Warm-Up 2/261.
J
10 β22β23=10 β 25=10 β32
Rigor:You will learn how to find the real and complex
zeros of polynomial functions.
Relevance:You will be able to use graphs and equations of
polynomial functions to solve real world problems.
2-4 Zeros of Polynomial Functions
Example 1a: List possible rational zeros and determine which, if any are zeros.
π (π₯)=π₯3+2π₯+1
Step 1 Identify possible rational zeros.
and
Step 2 Test possible rational zeros to determine if they are rational zeros.
ππ=Β± Factors 1
Factors 1=1ππβ1
There are no rational zeros.
Example 1b: List possible rational zeros and determine which, if any, are zeros.
π (π₯ )=π₯4+4 π₯3 β12π₯β 9Step 1 Identify possible rational zeros.
Step 2 Test possible rational zeros to determine if they are rational zeros.
ππ=Β± Factors 9
Factors 1=Β±1 , Β± 3 ,ππ Β± 9
There are two rational zeros at .
0
3
3
1 4 β 12
β
β 9
β 1 β 3
β 31 β 9
9
0
β 1 β 3
9
0
1 3 β 9
β β 3 0
β 31 0
β 3
π (π₯)=(π₯+1) (π₯+3 )(π₯2β 3)
π₯2β3=0π₯2=3π₯=Β±β3
Example 2: List possible rational zeros and determine which, if any, are zeros.
h (π₯ )=3 π₯3β 7 π₯2 β2 2π₯+8Step 1 Identify possible rational zeros.
Step 2 Test possible rational zeros to determine if they are rational zeros.
ππ=Β± Factors 8
Factors 3=Β±1 , Β± 2 , Β± 4 Β± 8 ,Β± 13 , Β± 2
3 , Β± 43 ,ππ Β± 8
3
There are 3 rational zeros at .
β 22
β 8
β 13
3 β 7 8
β β 6 26
43 0
β 2 4
β 1
3 β 13
β 12 β 4
03
4
h (π₯)=(π₯+2) (π₯β 4 )(3 π₯β1)3 π₯β1=0
3 π₯=1π₯=
13
Example 6: Write a polynomial function of least degree with real coefficients in standard form that has the given zeros.
y = (x + 2)(x β 4)[x β (3 β i)][x β (3 + i)]
y = (x + 2)(x β 4)[(x β 3) + i][(x β 3) β i]
y = (xΒ² β 4x + 2x β 8)[(x β 3)Β² β i(x β 3) + i(x β 3) β iΒ²]
y = (xΒ² β 2x β 8)[(x β 3)Β² + 1]
y = (xΒ² β 2x β 8)(xΒ² β 6x + 9 + 1)
y = (xΒ² β 2x β 8)(xΒ² β 6x + 10)
y = x4 β 6x3 + 10xΒ² β 2x3 + 12xΒ² β 20x β 8xΒ² + 48x β 80
y = x4 β 8x3 + 14xΒ² + 28x β 80
(x β c)3 + i
Example 7: Write function as (a) the product of linear and irreducible quadratic factors and (b) the product of linear factors. Then (c) list all zeros.π (π₯ )=π₯5 β18 π₯3+30π₯2 β19π₯+30
(a) the product of linear and irreducible quadratic factors
π(π₯)=(π₯+5) (π₯β2 )(π₯β 3)(π₯2+1)
π₯2+1
Example 7: Write function as (a) the product of linear and irreducible quadratic factors and (b) the product of linear factors. Then (c) list all zeros.π (π₯ )=π₯5 β18 π₯3+30π₯2 β19π₯+30
(b) the product of linear factors
π(π₯)=(π₯+5) (π₯β2 )(π₯β 3)(π₯2+1)
π₯2+1=0π₯2=β1π₯=Β±ββ1π₯=Β± π
π(π₯)=(π₯+5) (π₯β2 )(π₯β 3)(π₯+ π)(π₯βπ)(c) List all zerosThere are 5 zeros: .
Example 8: Use given zeros to find all complex zeros. Then write the linear factorization of the function.π (π₯ )=π₯4 β 6π₯3+20 π₯2β 22π₯β 13 given 2 β3 πas a zero of π .
20
24 + 3i
β 4 β 3i
1 β 6 β 22
β
β 13
2 β 3i β 17 + 6i
3 + 6i1 2 + 3i
13
0
2 β 3i
3 + 6i
β 2 β 3i
β 2
1 β 4 β 3i 2 + 3i
β 2 + 3i β 4 β 6i
β 11 0
2 + 3i
π (π₯ )= [π₯β (2 β3 π ) ] [π₯β (2+3 π ) ] (π₯2 β2 π₯β1)
π₯2β2 π₯β1
Example 8: Use given zeros to find all complex zeros. Then write the linear factorization of the function.π (π₯ )=π₯4 β 6π₯3+20 π₯2β 22π₯β 13 given 2 β3 πas a zero of π .
π (π₯ )= [π₯β (2 β3 π ) ] [π₯β (2+3 π )] (π₯2 β2 π₯β1)
π₯=βπΒ±βπ2 β 4ππ2π
π₯=2 Β±β4 β 4 (1)(β1)
2π₯=2 Β±β8
2
π₯=2 Β±2β22
π₯=1 Β±β2
π (π₯ )= [π₯β (2 β3 π ) ] [π₯β (2+3 π ) ] [π₯β ( 1+β2 ) ] [π₯β ( 1ββ2 ) ]
ββ1math!
2-4 Assignment: TX p127, 4-16 EOE & 32-52 EOE
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