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Complex Zeros of Polynomial Functions, Lecture notes of Algebra

Universal Technical Institute of California IncAlgebra

The process of finding complex zeros of polynomial functions using the given information and the fundamental theorem of algebra. It includes various examples with detailed steps and explanations.

Typology: Lecture notes

2021/2022

Uploaded on 09/12/2022

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Chapter 5
528
The Zeros of a Polynomial Function
5.4 Complex Zeros; Fundamental Theorem of Algebra
1. Since complex zeros appear in conjugate pairs,
4+i
, the conjugate of
4−i
, is the
remaining zero of f.
2. Since complex zeros appear in conjugate pairs,
3−i
, the conjugate of
3+i
, is the
remaining zero of f.
3. Since complex zeros appear in conjugate pairs,
−i
, the conjugate of
i
, and
1−i
, the
conjugate of
1+i
, are the remaining zeros of f.
4. Since complex zeros appear in conjugate pairs,
2−i
, the conjugate of
2+i
, is the
remaining zero of f.
5. Since complex zeros appear in conjugate pairs,
−i
, the conjugate of
i
, and −2i, the
conjugate of
2i
, are the remaining zeros of f.
6. Since complex zeros appear in conjugate pairs,
−i
, the conjugate of
i
, is the remaining zero
of f.
7. Since complex zeros appear in conjugate pairs,
−i
, the conjugate of
i
, is the remaining zero
of f.
8. Since complex zeros appear in conjugate pairs,
2+i
, the conjugate of
2−i
, and
i
, the
conjugate of
−i
, are the remaining zeros of f.
9. Since complex zeros appear in conjugate pairs,
2−i
, the conjugate of
2+i
, and
−3+i
, the
conjugate of
−3−i
, are the remaining zeros of f.
10. Since complex zeros appear in conjugate pairs,
−i
, the conjugate of
i
,
3+2i
, the conjugate
of
3−2i
, and −2−i, the conjugate of −2+i, are the remaining zeros of f.
pf3
pf4
pf5
pf8
pf9
pfa

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Chapter 5

The Zeros of a Polynomial Function

5.4 Complex Zeros; Fundamental Theorem of Algebra

  1. Since complex zeros appear in conjugate pairs, 4 + i , the conjugate of 4 − i , is the remaining zero of f.
  2. Since complex zeros appear in conjugate pairs, 3 − i , the conjugate of 3 + i , is the remaining zero of f.
  3. Since complex zeros appear in conjugate pairs, − i , the conjugate of i , and 1 − i , the conjugate of 1 + i , are the remaining zeros of f.
  4. Since complex zeros appear in conjugate pairs, 2 − i , the conjugate of 2 + i , is the remaining zero of f.
  5. Since complex zeros appear in conjugate pairs, − i , the conjugate of i , and − 2 i , the conjugate of 2 i , are the remaining zeros of f.
  6. Since complex zeros appear in conjugate pairs, − i , the conjugate of i , is the remaining zero of f.
  7. Since complex zeros appear in conjugate pairs, − i , the conjugate of i , is the remaining zero of f.
  8. Since complex zeros appear in conjugate pairs, 2 + i , the conjugate of 2 − i , and i , the conjugate of − i , are the remaining zeros of f.
  9. Since complex zeros appear in conjugate pairs, 2 − i , the conjugate of 2 + i , and − 3 + i , the conjugate of − 3 − i , are the remaining zeros of f.
  10. Since complex zeros appear in conjugate pairs, − i , the conjugate of i , 3 + 2 i , the conjugate of 3 − 2 i , and − 2 − i , the conjugate of − 2 + i , are the remaining zeros of f.

Section 5.4 Complex Zeros; Fundamental Theorem of Algebra

  1. Since 3 + 2 i is a zero, its conjugate 3 − 2 i is also a zero of f. Finding the function:

f ( x ) = ( x − 4)( x − 4) ( x − (3 + 2 i )) ( x − (3 − 2 i ))

= (^) ( x^2 − 8 x + (^16) ) (( x − 3) − 2 i ) (( x − 3) + 2 i ) = (^) ( x^2 − 8 x + (^16) ) ( x^2 − 6 x + 9 − 4 i^2 ) = (^) ( x^2 − 8 x + (^16) ) ( x^2 − 6 x + (^13) ) = x^4 − 6 x^3 + 13 x^2 − 8 x^3 + 48 x^2 − 104 x + 16 x^2 − 96 x + 208 = x^4 − 14 x^3 + 77 x^2 − 200 x + 208

  1. Since 1 + 2 i and i are zeros, their conjugates 1 − 2 i and − i are also zeros of f. Finding the function:

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

= ( x − i )( x + i ) (( x − 1)− 2 i ) (( x − 1) + 2 i )

= (^) ( x^2 − i^2 ) ( x^2 − 2 x + 1 − 4 i^2 ) = (^) ( x^2 + (^1) ) ( x^2 − 2 x + (^5) ) = x^4 − 2 x^3 + 5 x^2 + 1 x^2 − 2 x + 5 = x^4 − 2 x^3 + 6 x^2 − 2 x + 5

  1. Since − i is a zero, its conjugate i is also a zero, and since 1 + i is a zero, its conjugate 1 − i is also a zero of f. Finding the function:

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

= ( x − 2) (^) ( x^2 − i^2 ) (( x − 1) − i )( ( x − 1)+ i ) = ( x − 2) (^) ( x^2 + (^1) ) ( x^2 − 2 x + 1 − i^2 ) = (^) ( x^3 − 2 x^2 + x − (^2) )( x^2 − 2 x + (^2) ) = x^5 − 2 x^4 + 2 x^3 − 2 x^4 + 4 x^3 − 4 x^2 + x^3 − 2 x^2 + 2 x − 2 x^2 + 4 x − 4 = x^5 − 4 x^4 + 7 x^3 − 8 x^2 + 6 x − 4

  1. Since i is a zero, its conjugate − i is also a zero; since 4 − i is a zero, its conjugate 4 + i is also a zero; and since 2 + i is a zero, its conjugate 2 − i is also a zero of f. Finding the function:

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

= (^) ( x^2 − i^2 ) (( x − 4) − i ) (( x − 4) + i ) (( x − 2) − i ) (( x − 2 )+ i ) = (^) ( x^2 + (^1) ) ( x^2 − 8 x + 16 − i^2 ) ( x^2 − 4 x + 4 − i^2 ) = (^) ( x^2 + (^1) )( x^2 − 8 x + (^17) ) ( x^2 − 4 x + (^5) ) = (^) ( x^4 − 8 x^3 + 17 x^2 + x^2 − 8 x + (^17) ) ( x^2 − 4 x + (^5) ) = (^) ( x^4 − 8 x^3 + 18 x^2 − 8 x + (^17) )( x^2 − 4 x + (^5) ) = x^6 − 4 x^5 + 5 x^4 − 8 x^5 + 32 x^4 − 40 x^3 + 18 x^4 − 72 x^3 + 90 x^2 − 8 x^3

  • 32 x^2 − 40 x + 17 x^2 − 68 x + 85 = x^6 − 12 x^5 + 55 x^4 − 120 x^3 + 139 x^2 − 108 x + 85

Section 5.4 Complex Zeros; Fundamental Theorem of Algebra

  1. Since − 2 i is a zero, its conjugate 2 i is also a zero of f. x − 2 i and x + 2 i are factors of f.

Thus, ( x − 2 i )( x + 2 i ) = x^2 + 4 is a factor of f. Using division to find the other factor:

x^2 + 4 2 x^4 + 5 x^3 + 5 x^2 + 20 x − 12 2 x^4 + 8 x^2 5 x^3 − 3 x^2 + 20 x 5 x^3 + 20 x − 3 x^2 − 12 − 3 x^2 − 12

2 x^2 + 5 x − 3 )

2 x^2 + 5 x − 3 = ( 2 x − 1)( x + 3) are factors and the remaining zeros are 12 and − 3. The zeros of f are 2 i , − 2 i , − 3 , 12.

  1. Since 3 i is a zero, its conjugate − 3 i is also a zero of h. x − 3 i and x + 3 i are factors of h. Thus, ( x − 3 i )( x + 3 i ) = x^2 + 9 is a factor of h. Using division to find the other factor:

x^2 + 9 3 x^4 + 5 x^3 + 25 x^2 + 45 x − 18 3 x^4 + 27 x^2 5 x^3 − 2 x^2 + 45 x 5 x^3 + 45 x − 2 x^2 − 18 − 2 x^2 − 18

3 x^2 + 5 x − 2 )

3 x^2 + 5 x − 2 = (3 x − 1)( x + 2) are factors and the remaining zeros are 13 and − 2. The zeros of h are 3 i , − 3 i , − 2, 13.

  1. Since 3 − 2 i is a zero, its conjugate 3 + 2 i is also a zero of h. x − (3 − 2 i ) and x − (3 + 2 i ) are factors of h. Thus, ( x − (3 − 2 i ))( x − (3 + 2 i )) = (( x − 3) + 2 i )(( x − 3) − 2 i ) = x^2 − 6 x + 9 − 4 i^2 = x^2 − 6 x + 13 is a factor of h. Using division to find the other factor:

x^2 − 6 x + 13 x^4 − 9 x^3 + 21 x^2 + 21 x − 130 x^4 − 6 x^3 + 13 x^2 − 3 x^3 + 8 x^2 + 21 x − 3 x^3 + 18 x^2 − 39 x − 10 x^2 + 60 x − 130 − 10 x^2 + 60 x − 130

x^2 − 3 x − 10 )

x^2 − 3 x − 10 = ( x + 2)( x − 5) are factors and the remaining zeros are –2 and 5. The zeros of h are 3 − 2 i , 3 + 2 i , − 2, 5.

Chapter 5 The Zeros of a Polynomial Function

  1. Since 1 + 3 i is a zero, its conjugate 1 − 3 i is also a zero of f. x − (1+ 3 i ) and x − (1− 3 i ) are factors of f. Thus, ( x − (1+ 3 i ))( x − (1− 3 i )) = (( x − 1) − 3 i )(( x − 1) + 3 i ) = x^2 − 2 x + 1 − 9 i^2 = x^2 − 2 x + 10 is a factor of f. Using division to find the other factor:

x^2 − 2 x + 10 x^4 − 7 x^3 + 14 x^2 − 38 x − 60 x^4 − 2 x^3 + 10 x^2 − 5 x^3 + 4 x^2 − 38 x − 5 x^3 + 10 x^2 − 50 x − 6 x^2 + 12 x − 60 − 6 x^2 + 12 x − 60

x^2 − 5 x − 6 )

x^2 − 5 x − 6 = ( x + 1)( x − 6 ) are factors and the remaining zeros are –1 and 6. The zeros of f are 1 + 3 i , 1− 3 i , − 1 , 6.

  1. Since − 4 i is a zero, its conjugate 4 i is also a zero of h. x − 4 i and x + 4 i are factors of h.

Thus, ( x − 4 i )( x + 4 i ) = x^2 + 16 is a factor of h. Using division to find the other factor:

x^2 + 16 3 x^5 + 2 x^4 + 15 x^3 + 10 x^2 − 528 x − 352 3 x^5 + 48 x^3 2 x^4 − 33 x^3 + 10 x^2 2 x^4 + 32 x^2 − 33 x^3 − 22 x^2 − 528 x − 33 x^3 − 528 x − 22 x^2 − 352 − 22 x^2 − 352

3 x^3 + 2 x^2 − 33 x − 22 )

3 x^3 + 2 x^2 − 33 x − 22 = x^2 (3 x + 2) − 11(3 x + 2) = ( 3 x + 2 ) ( x^2 − 11) = (3 x + 2 )( x − (^11) ) ( x + (^11) ) are factors and the remaining zeros are − 32 , 11, and − 11. The zeros of h are 4 i , − 4 i , − 11, 11, − 32.

Chapter 5 The Zeros of a Polynomial Function

Using the quadratic formula to find the zeros of x^2 − 6 x + 13 = 0 :

x =

−(− 6) ± (− 6)^2 − 4(1)(13)

2(1) =^

2 =^

6 ± 4 i 2 =^3 ±^2 i^. The complex zeros are 2, 3 − 2 i , 3 + 2 i.

  1. f ( x ) = x^3 + 13 x^2 + 57 x + 85 Step 1: f ( x ) has 3 complex zeros. Step 2: By Descartes Rule of Signs, there are no positive real zeros. f (− x ) = (− x )^3 + 13(− x )^2 + 57(− x ) + 85 = − x^3 + 13 x^2 − 57 x + 85 ; thus, there are 3 or 1 negative real zeros. Step 3: Possible rational zeros: p = ± 1 , ± 5, ± 17, ± 85; q = ±1;

p q = ±^1 ,^ ±^5 ,^ ±^ 17,^ ±^85 Step 4: Using synthetic division: −5 1 13 57 85 − 5 − 40 − 85 1 8 17 0

)

Since the remainder is 0, x + 5 is a factor. The other factor is the quotient: x^2 + 8 x + 17.

Using the quadratic formula to find the zeros of x^2 + 8 x + 17 = 0 :

x =

− 8 ± 2 i 2 = −^4 ±^ i^. The complex zeros are − 5 , − 4 − i , − 4 + i.

  1. f ( x ) = x^4 + 5 x^2 + 4 = (^) ( x^2 + (^4) ) ( x^2 + (^1) ) = ( x + 2 i )( x − 2 i )( x + i )( x − i )

The zeros are: − 2 i , − i , i , 2 i.

  1. f ( x ) = x^4 + 13 x^2 + 36 = (^) ( x^2 + (^4) ) ( x^2 + (^9) ) = ( x + 2 i )( x − 2 i )( x + 3 i )( x − 3 i )

The zeros are: − 3 i , − 2 i , 2 i , 3 i.

  1. f ( x ) = x^4 + 2 x^3 + 22 x^2 + 50 x − 75 Step 1: f ( x ) has 4 complex zeros. Step 2: By Descartes Rule of Signs, there is 1 positive real zero. f (− x ) = (− x )^4 + 2(− x )^3 + 22(− x )^2 + 5 0 (− x ) − 75 = x^4 − 2 x^3 + 22 x^2 − 50 x − 75 thus, there are 3 or 1 negative real zeros. Step 3: Possible rational zeros: p = ± 1 , ± 3 , ± 5 , ± 15, ± 25, ± 75; q = ±1; p q

Section 5.4 Complex Zeros; Fundamental Theorem of Algebra

Step 4: Using synthetic division: −3 1 2 22 50 − 75 − 3 3 − 75 75 1 − 1 25 − 25 0

)

Since the remainder is 0, x + 3 is a factor. The other factor is the quotient: x^3 − x^2 + 25 x − 25 = x^2 ( x − 1) + 25( x − 1) = ( x − 1)( x^2 + (^25) ) = ( x − 1)( x + 5 i )( x − 5 i )

The complex zeros are − 3 , 1 , − 5 i , 5 i.

  1. f ( x ) = x^4 + 3 x^3 − 19 x^2 + 27 x − 252 Step 1: f ( x ) has 4 complex zeros. Step 2: By Descartes Rule of Signs, there are 3 or 1 positive real zeros. f (− x ) = (− x )^4 + 3(− x )^3 − 19(− x )^2 + 27(− x ) − 252 = x^4 − 3 x^3 − 19 x^2 − 27 x − 252 thus, there is 1 negative real zero. Step 3: Possible rational zeros: p = ± 1 , ± 2, ± 3, ± 4, ± 6 , ± 7, ± 9, ± 12, ± 14, ± 18, ± 21, ± 28, ± 36, ± 42, ± 63, ± 84, ± 126, ± 252; q = ±1; The possible rational zeros are the same as the values of p. Step 4: Using synthetic division: −7 1 3 − 19 27 − 252 − 7 28 − 63 252 1 − 4 9 − 36 0

)

Since the remainder is 0, x + 7 is a factor. The other factor is the quotient: x^3 − 4 x^2 + 9 x − 36 = x^2 ( x − 4 )+ 9( x − 4 ) = ( x − 4 )( x^2 + (^9) ) = ( x − 4 ) ( x + 3 i )( x − 3 i )

The complex zeros are −7, 4, − 3 i , 3 i.

  1. f ( x ) = 3 x^4 − x^3 − 9 x^2 + 159 x − 52 Step 1: f ( x ) has 4 complex zeros. Step 2: By Descartes Rule of Signs, there are 3 or 1 positive real zeros. f (− x ) = 3(− x )^4 − (− x )^3 − 9(− x )^2 + 159(− x ) − 52 = 3 x^4 + x^3 − 9 x^2 − 159 x − 52 thus, there is 1 negative real zero. Step 3: Possible rational zeros:

p = ± 1 , ± 2, ± 4, ± 13, ± 26, ± 52; q = ± 1 , ± 3; p q

Section 5.4 Complex Zeros; Fundamental Theorem of Algebra

  1. One of the remaining zeros must be 4 + i , the conjugate of 4 − i. The third zero is a real number. Thus the fourth zero must also be a real number in order to have a degree 4 polynomial.