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.
Rational Zeros Theorem
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.
Rational Zeros Theorem
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
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
Rational Zeros Theorem
Example: For the polynomial function defined byp p y y
(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:
Rational Zeros Theorem
Example: For the polynomial function defined byp p y y
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
Rational Zeros Theorem
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.
Fundamental Theorem of Algebrag
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
Fundamental Theorem of Algebrag
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
Fundamental Theorem of Algebrag
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
Fundamental Theorem of Algebrag
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
Fundamental Theorem of Algebrag
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.
Conjugate Zeros Theoremj g
Example: Find a polynomial function of least degree p p y ghaving only real coefficients and zeros –4 and 3 – i.
Conjugate Zeros Theoremj g
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
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
( )( )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
Putting It All Togetherg g
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
Putting It All Togetherg g
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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