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

• When does f(x) = 20?

10)( 2 xxxf

Example

33

22

11

00

22

11

00

)(

)(

xbxbxbxbxg

xaxaxaxf

Then

)()())((

)()())((

xgxfxgf

xgxfxgf

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 + 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 0 1 2 3 4 5 6 7 8 9 10

Y -420 -240 -90 0 24 0 -30 0 180 624 1470

Grapher demo

http://www.math.wvu.edu/~mays/AVdemo/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 2 3 4 5 6

Y -10608 -4059 6510 2494 -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 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)(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 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

21175)(

21175)(23

23

xxxxf

xxxxf

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 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. 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 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.

If we have reduced the original polynomial to degree two we can use the quadratic formula to find the remaining zeros.

)84)(3(4

1)( 2

xxxxP

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