3.3 the fundamental theorem of algebra x

3.3 the fundamental theorem of algebra

1. The Fundamental Theorem of Algebra

2. The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials.

3. The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials. From this theorem we also get the corresponding statements about factoring real polynomials.

4. The Fundamental Theorem of Algebra: The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials. From this theorem we also get the corresponding statements about factoring real polynomials.

5. The Fundamental Theorem of Algebra: I. ( Completeness Theorem ) Let P(x) be a degree n polynomial then P(x) has exactly n complex roots , counting orders. The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials. From this theorem we also get the corresponding statements about factoring real polynomials.

6. The Fundamental Theorem of Algebra: I. ( Completeness Theorem ) Let P(x) be a degree n polynomial then P(x) has exactly n complex roots , counting orders. II. If P(x) is a real polynomial , then its complex roots must be in conjugate pairs, i.e. if a + bi is a root then a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials. From this theorem we also get the corresponding statements about factoring real polynomials.

7. The Fundamental Theorem of Algebra: I. ( Completeness Theorem ) Let P(x) be a degree n polynomial then P(x) has exactly n complex roots , counting orders. II. If P(x) is a real polynomial , then its complex roots must be in conjugate pairs, i.e. if a + bi is a root then a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials. From this theorem we also get the corresponding statements about factoring real polynomials. The proof of part I of this theorem is beyond the scope of this class.

8. The Fundamental Theorem of Algebra: I. ( Completeness Theorem ) Let P(x) be a degree n polynomial then P(x) has exactly n complex roots , counting orders. II. If P(x) is a real polynomial , then its complex roots must be in conjugate pairs, i.e. if a + bi is a root then a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra The Fundamental Theorem of Algebra establishes the number of roots and the type of roots for polynomials. From this theorem we also get the corresponding statements about factoring real polynomials. The proof of part I of this theorem is beyond the scope of this class. We will show part II. But we need some rules about conjugates.

9. The Fundamental Theorem of Algebra II . If P(x) is a real polynomial , then its complex roots must be in conjugate pairs, i.e. if a + bi is a root then a – bi must also be a root and vice versa.

10. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i,

11. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, if z = 3, then z * = 3.

12. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative,

13. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically

14. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b.

15. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b.

16. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * *

17. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * *

18. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] = a + c – ( b + d) i * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * *

19. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] = a + c – ( b + d) i = * * z 1 + z 2 . * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * *

20. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * Similarly one may show part b. To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] = a + c – ( b + d) i = * * z 1 + z 2 . * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * *

21. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] = a + c – ( b + d) i = * * z 1 + z 2 . * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * * Similarly one may show part b. From part b, we get

22. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, ( z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] = a + c – ( b + d) i = * * z 1 + z 2 . * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * * (az) = a z where a is a real number since a = a. * * c. * Similarly one may show part b . From part b, we get

23. Let z * denote the conjugate operation of the complex number z. The Fundamental Theorem of Algebra For example, if z = 3 – 2i then z * = 3 + 2i, (z 1 + z 2 ) = z 1 + z 2 . * * * a. (z 1 z 2 ) = z 1 z 2 . * * * To see part a, let z 1 = a + bi and z 2 = c + di, then z 1 = a – bi and z 2 = c – di respectively. Hence ( z 1 + z 2 ) = [( a + bi ) + ( c + di)] = a + c – ( b + d) i = * * z 1 + z 2 . * * if z = 3, then z * = 3. The conjugation operation and algebraic operations are commutative, specifically b. * * (az) = a z where a is a real number since a = a. * * c. * (z ) n = (z n ) for all complex numbers z. * * d. Similarly one may show part b . From part b, we get

24. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that:

25. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof:

26. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x).

27. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0,

28. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a

29. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a = a n ( z n ) * + a n-1 (z n-1 ) * + … + a 1 z * + a by part d

30. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a = a n ( z n ) * + a n-1 (z n-1 ) * + … + a 1 z * + a by part d = (a n z n ) * + (a n-1 z n-1 ) * + … + (a 1 z ) * + a by part c

31. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a = a n ( z n ) * + a n-1 (z n-1 ) * + … + a 1 z * + a by part d = (a n z n ) * + (a n-1 z n-1 ) * + … + (a 1 z ) * + a by part c = (a n z n + a n-1 z n-1 + … + a 1 z + a ) * by part a

32. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a = a n ( z n ) * + a n-1 (z n-1 ) * + … + a 1 z * + a by part d = (a n z n ) * + (a n-1 z n-1 ) * + … + (a 1 z ) * + a by part c = (a n z n + a n-1 z n-1 + … + a 1 z + a ) * by part a = (P(z)) *

33. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a = a n ( z n ) * + a n-1 (z n-1 ) * + … + a 1 z * + a by part d = (a n z n ) * + (a n-1 z n-1 ) * + … + (a 1 z ) * + a by part c = (a n z n + a n-1 z n-1 + … + a 1 z + a ) * by part a = (P(z)) * = (0) * = 0.

34. II. If P(x) is a real polynomial , then the complex roots must be in conjugate pairs, i.e. if z = a + bi is a root then z * = a – bi must also be a root and vice versa. The Fundamental Theorem of Algebra Now we are ready to prove that: Proof: Let P(x) = a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 x + a be a real polynomial and z is a root of P(x). So if P(z) = a n z n + a n-1 z n-1 + … + a 1 z + a = 0, then P(z * ) = a n (z * ) n + a n-1 (z * ) n-1 + … + a 1 z * + a = a n ( z n ) * + a n-1 (z n-1 ) * + … + a 1 z * + a by part d = (a n z n ) * + (a n-1 z n-1 ) * + … + (a 1 z ) * + a by part c = (a n z n + a n-1 z n-1 + … + a 1 z + a ) * by part a = (P(z)) * = (0) * = 0. Hence z * is also a root. QED

35. The Fundamental Theorem of Algebra Making 2 nd -degree Real Equations Recall the most important algebra of a conjugate pair z = a + bi and z * = a – bi: z + z * = (a + bi) + (a – bi) = 2a z(z * ) = ( a + bi )( a – bi) = a 2 + b 2

36. The Fundamental Theorem of Algebra Making 2 nd -degree Real Equations Recall the most important algebra of a conjugate pair z = a + bi and z * = a – bi: z + z * = (a + bi) + (a – bi) = 2a z(z * ) = ( a + bi )( a – bi) = a 2 + b 2 One 2 nd -degree real polynomial with conjugate roots z = a + bi and z * = a – bi is ( x – z)(x – z * ) = x 2 – (z + z * )x + z * z * = x 2 – ( 2a )x + ( a 2 + b 2 )

37. The Fundamental Theorem of Algebra Making 2 nd -degree Real Equations Recall the most important algebra of a conjugate pair z = a + bi and z * = a – bi: z + z * = (a + bi) + (a – bi) = 2a z(z * ) = ( a + bi )( a – bi) = a 2 + b 2 One 2 nd -degree real polynomial with conjugate roots z = a + bi and z * = a – bi is ( x – z)(x – z * ) = x 2 – (z + z * )x + z * z * = x 2 – ( 2a )x + ( a 2 + b 2 ) Example A. a. Find a real 2 nd degree polynomial with roots z = 3 + 2i and z * = 3 – 2i.

38. The Fundamental Theorem of Algebra Making 2 nd -degree Real Equations Recall the most important algebra of a conjugate pair z = a + bi and z * = a – bi: z + z * = (a + bi) + (a – bi) = 2a z(z * ) = ( a + bi )( a – bi) = a 2 + b 2 One 2 nd -degree real polynomial with conjugate roots z = a + bi and z * = a – bi is ( x – z)(x – z * ) = x 2 – (z + z * )x + z * z * = x 2 – ( 2a )x + ( a 2 + b 2 ) Example A. a. Find a real 2 nd degree polynomial with roots z = 3 + 2i and z * = 3 – 2i. z + z * = 2a = 6 and z(z * ) = a 2 + b 2 = 13 So a polynomial having z and z * as roots is ( x – z)(x – z * ) = x 2 – (z + z * )x + z * z * = x 2 – 6 x + 13

39. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * .

40. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * . In any given context, in order to recover the specific 2 nd -deg polynomials, we need another piece of data.

41. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * . In any given context, in order to recover the specific 2 nd -deg polynomials, we need another piece of data. Example A. b. Find the real 2 nd degree polynomial Q(x) with roots z = 3 + 2i and z * = 3 – 2i and with Q(1) = 2. From part a.

42. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * . In any given context, in order to recover the specific 2 nd -deg polynomials, we need another piece of data. Example A. b. Find the real 2 nd degree polynomial Q(x) with roots z = 3 + 2i and z * = 3 – 2i and with Q(1) = 2. From part a. Q(x) = k(x – z)(x – z * ) = k(x 2 – 6x + 13)

43. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * . In any given context, in order to recover the specific 2 nd -deg polynomials, we need another piece of data. Example A. b. Find the real 2 nd degree polynomial Q(x) with roots z = 3 + 2i and z * = 3 – 2i and with Q(1) = 2. From part a. Q(x) = k(x – z)(x – z * ) = k(x 2 – 6x + 13) Since Q(1) = k(x 2 – 6x + 13) = k((1) 2 – 6(1) + 13) = 2

44. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * . In any given context, in order to recover the specific 2 nd -deg polynomials, we need another piece of data. Example A. b. Find the real 2 nd degree polynomial Q(x) with roots z = 3 + 2i and z * = 3 – 2i and with Q(1) = 2. From part a. Q(x) = k(x – z)(x – z * ) = k(x 2 – 6x + 13) Since Q(1) = k(x 2 – 6x + 13) = k((1) 2 – 6(1) + 13) = 2 Hence 8k = 2 or k = 1/4.

45. The Fundamental Theorem of Algebra The general 2 nd degree polynomials of the form k (x – z)(x – z * ) where k ≠ 0 is a real number, also have roots z and z * . In any given context, in order to recover the specific 2 nd -deg polynomials, we need another piece of data. Example A. b. Find the real 2 nd degree polynomial Q(x) with roots z = 3 + 2i and z * = 3 – 2i and with Q(1) = 2. From part a. Q(x) = k(x – z)(x – z * ) = k(x 2 – 6x + 13) Since Q(1) = k(x 2 – 6x + 13) = k((1) 2 – 6(1) + 13) = 2 Hence 8k = 2 or k = 1/4. So Q(x) = ¼ (x 2 – 6x + 13), this is the specific equation given that Q(1) = 2 .

46. The Fundamental Theorem of Algebra From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary.

47. The Fundamental Theorem of Algebra From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r).

48. The Fundamental Theorem of Algebra From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial.

49. The Fundamental Theorem of Algebra From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the:

50. The Fundamental Theorem of Algebra From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the: Factorization Theorem of Real Polynomials

51. The Fundamental Theorem of Algebra i.e. P(x ) = A(x – #)( x – #).. (x 2 + #x + #)( x 2 + #x + #)... From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the: Factorization Theorem of Real Polynomials Let P(x) = Ax n +… be a real polynomial, then P(x) = AL 1 (x)L 2 (x).. Q 1 (x)Q 2 (x)..

52. The Fundamental Theorem of Algebra i.e. P(x ) = A(x – #)( x – #).. (x 2 + #x + #)( x 2 + #x + #)... From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the: Factorization Theorem of Real Polynomials Let P(x) = Ax n +… be a real polynomial, then P(x) = AL 1 (x)L 2 (x).. Q 1 (x)Q 2 (x).. where each L i (x)’s is of the linear form (x – r i ) with r i a real number ,

53. The Fundamental Theorem of Algebra i.e. P(x ) = A(x – #)( x – #).. (x 2 + #x + #)( x 2 + #x + #)... From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the: Each gives a real root. Factorization Theorem of Real Polynomials Let P(x) = Ax n +… be a real polynomial, then P(x) = AL 1 (x)L 2 (x).. Q 1 (x)Q 2 (x).. where each L i (x)’s is of the linear form (x – r i ) with r i a real number ,

54. The Fundamental Theorem of Algebra i.e. P(x ) = A(x – #)( x – #).. (x 2 + #x + #)( x 2 + #x + #)... From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the: Each gives a real root. Factorization Theorem of Real Polynomials Let P(x) = Ax n +… be a real polynomial, then P(x) = AL 1 (x)L 2 (x).. Q 1 (x)Q 2 (x).. where each L i (x)’s is of the linear form (x – r i ) with r i a real number, and each Q i (x ) is an irreducible real quadratic,

55. The Fundamental Theorem of Algebra i.e. P(x ) = A(x – #)( x – #).. (x 2 + #x + #)( x 2 + #x + #)... From the Completeness Theorem a degree n real polynomial P has exactly n roots, real or imaginary. a. A real root r of P must come from a factor (x – r). b. A complex root z always comes with its conjugate z * , from an irreducible 2 nd degree real polynomial. Summarizing these, we have the: Each gives a real root. Each gives a conjugate pair as roots. Factorization Theorem of Real Polynomials Let P(x) = Ax n +… be a real polynomial, then P(x) = AL 1 (x)L 2 (x).. Q 1 (x)Q 2 (x).. where each L i (x)’s is of the linear form (x – r i ) with r i a real number, and each Q i (x ) is an irreducible real quadratic,

56. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots.

57. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64)

58. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8)

59. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8) = 3(x – 2)( x 2 + 2x +4 ) ( x + 2)(x 2 – 2x + 4)

60. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8) = 3(x – 2)( x 2 + 2x +4 ) ( x + 2)(x 2 – 2x + 4) (x – 2) and ( x + 2) give the real roots x = 2 and –2.

61. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8) = 3(x – 2)( x 2 + 2x +4 ) ( x + 2)(x 2 – 2x + 4) (x – 2) and ( x + 2) give the real roots x = 2 and –2. ( x 2 + 2x +4 ) and (x 2 – 2x + 4) are irreducible quadratics with conjugate complex roots.

62. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8) = 3(x – 2)( x 2 + 2x +4 ) ( x + 2)(x 2 – 2x + 4) (x – 2) and ( x + 2) give the real roots x = 2 and –2. ( x 2 + 2x +4 ) and (x 2 – 2x + 4) are irreducible quadratics with conjugate complex roots. The roots of x 2 + 2x + 4 are x = -2 ±  -12 2

63. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8) = 3(x – 2)( x 2 + 2x +4 ) ( x + 2)(x 2 – 2x + 4) (x – 2) and ( x + 2) give the real roots x = 2 and –2. ( x 2 + 2x +4 ) and (x 2 – 2x + 4) are irreducible quadratics with conjugate complex roots. The roots of x 2 + 2x + 4 are x = = -1 ± i  3 -2 ±  -12 2

64. The Fundamental Theorem of Algebra Example B. Factor 3x 6 – 192 completely into real factors and list all its roots. 3x 6 – 192 = 3(x 6 – 64) = 3(x 3 – 8) ( x 3 + 8) = 3(x – 2)( x 2 + 2x +4 ) ( x + 2)(x 2 – 2x + 4) (x – 2) and ( x + 2) give the real roots x = 2 and –2. ( x 2 + 2x +4 ) and (x 2 – 2x + 4) are irreducible quadratics with conjugate complex roots. The roots of x 2 + 2x + 4 are x = = -1 ± i  3 -2 ±  -12 2 The roots of x 2 – 2x + 4 are x = = 1 ± i  3 2 ±  -12 2

65. The Fundamental Theorem of Algebra Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

66. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), (2 + i ) Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

67. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

68. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

69. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

70. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, so the quadratic with roots ( 2 + i ), (2 – i ) is (x 2 – 4x + 5). Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

71. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, so the quadratic with roots ( 2 + i ), (2 – i ) is (x 2 – 4x + 5). Therefore P(x) = k(x – 1)( x 2 + 1)(x 2 – 4x + 5) for some constant k. Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

72. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, so the quadratic with roots ( 2 + i ), (2 – i ) is (x 2 – 4x + 5). Therefore P(x) = k(x – 1)( x 2 + 1)(x 2 – 4x + 5) for some constant k. But P(0) = 10 Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

73. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, so the quadratic with roots ( 2 + i ), (2 – i ) is (x 2 – 4x + 5). Therefore P(x) = k(x – 1)( x 2 + 1)(x 2 – 4x + 5) for some constant k. But P(0) = 10 = k(0 – 1)( + 1)(0 – + 5) = -5k, Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

74. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, so the quadratic with roots ( 2 + i ), (2 – i ) is (x 2 – 4x + 5). Therefore P(x) = k(x – 1)( x 2 + 1)(x 2 – 4x + 5) for some constant k. But P(0) = 10 = k(0 – 1)( + 1)(0 – + 5) = -5k, so k = -2. Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

75. The Fundamental Theorem of Algebra P(x) is a real polynomial so the complex roots are in conjugate pairs which are x = i , - i , (2 – i ), ( 2 + i ) i + (- i ) = 0, i * (- i ) = 1, so the quadratic with roots i , - i is ( x 2 + 1). ( 2 + i ) + (2 – i ) = 4, ( 2 + i ) * (2 – i ) = 5, so the quadratic with roots ( 2 + i ), (2 – i ) is (x 2 – 4x + 5). Therefore P(x) = k(x – 1)( x 2 + 1)(x 2 – 4x + 5) for some constant k. But P(0) = 10 = k(0 – 1)( + 1)(0 – + 5) = -5k, so k = -2. Hence P(x) = -2(x – 1)( x 2 + 1)(x 2 – 4x + 5). Example C. Given that x = 1, i , 2 – i are roots of a degree 5 real polynomial P(x) and that P(0 ) = 10. Find P(x).

76. The Fundamental Theorem of Algebra Exercise A. Factor the following polynomials completely into real factors and list all its real and complex roots . 1. x 3 – 1 2 . x 3 – 8 3. 8x 3 + 27 4. 27x 3 + 125 5. x 4 – 16 6. 16x 4 – 81 11. x 6 + 1 12. x 6 – 1 7. x 4 – x 2 – 2 8. 4x 4 + 3x 2 – 1 9. x 4 + 3x 2 + 2 10. 3 x 4 + 4x 2 + 1 B . Find the real polynomial P(x ) given its roots, the degree, and its value at one point (the initial condition). 1 . roots: x = 1 + i , degree 2 with P(0) = 5. 2. roots: x = 2 – i , degree 2 with P(0) = –2. 3. roots: x = 2, 1 + 3i, degree 3 with P(1) = –4. 4. roots: x = –1, 2 – i , degree 3 with P (–1) = 3. 5. roots: x = –2 + i , 1 + 2i, degree 4 with P(1) = –3. 6. roots: x = –1 – i , 3 + i , degree 4 with P (–1) = 1.

77. The Fundamental Theorem of Algebra B . Find the real polynomial P(x ) given its roots, the degree, and its value at one point (the initial condition). 7. roots: x = 0 ( ord = 2 ), i , degree 4, P(1) = 2. 8. roots: x = 1, 1 + i , ( ord = 2), degree 5, P(2) = 1. 9. roots: x = –1, 2, i – 2 ( ord = 2), degree 6, P(1) = 2. 10. roots: x = 1 ( ord = 2), i√2 ( ord = 2), degree 6, P (–1 ) = 2. 11. roots: x = 0, –1, –2 + i√3, degree 4, P(1) = 1. 12. roots: x = 0 ( ord = 2), 3 + i √5 ( ord = 2), degree 6, P(1) = 2. 13. What may we conclude from the F undamental Theorem of Algebra about the roots of polynomials with only even degrees of x’s? only odd degrees of x’s?

78. The Fundamental Theorem of Algebra (Answers to the odd problems) Exercise A. 1. (x – 1)(x 2 + x +1), x = 1 , (– 1± 3i)/2 3. (2x + 3)(4x 2 – 6x + 9), x = - , (1 – √3 i), (1 + √3 i) 5. (x – 2)(x + 2)(x 2 + 4), x = – 2, 2, –2i, 2i 11. ( x 2 +1 )( x 2 + √ 3x + 1)( x 2 – √3x + 1 ), x = – i, (– √3 ± i )/ 2, ( √ 3 ± i )/2, 7. (x 2 + 1)( x 2 – 2), x = – √2, √2, – i, i 9. (x 2 + 2)( x 2 +1), x = – √ 2 i, √ 2 i, – i, i Exercise B . 1 . ( x 2 – 2x + 2) 3. (x 3 – 4x 2 + 14x – 20) 5 . – ( x 4 + 2x 3 + 2x 2 + 10x + 25) 3 2 3 4 3 4 5 2 4 9 3 40 7. x 4 – x 2 9 . (– 1/10 )(x + 1)(x – 2)(x 2 + 4x + 5) 11. ( x 4 + 5x 3 + 11x 2 + 7x) 1 24