3.2 properties of division and roots
- 1. Properties of Division and Roots
- 2. Properties of Division and Roots * Consequences of the division algorithm * Remainders, roots and linear factors
- 3. Properties of Division and Roots The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root.
- 4. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root.
- 5. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root.
- 6. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root.
- 7. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. For example, a polynomial with roots x = 1, -2, each with order 1, is (x – 1)(x + 2) = x2 + x – 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root.
- 8. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. For example, a polynomial with roots x = 1, -2, each with order 1, is (x – 1)(x + 2) = x2 + x – 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root. In general, polynomials of the form k(x – c1)m (x – c2)m ..(x – cn)m where k is any non–zero number, have roots 1 2 n
- 9. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. For example, a polynomial with roots x = 1, -2, each with order 1, is (x – 1)(x + 2) = x2 + x – 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root. In general, polynomials of the form k(x – c1)m (x – c2)m ..(x – cn)m where k is any non–zero number, have roots x = c1 of order m1, 1 2 n
- 10. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. For example, a polynomial with roots x = 1, -2, each with order 1, is (x – 1)(x + 2) = x2 + x – 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root. In general, polynomials of the form k(x – c1)m (x – c2)m ..(x – cn)m where k is any non–zero number, have roots x = c1 of order m1, c2 of order m2, 1 2 n
- 11. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. For example, a polynomial with roots x = 1, -2, each with order 1, is (x – 1)(x + 2) = x2 + x – 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root. In general, polynomials of the form k(x – c1)m (x – c2)m ..(x – cn)m where k is any non–zero number, have roots x = c1 of order m1, c2 of order m2, .. , and cn of order mn. 1 2 n
- 12. Properties of Division and Roots The polynomial x5 + 2x4 + x3 = x3(x + 1)2 has two roots, x = 0 and x = -1. The order of x = 0 is 3 and the order of the root x = -1 is 2. It is easy to make up polynomials that have specified roots and orders. For example, a polynomial with roots x = 1, -2, each with order 1, is (x – 1)(x + 2) = x2 + x – 2. The order (or multiplicity) of a root c of a polynomial P(x) is the number of repetitions of c as a root. In general, polynomials of the form k(x – c1)m (x – c2)m ..(x – cn)m where k is any non–zero number, have roots x = c1 of order m1, c2 of order m2, .. , and cn of order mn. Next we will see that conversely if c is a root of a polynomial P(x), then (x – c) must be a factor of P(x). 1 2 n
- 13. Recall the division theorem from the last section. Properties of Division and Roots
- 14. Long Division Theorem: Using long division for P(x) ÷ D(x), we may obtain a quotient Q(x) and a reminder R(x) such that D(x) P(x) = Q(x) + D(x) R(x) with deg R(x) < deg D(x). Recall the division theorem from the last section. Properties of Division and Roots
- 15. Long Division Theorem: Using long division for P(x) ÷ D(x), we may obtain a quotient Q(x) and a reminder R(x) such that D(x) P(x) = Q(x) + D(x) R(x) with deg R(x) < deg D(x). Recall the division theorem from the last section. Multiply both sides of this identity by D(x), we get the multiplicative form P(x) = Q(x)D(x) + R(x). Properties of Division and Roots
- 16. Long Division Theorem: Using long division for P(x) ÷ D(x), we may obtain a quotient Q(x) and a reminder R(x) such that D(x) P(x) = Q(x) + D(x) R(x) with deg R(x) < deg D(x). Recall the division theorem from the last section. Multiply both sides of this identity by D(x), we get Hence D(x) is a factor of P(x) if and only if R(x) = 0. Properties of Division and Roots the multiplicative form P(x) = Q(x)D(x) + R(x).
- 17. Long Division Theorem: Using long division for P(x) ÷ D(x), we may obtain a quotient Q(x) and a reminder R(x) such that D(x) P(x) = Q(x) + D(x) R(x) with deg R(x) < deg D(x). Recall the division theorem from the last section. Multiply both sides of this identity by D(x), we get Hence D(x) is a factor of P(x) if and only if R(x) = 0. In the case that the divisor D(x) = (x – c), the remainder R(x) must be a number, say r. Properties of Division and Roots the multiplicative form P(x) = Q(x)D(x) + R(x).
- 18. Long Division Theorem: Using long division for P(x) ÷ D(x), we may obtain a quotient Q(x) and a reminder R(x) such that D(x) P(x) = Q(x) + D(x) R(x) with deg R(x) < deg D(x). Recall the division theorem from the last section. Multiply both sides of this identity by D(x), we get Hence D(x) is a factor of P(x) if and only if R(x) = 0. In the case that the divisor D(x) = (x – c), the remainder R(x) must be a number, say r. Hence P(x) = Q(x)(x – c) + r. Properties of Division and Roots the multiplicative form P(x) = Q(x)D(x) + R(x).
- 19. Long Division Theorem: Using long division for P(x) ÷ D(x), we may obtain a quotient Q(x) and a reminder R(x) such that D(x) P(x) = Q(x) + D(x) R(x) with deg R(x) < deg D(x). Recall the division theorem from the last section. Multiply both sides of this identity by D(x), we get Hence D(x) is a factor of P(x) if and only if R(x) = 0. In the case that the divisor D(x) = (x – c), the remainder R(x) must be a number, say r. Hence P(x) = Q(x)(x – c) + r. Set x = c on both sides, we get P(c) = 0 + r = r. We state this as a theorem. Properties of Division and Roots the multiplicative form P(x) = Q(x)D(x) + R(x).
- 20. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Properties of Division and Roots
- 21. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). This gives a way to find the value of P(c) via synthetic division. Properties of Division and Roots
- 22. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). This gives a way to find the value of P(c) via synthetic division. Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. Properties of Division and Roots
- 23. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). This gives a way to find the value of P(c) via synthetic division. Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½). Properties of Division and Roots
- 24. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). This gives a way to find the value of P(c) via synthetic division. Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½). Set up the synthetic division Properties of Division and Roots
- 25. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division.
- 26. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 4 Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division.
- 27. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 4 2 0 Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division.
- 28. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 4 2 0 0 0 Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division.
- 29. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 4 2 0 0 0 6 0 Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division.
- 30. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 4 2 0 0 0 6 0 2 3 Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division. the remainder r
- 31. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Example A. Find P(1/2) given P(x) = 4x4 – 2x3 + 6x – 1 using synthetic division. We want the remainder of P(x) / (x – ½) Set up the synthetic division 4 –2 0 6 –11/2 4 2 0 0 0 6 0 2 3 Since the remainder r is the same as P(1/2), therefore P(1/2) = 2. Properties of Division and Roots This gives a way to find the value of P(c) via synthetic division. the remainder r
- 32. If (x – c) is a factor of P(x) so that P(x) = Q(x)(x – c) then of course x = c is a root of P(x) since P(c) = 0. Properties of Division and Roots
- 33. Conversely if x = c is a root of P(x) so that P(c) = 0, by the Remainder Theorem that P(c) is the remainder of P(x) ÷ (x – c), If (x – c) is a factor of P(x) so that P(x) = Q(x)(x – c) then of course x = c is a root of P(x) since P(c) = 0. Properties of Division and Roots
- 34. Conversely if x = c is a root of P(x) so that P(c) = 0, by the Remainder Theorem that P(c) is the remainder of P(x) ÷ (x – c), we see that P(x) ÷ (x – c) = Q(x) + 0, or that P(x) = Q(x)(x – c). If (x – c) is a factor of P(x) so that P(x) = Q(x)(x – c) then of course x = c is a root of P(x) since P(c) = 0. Properties of Division and Roots
- 35. Conversely if x = c is a root of P(x) so that P(c) = 0, by the Remainder Theorem that P(c) is the remainder of P(x) ÷ (x – c), we see that P(x) ÷ (x – c) = Q(x) + 0, or that P(x) = Q(x)(x – c). Therefore (x – c) is a factor of P(x). If (x – c) is a factor of P(x) so that P(x) = Q(x)(x – c) then of course x = c is a root of P(x) since P(c) = 0. Properties of Division and Roots
- 36. Conversely if x = c is a root of P(x) so that P(c) = 0, by the Remainder Theorem that P(c) is the remainder of P(x) ÷ (x – c), we see that P(x) ÷ (x – c) = Q(x) + 0, or that P(x) = Q(x)(x – c). Therefore (x – c) is a factor of P(x). Factor Theorem: Let P(x) be a polynomial, then x = c is a root of P(x) if and only if (x – c) is a factor of P(x), i.e. P(x) = Q(x)(x – c) for some Q(x). If (x – c) is a factor of P(x) so that P(x) = Q(x)(x – c) then of course x = c is a root of P(x) since P(c) = 0. Properties of Division and Roots
- 37. Conversely if x = c is a root of P(x) so that P(c) = 0, by the Remainder Theorem that P(c) is the remainder of P(x) ÷ (x – c), we see that P(x) ÷ (x – c) = Q(x) + 0, or that P(x) = Q(x)(x – c). Therefore (x – c) is a factor of P(x). Factor Theorem: Let P(x) be a polynomial, then x = c is a root of P(x) if and only if (x – c) is a factor of P(x), i.e. P(x) = Q(x)(x – c) for some Q(x). If (x – c) is a factor of P(x) so that P(x) = Q(x)(x – c) then of course x = c is a root of P(x) since P(c) = 0. Properties of Division and Roots The Factor Theorem says that knowing “x = c is a root of a polynomial P(x), i.e. P(c) = 0” is same as knowing that “(x – c) is a factor of P(x), i.e. P(x) = Q(x)(x – c)”.
- 38. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. Properties of Division and Roots
- 39. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). Properties of Division and Roots
- 40. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Properties of Division and Roots
- 41. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Hence x = 1 is a root of P(x). Therefore (x – 1) is a factor of P(x). Properties of Division and Roots
- 42. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Hence x = 1 is a root of P(x). Therefore (x – 1) is a factor of P(x). Properties of Division and Roots Let P(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0
- 43. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Hence x = 1 is a root of P(x). Therefore (x – 1) is a factor of P(x). Properties of Division and Roots Let P(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 then P(1) = an + an-1 + an-2 + … + a1 + a0
- 44. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Hence x = 1 is a root of P(x). Therefore (x – 1) is a factor of P(x). Properties of Division and Roots Let P(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 then P(1) = an + an-1 + an-2 + … + a1 + a0 Therefore an + an-1 + an-2 + … + a1 + a0 = 0 means P(1) = 0
- 45. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Hence x = 1 is a root of P(x). Therefore (x – 1) is a factor of P(x). Properties of Division and Roots Let P(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 then P(1) = an + an-1 + an-2 + … + a1 + a0 Therefore an + an-1 + an-2 + … + a1 + a0 = 0 means P(1) = 0 therefore that (x – 1) is a factor of P(x).
- 46. Example B. Use the Factor Theorem to show that (x – 1) is a factor P(x) = 3x100 – 5x50 + 9x30 – 13x20 + 6. We check to see if x = 1 is a root of P(x). P(1) = 3 – 5 + 9 – 13 + 6 = 0. Hence x = 1 is a root of P(x). Therefore (x – 1) is a factor of P(x). Properties of Division and Roots Let P(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 then P(1) = an + an-1 + an-2 + … + a1 + a0 Therefore an + an-1 + an-2 + … + a1 + a0 = 0 means P(1) = 0 therefore that (x – 1) is a factor of P(x). Fact I: For P(x) = anxn + an-1xn-1 + … + a1x + a0, then (x – 1) is a factor P(x) if and only if an + an-1 + an-2 + … + a1 + a0 = 0.
- 47. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 Properties of Division and Roots
- 48. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, Properties of Division and Roots
- 49. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, and that Properties of Division and Roots (x – 1) is a not factor of 5x4 – 7x – 2 because 5 – 7 – 2 = -4 = 0.
- 50. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, and that Properties of Division and Roots (x – 1) is a not factor of 5x4 – 7x – 2 because 5 – 7 – 2 = -4 = 0. Suppose the degree n of P is even,
- 51. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, and that Properties of Division and Roots P(-1) = an – an-1 + an-2 – an-3 … (x – 1) is a not factor of 5x4 – 7x – 2 because 5 – 7 – 2 = -4 = 0. Suppose the degree n of P is even, then
- 52. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, and that Properties of Division and Roots P(-1) = an – an-1 + an-2 – an-3 … (x – 1) is a not factor of 5x4 – 7x – 2 because 5 – 7 – 2 = -4 = 0. Having P(-1) = an – an-1 + an-2 – an-3 … = 0 means an + an-2 … = an-1 + an-3 … Suppose the degree n of P is even, then
- 53. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, and that Properties of Division and Roots P(-1) = an – an-1 + an-2 – an-3 … (x – 1) is a not factor of 5x4 – 7x – 2 because 5 – 7 – 2 = -4 = 0. Having P(-1) = an – an-1 + an-2 – an-3 … = 0 means an + an-2 … = an-1 + an-3 … Hence (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms. Suppose the degree n of P is even, then
- 54. Example C. By the above fact it's easy to see that (x – 1) is a factor of 4x6 – 5x4 + 12x2 – 9x – 2 because 4 – 5 + 12 – 9 – 2 = 0, and that Properties of Division and Roots P(-1) = an – an-1 + an-2 – an-3 … (x – 1) is a not factor of 5x4 – 7x – 2 because 5 – 7 – 2 = -4 = 0. Having P(-1) = an – an-1 + an-2 – an-3 … = 0 means an + an-2 … = an-1 + an-3 … Hence (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms. The same holds true if n is odd. Suppose the degree n of P is even, then
- 55. Properties of Division and Roots Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms
- 56. Example D. Find k given that (x + 1) is a factor of x9 – 2x8 + 3x7 –4x6 + k. Properties of Division and Roots Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms
- 57. Example D. Find k given that (x + 1) is a factor of x9 – 2x8 + 3x7 –4x6 + k. Properties of Division and Roots We must have 1 + 3 = Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms
- 58. Example D. Find k given that (x + 1) is a factor of x9 – 2x8 + 3x7 –4x6 + k. Properties of Division and Roots We must have 1 + 3 = -2 – 4 + k Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms
- 59. Example D. Find k given that (x + 1) is a factor of x9 – 2x8 + 3x7 –4x6 + k. Properties of Division and Roots We must have 1 + 3 = -2 – 4 + k 4 = - 6 + k 10 = k Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms
- 60. Example D. Find k given that (x + 1) is a factor of x9 – 2x8 + 3x7 –4x6 + k. Properties of Division and Roots We must have 1 + 3 = -2 – 4 + k 4 = - 6 + k 10 = k Example E. Factor x4 – 2x3 – 3x2 + 2x + 2 completely into real factors. Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms
- 61. Example D. Find k given that (x + 1) is a factor of x9 – 2x8 + 3x7 –4x6 + k. Properties of Division and Roots Fact II: For any polynomial P(x), (x + 1) is a factor of P(x) if and only if the sum of the coefficients of the even degree terms = the sum of the coefficients of the odd degree terms We must have 1 + 3 = -2 – 4 + k 4 = - 6 + k 10 = k Example E. Factor x4 – 2x3 – 3x2 + 2x + 2 completely into real factors. By observation that 1 – 2 – 3 + 2 + 2 = 0, we see that (x – 1) is a factor.
- 62. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor.
- 63. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21
- 64. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0
- 65. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0–1
- 66. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0–1 1 –1 –2 2 –2 2 0
- 67. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0–1 1 –1 –2 2 –2 2 0 So x4 – 2x3 – 3x2 + 2x + 2 = (x – 1)(x +1)(x2 – 2x – 2)
- 68. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0–1 1 –1 –2 2 –2 2 0 So x4 – 2x3 – 3x2 + 2x + 2 = (x – 1)(x +1)(x2 – 2x – 2) (x2 – 2x – 2) is irreducible,
- 69. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0–1 1 –1 –2 2 –2 2 0 So x4 – 2x3 – 3x2 + 2x + 2 = (x – 1)(x +1)(x2 – 2x – 2) (x2 – 2x – 2) is irreducible, its roots are 1 ± 3 by the quadratic formula.
- 70. Properties of Division and Roots At the same time, 1 – 3 + 2 = – 2 + 2 = 0, we see that (x + 1) is a factor. Use synthetic division twice 1 –2 –3 2 21 1 1 –1 –1 –4 –4 –2 –2 0–1 1 –1 –2 2 –2 2 0 So x4 – 2x3 – 3x2 + 2x + 2 = (x – 1)(x +1)(x2 – 2x – 2) (x2 – 2x – 2) is irreducible, its roots are 1 ± 3 by the quadratic formula. So x4 – 2x3 – 3x2 + 2x + 2 factors completely as (x – 1)(x +1)(x – (1 + 3))(x – (1 – 3)).
- 71. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Exercise A. Find the remainder r of P(x) ÷ (x – c) by evaluating P(c). Properties of Division and Roots 1. x – 1 –2x + 3 4. x2 – 9 7. x + 3 x2 – 2x + 3 2. x + 1 3x + 2 3. x – 2 3x + 1 8. x – 3 2x2 – 2x + 1 9. x + 2 –2x2 + 4x + 1 5. x + 2 x2 + 4 6. x – 3 x2 + 9 10. x3 – 2x + 3 11. x – 3 2x3 – 2x + 1 12. x + 2 –2x4 + 4x + 1 x – 3 x – 2
- 72. Remainder Theorem: Let P(x) be a polynomial, the remainder of P(x) ÷ (x – c) is r = P(c). Properties of Division and Roots B. Find P(c) by finding the remainder of P(x) ÷ (x – c). 1. Find P(1/2) given P(x) = 2x3 – 5x2 + 6x – 1 2. Find P(1/3) given P(x) = 3x3 – 4x2 + 7x – 1 3. Find P(1/4) given P(x) = 4x4 – 5x3 + 2x – 1 4. Find P(1/5) given P(x) = 5x4 + 4x3 + x – 1 5. Find P(1/6) given P(x) = 12x5 – 2x4 + 60x – 1 6. Find P(1/7) given P(x) = 7x20 – x19 + x – 1 7. Find P(1/8) given P(x) = 64x200 – x198 + x – 1 8. Find P(1/10) given P(x) = 20x200 – 2x199 – 100x19 + 10x18 – 1
- 73. C. Use the Factor Theorem to show that (x – c) is a factor of P(x) by checking that c is a root. Properties of Division and Roots 1. (x – 1) is a factor of P(x) = 2x3 – 5x2 + 4x – 1 2. (x – 1) is a factor of P(x) = 9x3 – 7x2 + 6x – 8 3. (x + 1) is a factor of P(x) = 100x3 + 99x2 – 98x – 97 4. (x + 1) is a factor of P(x) = 10x71 – 9x70 – 9x61 + 10x60 + 98x7 + 98 5. Check that (x + 2) is a factor of P(x) = x3 – 7x – 6, then factor P(x) completely. 7. Check that (x + 3) and (x + 4) are factors of P(x) = x4 + x3 –16x2 – 4x + 48 then factor it completely. 6. Check that (x + 3) is a factor of P(x) = x3 + 3x2 – 4x –12 then factor P(x) completely.
- 74. Properties of Division and Roots 10. Create a 4th degree polynomial in the expanded with (x + 1) as a factor. 11. Create a 5th degree polynomial in the expanded with (x + 1) as a factor. 8. Create a 4th degree polynomial in the expanded with (x – 1) as a factor. 9. Create a 5th degree polynomial in the expanded with (x – 1) as a factor. 12. Solve for k if (x + 1) is a factor of 2x3 – 5x2 + 4x + k. 13. Solve for k if (x + 2) is a factor of 2x3 – kx2 + 4x + k. 14. Solve for k if (x + 2) is a factor of kx3 – x2 + 3kx + k.
- 75. (Answers to odd problems) Exercise A. 1. 1 7. 18 3. 7 9. −15 5. 8 11. 49 Exercise B. 1. 1 3. −9/16 5. 9 7. 0 Exercise C. 5. 𝑃(𝑥) = (𝑥 + 2)(𝑥 + 1)(𝑥 − 3) 7. 𝑃(𝑥) = (𝑥 − 3)(𝑥 + 4)(𝑥 − 2)(𝑥 + 2) 11. 𝑃 𝑥 = 𝑥5 + 𝑥 = 𝑥4(𝑥 + 1) 9. 𝑃 𝑥 = 𝑥5 − 𝑥 = 𝑥4(𝑥 − 1) 13. 𝑘 = 8 Properties of Division and Roots