached closely at the far left and right of a graph.
Divide, and discard the remainder. y = quotient will be the slant asymptote.
If n = m + 2, a quadratic type curve serves as the end behavior, and so on.
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Algebra II Honors
Graphing, it helps to start at a known point like the y-intercept, if not zero.
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