3-2 Zeros of Polynomial Functions (Presentation)

3‐2 Zeros of Polynomial Functions

Unit 3 Quadratic and Polynomial Functions

Concepts and Objectivesp j

Objective #10jFind rational zeros of a polynomial functionUse the Fundamental Theorem of Algebra to find a f ti th t ti fi i ditifunction that satisfies given conditionsFind all zeros of a polynomial function

Factor Theorem

The polynomial x – k is a factor of the polynomialThe polynomial x k is a factor of the polynomial f(k) if and only if f(k) = 0.

Example: Determine whether x + 4 is a factor of ( ) = − + +4 23 48 8 32f x x x x

Factor Theorem

The polynomial x – k is a factor of the polynomialThe polynomial x k is a factor of the polynomial f(k) if and only if f(k) = 0.

Example: Determine whether x + 4 is a factor of( ) = − + +4 23 48 8 32f x x x x

Yes it is

− −4 3 0 48 8 32048–12 –32

Yes, it is. –123 0 08

Rational Zeros Theorem

If p/ i ti l b itt i l tIf p/q is a rational number written in lowest terms, and if p/q is a zero of f, a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.

In other words, the numerator is a factor of the last number and the denominator is a factor of the firstnumber and the denominator is a factor of the first coefficient.


Example: For the polynomial function defined byp p y y

(a) List all possible rational zeros( ) = − − + +4 3 28 26 27 11 4f x x x x x

(b) Find all rational zeros and factor f(x) into linear f tfactors.



(a) List all possible rational zeros( ) = − − + +4 3 28 26 27 11 4f x x x x x

For a rational number to be zero, pmust be a f t f 4 d t b f t f 8

pq

factor of 4 and qmust be a factor of 8:{ }∈ ± ± ±1, 2, 4p

⎧ ⎫1 1 1{ }∈ ± ± ± ±, 1, 2, 4, 8q

⎧ ⎫∈ ± ± ± ± ± ±⎨ ⎬⎩ ⎭

1 1 11, 2, 4, , ,2 4 8

pq



(b) Find all rational zeros and factor f(x) into linear ( ) = − − + +4 3 28 26 27 11 4f x x x x x

factors.

L k t th h f f( ) t j d h itLook at the graph of f(x) to judge where it crosses the x‐axis:



Use synthetic division to show that –1 is a zero:( ) = − − + +4 3 28 26 27 11 4f x x x x x

− − −1 8 26 27 11 4–8 34 –7 –4

8 –34 7 4 0

( ) ( )( )= + − + +3 21 8 34 7 4f x x x x x


Example, cont.p ,

Now, we can check the remainder for a zero at 4:( ) ( )( )= + − + +3 21 8 34 7 4f x x x x x

−4 8 34 7 432 –8 –4

8 –2 –1 0( ) ( )( )( )= + − − −21 4 8 2 1f x x x x x

zeros are at –1, 4,

( ) ( )( )( )( )= + − + −1 4 4 1 2 1f x x x x x

−1 1,zeros are at 1, 4, , 4 2

Fundamental Theorem of Algebrag

Every function defined by a polynomial of degreeEvery function defined by a polynomial of degree 1 or more has at least one complex zero.

A function defined by a polynomial of degree nhas at most n distinct zeros.

The number of times a zero occurs is referred to as theThe number of times a zero occurs is referred to as the multiplicity of the zero.


Example: Find a function f defined by a polynomial of p f y p ydegree 3 that satisfies the following conditions.(a) Zeros of –3, –2, and 5; f(–1) = 6

(b) 4 is a zero of multiplicity 3; f(2) = –24


Example: Find a function f defined by a polynomial of p f y p ydegree 3 that satisfies the following conditions.(a) Zeros of –3, –2, and 5; f(–1) = 6

Since f is of degree 3, there are at most 3 zeros, so these three must be it Therefore f(x) has the formthree must be it. Therefore, f(x) has the form

( ) ( )( )( )+ +3 2 5f x a x x x

( ) ( )( ) ( )( )( )= − − − − −3 2 5f x a x x x

( ) ( )( )( )= + + −3 2 5f x a x x x


Example, cont.p ,We also know that f(–1) = 6, so we can solve for a:

( ) ( )( )( )− −= −− −+ +31 1 1 2 51f a

( )( )( )= − = −6 2 1 6 12a a

= −1

a

Therefore, or2

a

( ) ( )( )( )= − + + −1 3 2 52

f x x x x

( ) = − + +31 19 152 2

f x x x


Example: Find a function f defined by a polynomial of p f y p ydegree 3 that satisfies the following conditions.(b) 4 is a zero of multiplicity 3; f(2) = –24

This means that the zero 4 occurs 3 times:( ) ( )( )( )4 4 4f

or( ) ( )( )( )= − − −4 4 4f x a x x x

( ) ( )=34f x a x( ) ( )= −4f x a x


Example, cont.p ,

Since f(2) = –24, we can solve for a:( ) ( )= −

342 2f a

( )− = − = −324 2 8a a

Therefore, or=3a

( ) ( )= −33 4f x x

( ) 3 2f ( ) = − + −3 23 36 144 192f x x x x

Conjugate Zeros Theoremj g

If f(x) defines a polynomial function having onlyIf f(x) defines a polynomial function having only real coefficients and if z = a + bi is a zero of f(x), where a and b are real numbers, then z = a – biis also a zero of f(x).

This means that if 3 + 2i is a zero for a polynomial function with real coefficients, then it also has 3 – 2i as a zero.


Example: Find a polynomial function of least degree p p y ghaving only real coefficients and zeros –4 and 3 – i.


Example: Find a polynomial function of least degree p p y ghaving only real coefficients and zeros –4 and 3 – i.The complex number 3 + imust also be a zero, so the l i l h t l t th d h t b t l tpolynomial has at least three zeros and has to be at least

degree 3. We don’t know anything else about the function, so we will let a = 1.

( ) ( )( ) ( )( ) ( )( )= − − − − − +4 3 3f x x x i x i

( ) ( ) ( ) ( ) ( )( )( )= + − + − − + − +24 3 3 3 3f x x x i x i x i i

( ) ( )( )2 24 3 3 9f x x x x ix x ix i= + − − − + + −

( ) ( )( )= + − + = − − +2 3 24 6 10 2 14 40f x x x x x x x

Putting It All Togetherg g

Example: Find all zeros of ( ) = − − + −4 3 217 55 50f x x x x xpgiven that 2 + i is a zero.

( )f


Example: Find all zeros of ( ) = − − + −4 3 217 55 50f x x x x xpgiven that 2 + i is a zero.

( )f

( )( )First, we divide the function by :( )( )− +2x i

+ − − −2 1 1 17 55 50i

1

2+i

1+i

1+3i

–16+3i

–35–10i

20–10i

50

0

( ) ( )( ) ( ) ( ) ( )( )= − + + + + − + + −3 22 1 16 3 20 10f x x i x i x i x i


Example, cont.p ,( ) ( )( ) ( ) ( ) ( )( )= − + + + + − + + −3 22 1 16 3 20 10f x x i x i x i x i

We also know that the conjugate, 2 – i is a zero, so we can divide the remainder by this:

− + − + −2 1 1 16 3 20 10i i i i

12–i3

6–3i10

–20+10i01 3 –10 0

( ) ( )( ) ( )( )( )= − + − − + −22 2 3 10f x x i x i x x


Example, cont.p ,

( ) ( )( ) ( )( )( )= − + − − + −22 2 3 10f x x i x i x x

Lastly, we can factor or use the quadratic formula to find our remaining zeros:

( )( )+ +2 3 10 5 2x x x x( )( )+ − = + −2 3 10 5 2x x x x

( ) ( )( ) ( )( )( )( )= − + − − + −2 2 5 2f x x i x i x x

So, our zeros are at 2+i, 2–i, –5, and 2.

Homework

College Algebrag gPage 337: 5‐60 (×5s)

Turn In: 20, 30, 40a, 50, 60

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3-2 Zeros of Polynomial Functions (Presentation)