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