Transcript Document

Polynomial Function
• f(x) = ao xo + a1x1+ a2x2 + …. + anxn
• Equality: f (x) = g (x) if for all ai = bi that is
the coefficients are equal.
• Linear: f(x) = ao xo + a1x1 = mx+b
• Quadratic: f(x) = ao xo + a1x1+ a2x2 = a x2 +
bx+ c
Example
f ( x)  x  x  10
2
• When does f(x) = 20?
Example
f ( x)  a0 x  a1 x  a2 x
0
1
2
g ( x)  b0 x  b1 x  b2 x  b3 x
0
1
2
Then
( f  g )(x)  f ( x)  g ( x)

( f  g )(x)  f ( x)  g ( x)

3
Polynomial Function
Two branches of studying polynomial
functions such as:
f(x) = a0 x0 + a1x1+ a2x2 + … + anxn
• Modern Algebra view - theory of solving
polynomials by factoring
• Modeling view – solving real world
problems modeled by polynomial functions
which almost never factor
Solving Polynomial EquationsAlgebraic Method
• Set Polynomial Equation equal to zero.
• Factor the resulting polynomial expression
into a product of linear expressions.
• Reduce the equation to a series of linear
equations.
This is a classic example of analytic
reasoning – reducing a more complex
problem to one we already know how to
solve.
Question
• For any given polynomial equation
anxn + an-1xn-1+ an-2xn-2 + ... + a2x2 + a1x1 + a0 = 0
What are the problems with applying the
algebraic method to find solutions to the
equations?
History - Tartaglia
• Niccolò Tartaglia (1499--1557) was
wounded during a battle as a young
child and was left with a speech
impediment. He was given the
nickname Tartaglia which means
“stammerer” and he kept that name for
the rest of his life.
• He worked on solving for the roots of
cubic polynomials. He was able to
defeat a rival mathematician, Fior, in a
mathematical problem-solving contest
through his knowledge of solving
certain types of cubics. Through his
work, Tartaglia knew how to solve most
cubics.
History - Cardano
• Girolamo Cardano was a doctor and teacher of
mathematics in Milan. Within a few years
Cardano became the city's most famous
physician. In 1539, while awaiting the
publication of Practica arithmetica, his first book
on mathematics, Cardano learned that Tartaglia
knew the procedure for solving cubic equations.
• Cardano convinced Tartaglia to share his secrets for solving cubics.
Tartaglia did, on the condition that Cardano not publish them until after
Tartaglia published his results. Cardano continued to refine the
method and soon knew how to solve all cubic equations. In 1545,
Cardano published the solution method, giving Tartaglia proper credit
but reneging on his promise to keep it a secret. Tartaglia was outraged
and verbally attacked Cardano, but to no avail since Cardano had the
popular support behind him.
History - Quartics
• After Tartaglia had shown Cardano how to solve cubics, Cardano
encouraged his own student, Lodovico Ferrari, to examine quartic
equations. Ferrari managed to solve the quartic with perhaps the
most elegant of all the methods that were found to solve this type of
problem. Cardano published all 20 cases of quartic equations in Ars
Magna.
• Here, again in modern notation, is Ferrari's solution of the case: x4 +
px2 + qx + r = 0. First complete the square to obtain
x4 + 2px2 + p2 = px2 - qx - r + p2
i.e.
(x2 + p)2 = px2 - qx - r + p2
• Now the clever bit. For any y we have
(x2 + p + y)2 = px2 - qx - r + p2 + 2y(x2 + p) + y2
= (p + 2y)x2 - qx + (p2 - r + 2py + y2) (*)
Quartics (continued)
• Now the right hand side is a quadratic in x and we can choose y so
that it is a perfect square. This is done by making the discriminant
zero, in this case
(-q)2 -4(p + 2y)(p2 - r + 2py + y2) = 0.
• Rewrite this last equation as
(q2 - 4p3 + 4 pr) + (-16p2 + 8r)y - 20 py2 - 8y3 = 0
to see that it is a cubic in y.
• Now we know how to solve cubics, so solve for y. With this value of
y the right hand side of (*) is a perfect square so, taking the square
root of both sides, we obtain a quadratic in x. Solve this quadratic
and we have the required solution to the quartic equation.
Algebraic solution
Let P(x) = 2x3 - 3x2 - 8x + 12
Find solutions to P(x) = 0.
Synthetic Division:
2
2 -3 -8 12
4
2 12
2 1 -6
0
Method 1 (continued)
• Since remainder = 0, (i.e. P(2) = 0), x = 2
is a solution.
• P(x) = (x-2)(2x2 + x - 6).
• The problem is now reduced to a quadratic
equation, so apply the quadratic formula to
solve for remaining solutions.
Table Method Of Solving
Polynomial Equations
• Solve: x4 - 11x3 + 11x2 + 179x - 420 = 0
• Form Related Polynomial Function
y = x4 - 11x3 + 11x2 + 179x - 420
Create a table of values using the TI -73 or a
CAS such as Derive
X
Y
0
1
-420 -240
2
3
4
5
6
7
8
9
10
-90
0
24
0
-30
0
180
624
1470
Grapher demo
http://www.math.wvu.edu/~mays/A
Vdemo/Labs/Lab01/Lab01-04.htm
• Question: What happens to the y-values
as the table approaches a solution to the
polynomial equation?
• Question: For a general polynomial
equation, what are some difficulties in
using the table method for finding
solutions?
Table Method: Example 2
• Solve
1000x4 - 9100x3 + 22310x2 - 7661x - 10608 =0
Related Function:
y =1000x4 - 9100x3 + 22310x2 - 7661x - 10608
Create Table:
X
0
1
Y
-10608
-4059
2
6510
3
2494
4
5
6
-10692
-3663
76986
Questions
• Where are the solutions occurring?
• How can we be sure there are solutions on
the intervals where signs change?
• Can we be sure no solutions exist on
intervals where the signs do not change?
Intermediate Value Theorem
Zeros Case
• If P is a polynomial function with real
coefficients and P(a) and P(b) are of
opposite signs, then for some intermediate
value c  (a,b) we have P(c) =0
Question
• Must there be only one solution when the
Intermediate Value Theorem is satisfied?
• If P(a) and P(b) have the same sign, could
there still be zeros in (a,b)?
Intermediate Value Theorem
(IMT)
• If the hypothesis of the IMT is satisfied,
must there be only one unique solution on
the interval (a,b)?
• If the hypothesis of the IMT is not satisfied
does it imply that no solution exists on the
interval?
IMT – Multiple Solutions
IMT not Satisfied – Solution
Explore
• Let f(x) = (x-1)(x+3)(x-7) which expands to
f(x) = x3 - 5x2 - 17x + 21.
• So we know f(x) has an upper bound of 7
on its zeros.
• Use synthetic division to search for
patterns in finding an upper bound on the
zeros of a polynomial.
Upper Bound Theorem
• For zeros of a Polynomial Equation
• Case 1: If P is an n degree polynomial with
real coefficients and an > 0 is divided by x - r
using synthetic division, and the
quotient/remainder row of that division is all
non-negative numbers, then r is an upper
bound of the real zeros of P.
• Case 2: If an< 0, Case 1 hold except the
quotient/remainder row must be all nonpositive numbers.
Proof Of UB Theorem
1.
2.
3.
4.
P(x) = (x-r) • q(x) +R
Coefficients of q(x)  0 and R  0.
 a> r, (a-r) q(a) + R >0.
Thus P(a) > 0, so P(x) can never cross
the x - axis to the right of x = r.
Lower Bound Theorem
Lower Bound Theorem for Zeros of a
Polynomial Equation
• Reflect y = f(x) in the y-axis — y = f(-x)
• Apply the Upper Bound Theorem.
Example
f ( x)  x  5 x  17x  21
3
2
f ( x)   x  5 x  17x  21
3
2
Graph Method for solving a
Polynomial Equation
1. Determine the Related Polynomial
Function.
2. Plot the Complete Graph of the related
function
• End Behavior – 4 types
• Intermediate Behavior: turns,
x-intercepts, y-intercepts
3 . Use Trace Function and Zoom to
appropriate solutions.
Solve:
x5 - 3x4 - x3 + 3x2 - 2x + 6 = 0
1. Related Function:
2. Upper Bound:
Lower Bound:
3. Plot related function at appropriate scale
and zoom in to approximate solutions
within an error of 0.01
y = x5 - 3x4 - x3 + 3x2 - 2x + 6
Zoom-in to find solution on (1,2)
Algebraic Method For Finding
Zeros
• Divide the polynomial function P by x-r,
If the remainder is zero, then x = r is a
zero of the polynomial function.
Remainder Theorem
• If a polynomial function P with complex
coefficients is divided by x - a, then the
remainder is P(a).
P(x) = (x-a)q(x) + P(a)
where deg (q(k)) = deg(P(x) - 1)
Proof Of Remainder Theorem
1.
2.
3.
4.
P(x) = (x - a) q(x) + R
P(a) = (a - a) q(a) + R
P(a) = 0 • q(a) + R
P(a) = R
Factor Theorem
• If a is a zero of the polynomial function P,
then x - a is a factor of P .
• Also, if x - a is a factor of P, then a is a
zero of P.
Example
Is x+1 a factor f(x) = x19 + 1 ?
• Use factor & Remainder Theorems:
• Use Synthetic Division:
Which is better for this question?
Rational Zeros Theorem
•
If a polynomial function has integer coefficients,
every rational zero that it might possibly have is of
the form p / q where:
1. p and q are integers with no common factors
2. The numerator p is a factor of the constant term a0.
3. The denominator q is a factor of the leading
coefficient an.
Proof Of Rational Zeros Theorem
Justify that the denominator q must be a
factor of the leading coefficient an .
1. Let P(x) =anxn + an-1xn-1+ an -2xn-2 + ... +
a2x2 + a1x1 + a0
and let x = p /q be a zero.
2. Then P(p/q) = 0 giving
an(p /q)n + an-1(p /q)n-1 ... + a2(p /q)2 + a1(p
/q)1 + a0 = 0
3.
4.
5.
6.
7.
8.
anpn + an-1pn-1 q +...+ a1pqn-1+ a0qn = 0
an-1pn-1q + ... + a1p qn-1 + a0 qn = -anpn
q [an-1pn-1 + ... + a1p qn-2 + a0 qn-1] = -anpn
So q | (-anpn )
But q | pn
q | an
Find the zeros of P(x) = 4x4 – 11x3 – 11x2 + 22x +6
STEP 1: Apply the Rational Zeros Theorem:
an = 4 so q | 4, q = 1, 2, 4
a0 = 6 so p | 6, p = 1, 2, 3, 6
Possible Rational Zeros p|q:
1, 1/2, 1/4, 2, 3, 3/2, 3/4, 6
16 cases to check.
STEP 2. Reduce cases using UB Theorem
4 -11 -11 22 6
LB Theorem
4 11 -11
-22
6
UB and LB yield (-2,4)
What possible rational zeros are now
eliminated?
Possible Rational Zeros p|q:
1,
1/2, 1/4,
2,
3, 3/2, 3/4,
6
-1, -1/2, -1/4, -2, -3, -3/2, -3/4, -6
STEP 3. Graph P and find complete graph.
Which of the remaining possible rational
zeros are most plausible now?
1, 1/2, 1/4, 2, 3, 3/2, 3/4
-1, -1/2, -1/4,
-3/2, -3/4
STEP 4. Apply Remainder & Factor Theorems or
synthetic division to check zeros.
• Use the Remainder and Factor Theorems
to determine if x = 3 is a zero.
P(3) =
• Use synthetic division to determine if
x = –1/4 is a zero.
4 -11 -11 22 6
STEP 5. Factor out known factors and search for
remaining zeros.
1

2
P( x )   x  ( x  3)(4 x  8)
4

If we have reduced the original polynomial to
degree two we can use the quadratic
formula to find the remaining zeros.
Question
• Can we always factor a polynomial
completely into linear factors?
N Zeros Factorization Theorem
Every n-degree polynomial
P(x) = anxn + an-1xn-1 +...+ a1x + a0
with complex coefficients ai can be
factored completely into:
P(x) = an (x - r1 ) (x - r2 )…(x - rn)
Where r are the N zeros of P, including
complex roots and repeated roots
The Fundamental Theorem Of
Algebra
Every polynomial function P of degree n  1
has at least one zero, possibly complex.
Gauss
• Using the
Fundamental
Theorem of Algebra,
which was proven by
Gauss at the age of
20, we can prove the
remarkable N Zeros
Theorem.
Verification Of N Zeros
Factorization Theorem
1. P(x) = anxn + an-1xn-1 +...+ a1x + a0
2. There exists a complex zero r1 for P(b)
3. So x - r1 is a factor of P(x)
4. P(x) = ( x - r1 ) • Q1(x) where Q1(x) is a
polynomial function.
5. Q1(x) = ( x - r2 ) • Q2(x) where
deg Q2(x) < deg Q1(b)
6. P(x) = ( x - r1 ) ( x - r2 ) • Q2(x)
7. Repeat this process until completely
factored.
Complex Conjugates Zeros
Theorem
• Let P be a polynomial function with real
coefficients and a real constant.
• If a + bi is a complex zero of P, then
a - bi is as well.
VERIFY:
Question
• Must each of the following have a complex
conjugate pair of zeros if x = - 2i is a zero?
• f(x) = x2 + 4
• g(x) = x2 + 5i x - 6
Factoring Website
• http://wims.unice.fr/wims.cgi
Select Factoris