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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 mxb
- Quadratic f(x) ao xo a1x1 a2x2 a x2

bx c

Example

- When does f(x) 20?

Example

Then

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 Equations-Algebraic 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 p2i.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

Grapher demo

- http//www.math.wvu.edu/mays/AVdemo/Labs/Lab01/La

b01-04.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

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 TheoremZeros 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)(x3)(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 anquotient/remainder row must be all non-positive

numbers.

Proof Of UB Theorem

- 1. P(x) (x-r) q(x) R
- 2. Coefficients of q(x) ? 0 and R ? 0.
- 3. ? a r, (a-r) q(a) R 0.
- 4. 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

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. P(x) (x - a) q(x) R
- 2. P(a) (a - a) q(a) R
- 3. P(a) 0 q(a) R
- 4. 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 x1 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. anpn an-1pn-1 q ... a1pqn-1 a0qn 0
- 4. an-1pn-1q ... a1p qn-1 a0 qn -anpn
- 5. q an-1pn-1 ... a1p qn-2 a0 qn-1

-anpn - 6. So q (-anpn )
- 7. But q pn
- 8. 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 pq
- ?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 pq
- 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.

- 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)
- 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

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