Unit 3 Polynomial Functions

Unit 3Polynomial Functions

Friday, January 31, 2020

Polynomial Function

A polynomial function is define by a polynomial expression in x given in the standard form;

y = anxn + an-1x

n-1 + … +a1x + a0

The value of n must be a positive integer

an-1, ..., a1, a0 are called coefficients and these

are real numbers

Kinds of Polynomial FunctionPolynomial can be classified by the degree of the polynomial as shown below

Polynomial Function in Standard Form Degree

Name of Function

y = mx + b 1 Linear

y = ax2 + bx + c 2 Quadratic

Y = ax3 + bx2 + cx + d 3 Cubic

Y = ax4 + bx3 + cx2 + dx + e 4 Quartic

Y = ax5 + bx4 + cx3 + dx2 + ex 5 Quintic

The largest exponent within the polynomial determines the degree of the polynomial.

Leading CoefficientThe leading coefficient is the coefficient of the first term in a polynomial when the terms are written in descending order by degrees.

For example, the quartic function f(x) = -2x4 + x3 – 5x2 – 10 has a leading coefficient of -2.

Graphs of Polynomial Function

Exploring the graphs of polynomial function

Linear

Function

Quadratic

Function

Cubic

Function

Quartic

Function

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

-5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 9 10

-60

-55

-50

-45

-40

-35

-30

-25

-20

-15

-10

-5

5

10

Cubic PolynomialsLook at the two graphs and discuss the questions given

below.

1. How can you check to see if both graphs are functions?

3. What is the end behaviour for each graph?

4. Which graph do you think has a positive leading coeffient? Why?

5. Which graph do you think has a negative leading coefficient? Why?

2. How many x-intercepts do graphs A & B have?

Graph BGraph A

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10

Cubic Polynomials

Equation

Factored form & Standard form

X-Intercepts

Sign of Leading

Coefficient

End Behaviour

Domain and Range

Factoredy=(x+1)(x+4)(x-2)

Standardy=x3+3x2-6x-8

-4, -1, 2 Positive

As x, y and x-,

y-

Domain

{x| x Є R}

Range

{y| y Є R}

Factoredy=-(x+1)(x+4)(x-2)

Standardy=-x3-3x2+6x+8

-4, -1, 2Negativ

e

As x, y-

and

x-, y

Domain

{x| x Є R}

Range

{y| y Є R}

The following chart shows the properties of the graphs on the left.

-5 -4 -3 -2 -1 1 2 3 4 5

-12

-10

-8

-6

-4

-2

2

4

6

8

10

12

-5 -4 -3 -2 -1 1 2 3 4 5

-12

-10

-8

-6

-4

-2

2

4

6

8

10

12

Quartic PolynomialsLook at the two graphs and discuss the questions given below.

1. How can you check to see if both graphs are functions?

3. What is the end behaviour for each graph?

4. Which graph do you think has a positive leading coeffient? Why?

5. Which graph do you think has a negative leading coefficient? Why?

2. How many x-intercepts do graphs A & B have?

Graph BGraph A

-5 -4 -3 -2 -1 1 2 3 4 5

-10

-8

-6

-4

-2

2

4

6

8

10

12

14

-5 -4 -3 -2 -1 1 2 3 4 5

-14

-12

-10

-8

-6

-4

-2

2

4

6

8

10

Quartic Polynomials

Equation

Factored form & Standard form

X-Intercepts

Sign of Leading

Coefficient

End Behaviour

Domain and Range

Factoredy=(x-4)2(x-1)(x+1)

Standardy=x4-8x3+15x2+8x-16

-1,1,4 Positive

As x, y and x-, y

Domain

{x| x Є R}

Range

{y| y Є R,

y ≥ -16.95}

Factoredy=-(x-4)2(x-1)(x+1)

Standardy=-x4+8x3-15x2-8x+16

-1,1,4Negativ

e

As x, y-

and

x-, y-

Domain

{x| x Є R}

Range

{y| y Є R,

y ≤ 16.95}

The following chart shows the properties of the graphs on the left.

-5 -4 -3 -2 -1 1 2 3 4 5

-15

-12

-9

-6

-3

3

6

9

12

15

18

-5 -4 -3 -2 -1 1 2 3 4 5

-18

-15

-12

-9

-6

-3

3

6

9

12

15

Polynomial Equation

If y = f(x) is a polynomial function then f(x) = 0 is a polynomial equation. Example

1.) F(x) = 2x4 ‒ 3x3 + x ‒ 4

2.) F(x) = 4x4 + 3x3 + 2x + 1

2x4 ‒ 3x3 + x ‒ 4 = 0

4x4 + 3x3 + 2x + 1 = 0

polynomial function polynomial equation

Note: The solution to a polynomial equation is called the zero of the polynomial function.

Remainder TheoremIf a polynomial function P(x) is divided by x –a, then the remainder is number P(a) where the function is evaluated at x = a

1.) F(x) = 2x4 ‒ 3x3 + x ‒ 4 divided by (x – 2)

3.) F(x) = 2x³ + 1 ‒ x + 3x² divided by (x + 2)

4.) F(x) = 2x³ ‒ 3x² ‒ x + 2 divided by (x – 1)

2.) F(x) = 4x4 + 3x3 + 2x + 1 divided by (x + 1)

Find the remainder of each

R = 6

R = 0

R = -1

R = 0

Factor TheoremIf a polynomial function P(x) is divided by x – a, and the remainder P(a) = 0 if and only if x – a is a factor of P(x).

Or in other words,• – If f(a) = 0, then x – a is a factor of f(x).• – If x – a is a factor of f(x), then f(a) = 0

• If we know a factor, we know a zero!• If we know a zero, we know a factor!

Fundamental Theorem of Algebra

Every polynomial function of degree n≥1 has at least one complex zero.Every polynomial function of degree n≥1 has exactly n complex zeros counting multiplicities.Example:1.) F(x) = 2x4 ‒ 3x3 + x ‒ 4 Degree 4

has exactly 4 complex zeros or solution

2.) F(x) = 2x³ + 1 ‒ x + 3x² Degree 3

has exactly 3 complex zeros or solution

Solving Polynomial Equation

• Solving Polynomial Equation in Factored Form

• Example:

1.) (x – 2)(2x – 1)(x + 1) = 0

x – 2 = 0

x = 2

2x – 1 = 0

2x = 1

2 2

x =1

2

x + 1 = 0

x = -1


• Solving Polynomial Equation in Factored Form

• Example:

2.) (x + 3)(3x + 2)(x - 5) = 0

x + 3 = 0

x = -3

3x + 2 = 0

3x = -2

3 3

x =-2

3

x - 5 = 0

x = 5

Finding Polynomial Function or Equation

Finding a Polynomial Function or Equation given its Roots, Zeros or Solution.

Example:

1. Find a Polynomial function whose zeros are x = 1 and x = -3

Solution:

f(x) =

(x – 1) (x + 3)

f(x) = x2 + 3x - 1x - 3

f(x) = x2 + 2x- 3 or x2 + 2x – 3 = 0

Finding Polynomial Function or Equation

Example:

2. Find a Polynomial function whose zeros are x = -2, x = -1 and x = 2

Solution:

f(x) =

(x + 2) (x + 1)

f(x) = (x2 + 1x + 2x + 2)

f(x) = (x2 + 3x + 2) (x – 2)

(x - 2)

(x - 2)

= x3 - 2x2+ 3x2- 6x + 2x - 4

= x3 + 1x2 - 4x - 4

f(x)

f(x)

Unit 3Solving Polynomial

Functions / Equations



Solving for the solutions or zeroes of a polynomial equation or function can be done by:

1. Graphing calculator, Synthetic Division and Factoring.

2. Rational Root Theorem, Synthetic Division and Factoring.


Solving by Graphing Calculator, Synthetic Division and Factoring

Example: Find the zeroes or solution

1. f(x) = 3x3 + x2 – 3x – 1

Step 1: Use graphing calculator to get as many exact zero as possible by doing y = 3x3

+ x2 – 3x - 1 and do 2nd TRACE ZERO.

Step 2: Use synthetic Division to reduced the function to a lower degree.

Rational Root/ Zero Theorem

The Rational Root/ Zero Theorem gives a list of possible rational zeros of a polynomial function. Equivalently, the theorem gives all possible rational roots of a polynomial equation.

The Rational Zero Theorem

If f(x) = anxn + an-1x

n-1 +…+ a1x + a0 has integer

coefficients and (where is reduced) is a

rational zero, then p is a factor of the constant term

a0 and q is a factor of the leading coefficient an.

p

q

p

q

Using the Rational Root/ Zero Theorem

List all possible rational zeros of f (x) = 15x3 + 14x2 - 3x – 2.

Solution The constant term is –2 and the leading coefficient is 15.

1 2 1 2 1 25 53 3 15 15

Factors of the constant term, 2Possible rational zeros

Factors of the leading coefficient, 15

1, 2

1, 3, 5, 15

1, 2, , , , , ,

-=

=

=

Divide 1

and 2

by 1.

Divide 1

and 2

by 3.

Divide 1

and 2

by 5.

Divide 1

and 2

by 15.

There are 16 possible rational zeros. The actual solution set to f(x) = 15x3 +

14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions.

Solve: x4 - 6x2 - 8x + 24 = 0.

Solution Because we are given an equation, we will use the

word "roots," rather than "zeros," in the solution process. We

begin by listing all possible rational roots.

Factors of the constant term, 24Possible rational zeros

Factors of the leading coefficient, 1

1, 2 3, 4, 6, 8, 12, 24

1

1, 2 3, 4, 6, 8, 12, 24

=

=

=

EXAMPLE: Solving a Polynomial Equation

Solve: x4 - 6x2 - 8x + 24 = 0.

Solution The graph of f(x) = x4 - 6x2 - 8x + 24 is shown the

figure below. Because the x-intercept is 2 it is a zero or root, we

will test 2 by synthetic division and show that it is a root of the

given equation.

x-intercept: 2

The zero remainder

indicates that 2 is a root

of x4 - 6x2 - 8x + 24 = 0.

2 1 0 -6 -8 24

2 4 -4 -24

1 2 -2 -12 0



Solve: x4 - 6x2 - 8x + 24 = 0.

Solution Now we can rewrite the given equation in

factored form.

(x – 2)(x3 + 2x2 - 2x - 12) = 0 This is the result obtained from

the synthetic division.

x – 2 = 0 or x3 + 2x2 - 2x - 12 = 0 by zero property.

x4 - 6x2 + 8x + 24 = 0 This is the given equation.

we must continue by factoring

x3 + 2x2 - 2x - 12 = 0


Solve: x4 - 6x2 - 8x + 24 = 0.

Solution Because the graph turns around at 2, this

means that 2 is a root of even multiplicity. Thus, 2 must

also be a root of x3 + 2x2 - 2x - 12 = 0. We use 2 again

for synthetic division

x-intercept: 2

2 1 2 -2 -12

2 8 12

1 4 6 0

These are the coefficients

of x3 + 2x2 - 2x - 12 = 0.

The zero remainder indicates that 2 is a root ofx3 + 2x2 - 2x - 12 = 0.


Solve: x4 - 6x2 - 8x + 24 = 0.

Solution Now we can solve the original equation as

follows.

(x – 2)(x3 + 2x2 - 2x - 12) = 0 This was obtained from the first

synthetic division.

x4 - 6x2 + 8x + 24 = 0 This is the given equation.

(x – 2)(x – 2)(x2 + 4x + 6) = 0 This was obtained from the

second synthetic division.

x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = 0 by zero property.

x = 2 x = 2 x2 + 4x + 6 = 0 Solve.


Solve: x4 - 6x2 - 8x + 24 = 0.

Solution We can use the quadratic formula to solve x2 + 4x + 6 = 0.

Let a = 1, b = 4, and c = 6.

24 4 4 1 6

2 1

- -=

We use the quadratic formula because x2 + 4x + 6 = 0

cannot be factored.

2 4

2

b b acx

a

- -=

Simplify.2 2i= -

Multiply and subtract under the radical. 4 8

2

- -=

4 2 2

2

i- = - = - =8 4(2)( 1) 2 2i

The solution set of the original equation is {2, -2 - i -2 + i }.2,i 2i

If f (x) = anxn + an-1xn-1 + … + a2x

2 + a1x + a0 be a

polynomial with real coefficients.

1. The number of positive real zeros of f is either equal to

the number of sign changes of f (x) or is less than that

number by an even integer. If there is only one variation in

sign, there is exactly one positive real zero.

2. The number of negative real zeros of f is either equal to

the number of sign changes of f (-x) or is less than that

number by an even integer. If f (-x) has only one variation in

sign, then f has exactly one negative real zero.

Descartes' Rule of Signs

Descartes' Rule of Signs

Example: State the possible positive and negative zeros of the polynomial function

1.) f(x) = 3x4 – 5x3 – 12x2 + 20x

Positive: 2 or 0 positive zeros

+ - - +

f(-x) = 3(-x)4 – 5(-x)3 – 12(-x)2 + 20(-x)

f(-x) = 3x4 + 5x3 – 12x2 – 20x

+ + - -Negative: 1 negative zeros

Unit 3Theorem on Roots of Polynomial

Equations


•The Conjugate Root Theorem

•The Irrational Root Theorem

Conjugate Root Theorem

If the complex number a + bi is a root of a

polynomial equation, then its conjugate a - bi is also a root

Note conjugate root always come in pairs

Example

i

i

23

23

-

+

i

i

- i

i

3

3

-


Example Problem

Write a polynomial function with the given zeros: 2, i,

1.)

f(x) =

(x - 2) (x - i)

f(x) = (x2 + 1)

(x + i)

(x - 2)

= x3 - 2x2 + 1x - 2f(x)

Solution


Example Problem

Write a polynomial function with the given zeros: 3, 4i

2.)

= x3 - 3x2 + 16x - 48f(x)

Solution


Example Problem

Write a polynomial function with the given zeros: 4, 2 - 3i

3.)

f(x) =

(x - 4) (x – 2 + 3i)

f(x) = (x2 -4x + 13)

(x - 2 – 3i)

(x - 4)

= x3 - 4x2 + 13xf(x)

Solution

- 4x2 + 16x- 52

= x3 - 8x2 + 29xf(x) - 52


Example Problem

Write a polynomial function with the given zeros: 3, 1 + 2i

4.)

= x3 - 5x2 + 11xf(x) - 15

Irrational Root Theorem

a b+If the irrational number is a root

of a polynomial equation, then a b- is

also a root

Note

Irrationals root always come in pairs

Unit 3Polynomial Inequalities


Polynomial Inequalities

They can be written in the forms:

0f x 0f x 0f x 0f x

To solve the inequality f (x) > 0 is to find the values of x that make f (x) positive

To solve the inequality f (x) < 0 is to find the values of x that make f (x) negative

If the expression f (x) is a product, we can find its sign by looking at the sign of each of its factors.


Determine the real number values of x that cause the given function to be (a) zero, (b) positive, (c) negative.

223 1 4f x x x x= + + -

Real zeros: x = –3, 4

We use a sign chart to find the sings of f(x) at any values of x:

–3 4

2

- + - 2

+ + - 2

+ + +

Negative Positive Positive

The function is positive on 3,4 4,-

The function is negative on , 3- -

Polynomial InequalitiesFor the same function, give the solution of each of the following:

223 1 4f x x x x= + + -

3,4 4, - 223 1 4 0x x x+ + -

3, - 223 1 4 0x x x+ + -

, 3 - - 223 1 4 0x x x+ + -

, 3 4 - - 223 1 4 0x x x+ + -

Polynomial InequalitiesSolve the given inequality analytically.

3 22 7 10 24 0x x x- - +

Let 3 22 7 10 24f x x x x= - - +

Use the RZT to list the possible rational zeros!!!

1 31, 2, 3, 4, 6, 8, 12, 24, ,

2 2

Look at the graph x = 4 seems like a zero of f(x)


Solve the given inequality analytically.

3 22 7 10 24 0x x x- - +

4 2 –7 –10 24

8 4 –24

2 1 –6 0

24 2 6f x x x x= - + -

4 2 3 2f x x x x= - - +

Zeros: 4, 3

2, - 2

Polynomial InequalitiesSolve the given inequality analytically.

3 22 7 10 24 0x x x- - +

4 2 3 2f x x x x= - - +

Now, for the sign chart:

–2 Positive Negative x3/2 4 PositiveNegative

Solution: 2,3 2 4,-


Solve the given inequality analytically.

1.) (x – 2)(2x – 1)(x + 1) > 0

2.) (x + 3)(3x + 2)(x - 5) ≤ 0

3.) 3x3 – 13x2 + 13x - 3 ≥ 0

x = 1, x = 3 and x = 1/3

Polynomial Inequalities with Unusual Answers…

•Give the solution to each of the following:

2 27 2 1 0x x+ + ,-

2 27 2 1 0x x+ + ,-

•The graph???

2 27 2 1 0x x+ +

2 27 2 1 0x x+ +

•No Solution

•No Solution

Polynomial Inequalities with Unusual Answers…

•Give the solution to each of the following:

22 3 3 2 5 0x x x- + + , 5 2 5 2,- - -

•The graph???

22 3 3 2 5 0x x x- + + ,-

22 3 3 2 5 0x x x- + + •No Solution

22 3 3 2 5 0x x x- + + 5 2x = -

Documents

Unit 3 Polynomial Functions