The document discusses synthetic division and factoring polynomials. It provides an example of using synthetic division to divide one polynomial by another. Key points covered include: the remainder theorem, which states that the remainder of dividing a polynomial P(x) by (x-c) is equal to the value of P(c); and the factor theorem, which states that c is a zero of P(x) if and only if (x-c) is a factor of P(x). The document then provides an example of using the fact that P(1)=0 to factor a polynomial completely.

1. 3.2 Dividing Polynomials
Psalm 32:8 I will instruct you and teach you in
the way you should go; I will counsel you with
my eye upon you.

2. Review of Synthetic Division

3. Review of Synthetic Division
works only if dividing by (x − a)

4. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2

5. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
1 2 -3 1

6. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1

7. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1

8. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
1

9. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2
1

10. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2
1 0

11. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0
1 0

12. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0
1 0 -3

13. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0 6
1 0 -3

14. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0 6
1 0 -3 7

15. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0 6
1 0 -3 7

16. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0 6
1 0 -3 7
remainder

17. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1
Example :
x+2
-2 1 2 -3 1
-2 0 6
1 0 -3 7
quotient remainder

18. Review of Synthetic Division
works only if dividing by (x − a)
3 2
x + 2x − 3x + 1 2 7
Example : x − 3+
x+2 x+2
-2 1 2 -3 1
-2 0 6
1 0 -3 7
quotient remainder

19. Same problem using Polynomial Division

20. Same problem using Polynomial Division
3 2
x+2 x + 2x − 3x + 1

21. Same problem using Polynomial Division
2
x
3 2
x+2 x + 2x − 3x + 1

22. Same problem using Polynomial Division
2
x
3 2
x+2 x + 2x − 3x + 1
3 2
x +2x

23. Same problem using Polynomial Division
2
x
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract

24. Same problem using Polynomial Division
2
x
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x

25. Same problem using Polynomial Division
2
x +0
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x

26. Same problem using Polynomial Division
2
x +0
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0

27. Same problem using Polynomial Division
2
x +0
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract

28. Same problem using Polynomial Division
2
x +0
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1

29. Same problem using Polynomial Division
2
x +0 −3
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1

30. Same problem using Polynomial Division
2
x +0 −3
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1
−3x − 6

31. Same problem using Polynomial Division
2
x +0 −3
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1
−3x − 6 subtract

32. Same problem using Polynomial Division
2
x +0 −3
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1
−3x − 6 subtract
7

33. Same problem using Polynomial Division
2
x +0 −3
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1
−3x − 6 subtract
7
2 7
x − 3+
x+2

34. Same problem using Polynomial Division
2
x +0 −3
3 2
x+2 x + 2x − 3x + 1
3
x +2x 2
subtract
0 − 3x
0+0 subtract
−3x + 1
−3x − 6 subtract
7
7
or
2
x − 3+ x 2 − 3 R7
x+2

35. If the divisor is in the form of
(x-a)
synthetic division is faster

36. 4 3
6x + 2x − x + 2
Divide: 2
x − 2x + 2

37. 4 3
6x + 2x − x + 2
Divide: 2
x − 2x + 2
2
6x + 14x + 16
2 4 3 2
x − 2x + 2 6x + 2x + 0x − x + 2
6x 4 − 12x 3 + 12x 2
14x 3 − 12x 2 − x
3 2
14x − 28x + 28x
2
16x − 29x + 2
2
16x − 32x + 32
3x − 30

38. 4 3
6x + 2x − x + 2
Divide: 2
x − 2x + 2
2
6x + 14x + 16
2 4 3 2
x − 2x + 2 6x + 2x + 0x − x + 2
6x 4 − 12x 3 + 12x 2
14x 3 − 12x 2 − x
3 2
14x − 28x + 28x
2
16x − 29x + 2
2
16x − 32x + 32
3x − 30
2
Q(x) = 6x + 14x + 16
R(x) = 3x − 30

39. Factor & Remainder Theorems

40. Factor & Remainder Theorems
Remainder Theorem:
If the polynomial P(x) is divided by (x-c)
then the remainder is the value of P(c).

41. Factor & Remainder Theorems
Remainder Theorem:
If the polynomial P(x) is divided by (x-c)
then the remainder is the value of P(c).
3 2
x + 2x − 3x + 1
Example :
x+2

42. Factor & Remainder Theorems
Remainder Theorem:
If the polynomial P(x) is divided by (x-c)
then the remainder is the value of P(c).
3 2
x + 2x − 3x + 1 we already did this ...
Example :
x+2 R=7

43. Factor & Remainder Theorems
Remainder Theorem:
If the polynomial P(x) is divided by (x-c)
then the remainder is the value of P(c).
3 2
x + 2x − 3x + 1 we already did this ...
Example :
x+2 R=7
3 2
P(−2) = (−2) + 2(−2) − 3(−2) + 1

44. Factor & Remainder Theorems
Remainder Theorem:
If the polynomial P(x) is divided by (x-c)
then the remainder is the value of P(c).
3 2
x + 2x − 3x + 1 we already did this ...
Example :
x+2 R=7
3 2
P(−2) = (−2) + 2(−2) − 3(−2) + 1
P(−2) = (−8) + (8) − (−6) + 1

45. Factor & Remainder Theorems
Remainder Theorem:
If the polynomial P(x) is divided by (x-c)
then the remainder is the value of P(c).
3 2
x + 2x − 3x + 1 we already did this ...
Example :
x+2 R=7
3 2
P(−2) = (−2) + 2(−2) − 3(−2) + 1
P(−2) = (−8) + (8) − (−6) + 1
P(−2) = 7

46. Key Points to remember:

47. Key Points to remember:
If P(c)=0, then we know

48. Key Points to remember:
If P(c)=0, then we know
1) c is a root

49. Key Points to remember:
If P(c)=0, then we know
1) c is a root
2) c is a zero

50. Key Points to remember:
If P(c)=0, then we know
1) c is a root
2) c is a zero
3) c is a solution to P(x)=0

51. Key Points to remember:
If P(c)=0, then we know
1) c is a root
2) c is a zero
3) c is a solution to P(x)=0
4) c is an x-intercept

52. Factor Theorem:
c is a zero of P(x) iff (x-c) is a factor of P(x).
which means that P(c)=0

53. Let P(x) = x + 21x − 157x + 135 . Use the fact
3 2
that P(1) = 0 to factor P(x) completely.

54. Let P(x) = x + 21x − 157x + 135 . Use the fact
3 2
that P(1) = 0 to factor P(x) completely.
1 1 21 -157 135
1 22 -135
1 22 -135 0

55. Let P(x) = x + 21x − 157x + 135 . Use the fact
3 2
that P(1) = 0 to factor P(x) completely.
1 1 21 -157 135
1 22 -135
1 22 -135 0
2
x + 22x − 135

56. Let P(x) = x + 21x − 157x + 135 . Use the fact
3 2
that P(1) = 0 to factor P(x) completely.
1 1 21 -157 135
1 22 -135
1 22 -135 0
2
x + 22x − 135 use Quadratic Formula

57. Let P(x) = x + 21x − 157x + 135 . Use the fact
3 2
that P(1) = 0 to factor P(x) completely.
1 1 21 -157 135
1 22 -135
1 22 -135 0
2
x + 22x − 135 use Quadratic Formula
x = 5, − 27

58. Let P(x) = x + 21x − 157x + 135 . Use the fact
3 2
that P(1) = 0 to factor P(x) completely.
1 1 21 -157 135
1 22 -135
1 22 -135 0
2
x + 22x − 135 use Quadratic Formula
x = 5, − 27
∴ (x − 1)(x − 5)(x + 27)

59. Find a polynomial of lowest degree that has
zeros of 0, 2 and 5.

60. Find a polynomial of lowest degree that has
zeros of 0, 2 and 5.
By the Zero Product Property:
x=0 x−2=0 x−5=0

61. Find a polynomial of lowest degree that has
zeros of 0, 2 and 5.
By the Zero Product Property:
x=0 x−2=0 x−5=0
x(x − 2)(x − 5) = 0

62. Find a polynomial of lowest degree that has
zeros of 0, 2 and 5.
By the Zero Product Property:
x=0 x−2=0 x−5=0
x(x − 2)(x − 5) = 0
2
x(x − 7x + 10) = 0

63. Find a polynomial of lowest degree that has
zeros of 0, 2 and 5.
By the Zero Product Property:
x=0 x−2=0 x−5=0
x(x − 2)(x − 5) = 0
2
x(x − 7x + 10) = 0
3 2
x − 7x + 10x = 0

64. Find a polynomial of lowest degree that has
zeros of 0, 2 and 5.
By the Zero Product Property:
x=0 x−2=0 x−5=0
x(x − 2)(x − 5) = 0
2
x(x − 7x + 10) = 0
3 2
x − 7x + 10x = 0
graph to verify ...

65. HW #2
“Great results cannot be achieved at once; and
we must be satisfied to advance in life as we
walk, step by step.” Samuel Smiles