1. Real Zeros of Polynomial Functions
Section 2.3

2. Objectives
Use long division to divide polynomials by other polynomials.
Use synthetic division to divide polynomials by binomials of the form (x-k).
Use the Remainder and Factor Theorems.
Use the Rational Zero Test to determine possible rational zeros of polynomial functions.
Use Descarte’s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials.

3. Dividing Polynomials
Dividing a polynomial by a polynomial uses a “long division” technique that is similar to the process known as long division in dividing two numbers, which is reviewed on the next slide.

4. Dividing Polynomials Divide 43 into 72. Multiply 1 times 43.
Subtract 43 from 72.
Bring down 5.
Divide 43 into 295.
Multiply 6 times 43.
Subtract 258 from 295.
Bring down 6.
Divide 43 into 376.
Multiply 8 times 43.
We then write our result as
Subtract 344 from 376.
Nothing to bring down.

5. Dividing Polynomials
As you can see from the previous example, there is a pattern in the long division technique.
Divide
Multiply
Subtract
Bring down
Then repeat these steps until you can’t bring down or divide any longer.
We will incorporate this same repeated technique with dividing polynomials.

6. Division of Polynomials

7. Dividing Polynomials
Long division of polynomials is similar to long division of whole numbers.
When you divide two polynomials you can check the answer using the following:
dividend = (quotient • divisor) + remainder
The result is written in the form:
quotient +

8. Long Division of Polynomials
Procedure:
Arrange the terms of both polynomials in descending order of the exponents of the variable.
If either the dividend or the divisor has missing terms, insert these terms with coefficients of 0.
Divide
Check, by multiplying the divisor times the quotient and adding the remainder.
If the remainder is zero, then the divisor is a factor of the dividend.

9. Example: Divide x2 + 3x – 2 by x – 1 and check the answer.1.
+ 2 2.
x x 3. 2x
– 2
2x + 2 4.
– 4 5.
remainder 6.
Answer: x + 2 +
– 4
Check:
(x + 2)
quotient
(x + 1)
divisor
+ (– 4)
remainder
= x2 + 3x – 2
dividend
correct

10. Example: Divide 4x + 2x3 – 1 by 2x – 2 and check the answer.
+ 3
Write the terms of the dividend in descending order.
2x3 – 2x2
Since there is no x2 term in the dividend, add 0x2 as a placeholder.
2x2
+ 4x
2x2 – 2x 1. 2. 6x
– 1 4. 3.
6x – 6 5 5.
Answer: x2 + x + 3 5 6. 7. 8.
Check: (x2 + x + 3)(2x – 2) = 4x + 2x3 – 1 9.

11. Example: Divide x2 – 5x + 6 by x – 2.
– 3
x2 – 2x
– 3x
+ 6
– 3x + 6
Answer: x – 3 with no remainder.
Check: (x – 2)(x – 3)
= x2 – 5x + 6

12. Example: Divide x3 + 3x2 – 2x + 2 by x + 3 and check the answer.
– 2
Note: the first subtraction eliminated two terms from the dividend.
x x2
0x2
– 2x
+ 2
– 2x – 6
Therefore, the quotient skips a term. 8
Answer: x2 – 2 + 8
Check: (x + 3)(x2 – 2) + 8
= x3 + 3x2 – 2x + 2

13. Your Turn: Divide by . x + 2 ( ) x2 + x 2x + 2 ( ) 2x + 2 CHECK:
( )
x2 + x 2x
+ 2
( )
2x + 2
CHECK:
(x + 1)(x + 2)
= x2 + 2x + x + 2
= x2 + 3x + 2

14. Your Turn: Divide by . x3 - x2 + 2x - 2 ( ) x4 + x3 -x3 + x2 ( )
( )
x4 + x3
-x3
+ x2
( )
-x3 – x2
2x2
+ 0x
( )
2x2 + 2x
-2x
+ 1
( x - 2) -2 3

15. Polynomial Long Division:
Example:
2x4 + 3x3 + 5x – 1
x2 – 2x + 2

16. 2x2
2x4 = 2x2 x2

17. -( ) 2x2 +7x +10 2x4 -4x3 +4x2 7x3 - 4x2 +5x -( ) 7x3 - 14x2 +14x
-( )
2x4
-4x3
+4x2
7x3
- 4x2
+5x
-( )
7x x2 +14x
10x2 - 9x -1
7x3 = 7x x2
-( )
10x2 - 20x +20
11x - 21
remainder

18. The solution is written:
Quotient + Remainder over Divisor

19. Your Turn:
y4 + 2y2 – y y2 – y + 1
Answer: y2 + y y2 – y + 1

20. Your Turn: (5x3 – 4x2 + 7x  2) ÷ (x2 + 1). Solution:
Quotient is 5x – 4 with a remainder of 2x + 2.

21. Division Algorithm for Polynomials
Let f(x) and d(x) be two polynomials, with degree of d(x) greater than zero and less than the degree of f(x). Then there exists unique polynomials q(x) and r(x) such that
f(x) = d(x) • q(x) r(x)
Dividend = Divisor • Quotient + Remainder
where either r(x) = 0 or the degree of r(x) is less than the degree of d(x). The polynomial r(x) is called the remainder.

22. Synthetic Division
A Quick method of dividing polynomials by a divisor of the form (x-k).
Procedure:
Arrange the terms of the dividend in descending order and represent each term by it’s coefficient only, use zeros to represent the coefficients of any missing terms.
The divisor must be of the form (x-k) or rewritten in a form equivalent to (x-k).
Divide, by multiplying and adding.
Synthetic division yields the quotient and the remainder.

23. Synthetic Division
2x3+7x2+10x+15 / x+2

24. Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 1: Write the coefficients of the polynomial, and the k-value of the divisor on the left 3 1 -3 1 6

25. Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 2: Draw a line and write the first coefficient under the line. 3 1 -3 1 6 1

26. Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 3: Multiply the k-value, 3, by the number below the line and write the product below the next coefficient. 3 1 -3 1 6 3 1

27. Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Step 4: Write the sum of -3 and 3 below the line. 3 1 -3 1 6 3 1

28. Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Repeat steps 3 and 4. 3 1 -3 1 6 3 1 1

29. Synthetic Division
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
Repeat steps 3 and 4. 3 1 -3 1 6 3 3 1 1 9

30. Synthetic Division 3 1 -3 1 6 3 3 1 1 9 x2 + 1 + x – 3 9
Use synthetic division to find the quotient:
(x3 – 3x2 + x + 6) ÷ (x – 3)
The remainder is 9 and the resulting numbers are the coefficients of the quotient. 3 1 -3 1 6 3 3 1 1 9 x
x – 3 9

31. Synthetic Division (x3 – 3x2 + 4) ÷ (x – 2) Method 1 Method 2 x2 - x
x – 2
x3 – 3x2 + 0x + 4 2 -2 -4
- (x3 – 2x2) 1 -1 -2
-x2 + 0x
- (-x2 + 2x)
(x2 – x – 2)
–2x + 4
- (–2x + 4)

32. Synthetic Division
Synthetic division is derived from the recursive nature of the algorithm. (Recursive – specifies a beginning value and a way to calculate each successive value)
This method relies on the relationships between the coefficients of x in the product of the quotient and divisor.

33. Example
Use synthetic division to find the quotient and remainder.
The quotient is – 4x4 – 7x3 – 8x2 – 14x – 28 and the remainder is –6. 2 –4 1 6 2 50
Note: We must write a for the missing term. –8
–14
–16
–28
–56 –4 –7 –8
–14
–28 –6

34. Your Turn: Use synthetic division to divide 5x3+8x2-x+6 by x+2.

35. Synthetic Division Divide 6x3+7x2+x+1 by 2x+3 2x+3 = 2(x+3/2)
(6x3+7x2+x+1)/(2x+3) = (1/2)[(6x3+7x2+x+1)/(x+3/2)]

36. Synthetic Division

37. Remainder Theorem

38. Let f(x) = 2x4 + x3 – 3x2 – 5
What is f(2)?
f(2) = 2(2)4 + (2)3 – 3(2)2 – 5
f(2) = 2(16) + 8 – 3(4) – 5
f(2) = – 12 – 5
f(2) = 23
This is the same as the remainder when divided by (x – 2):
– –5 4 10 14 28 2 5 7 14 23
f(2) = 23

39. The Remainder Theorem There are two parts of the remainder theorem:
If the polynomial f(x) is divided by (x – a), the remainder will be a number that is equal to f(a).
i.e.. If f(x) is divided by x – 4, f(4) will give the value of the remainder.
Dividend = (quotient ∙ divisor) + remainder
Also can see this as f(x) = [q(x) ∙ (x – a)] + f(a).
The quotient is always a polynomial with one degree less than f(x).
Synthetic division is helpful in solving these problems (this can also be called synthetic substitution).
The quotient may also be called a depressed polynomial because it has one less degree than the original polynomial.

40. The Remainder Theorem That it means: Used to find the remainder.
When dividing a polynomial by a divisor of the form (x-k), the Remainder Theorem gives us two ways of calculating the remainder.
Use synthetic division to divide by (x-k).
Evaluate the function at f(k).

41. The Remainder Theorem
Given P(x) = 3x3 – 4x2 + 9x + 5 is divided by x – 6, find the remainder.
Method 1
Method 2 6
P(6) = 3(6)3 – 4(6)2 + 9(6) + 5 18 84
558
= 3(216) – 4(36) 3 14 93
563
= 648 –
= 563

42. The Remainder Theorem
Use synthetic division and direct substitution to find f(4) when f(x) = x4 – 6x3 + 8x2 + 5x + 13. OR
f(4) = 44 – 6(4)3 + 8(4)2 + 5(4) + 13
=256 –
f(4) = 33

43. The Remainder Theorem
Give the factors of x3 – 11x2 + 36x – 36 if one factor is x – 6.
So, after we divide the polynomial by x – 6
we are left with x2 – 5x + 6 which we can
solve by factoring into (x – 3)(x - 2).
This means the factors are (x – 6), (x – 3), and (x - 2).
This can also be written in the f(x) = quotient ∙ divisor + remainder.
This would look like f(x) = (x2 – 5x + 6)(x – 6) + 0
or f(x)=(x-2)(x-3)(x-6)..

44. The Remainder Theorem
It is usually easiest to evaluate a polynomial for a specific value of “x” by using synthetic division and the remainder theorem
Example: Given
Find:

45. Your Turn:
Find the remainder when f(x) =4x3+10x2-3x-8 is divided by (x+1).
Solution: f(-1)=1, the remainder is 1.

46. The Remainder Theorem Problems
Use synthetic division to do (4x3 – 9x2 – 10x – 2) divided by (x – 3). Then write the answer in the form f(x) = [quotient ∙ divisor] + remainder.
Given f(x) = 4x2 + 6x – 7, find f(-5) by synthetic division or direct substitution.
Find the factors of x3 + 6x2 – x – 30 if one factors is (x + 5).
Answers:
1) (4x2 + 3x – 1)(x – 3) – 5
2) f(-5) = 63
3) (x + 5), (x – 2), and (x + 3)

47. The Remainder Theorem Problem
Show that x=1 is a solution of the equation 2x3-x2+8=7x+2, and find any further solutions.
The first part is simple substitution, but you must show the result.
2(1)3-(1)2+8=7(1)+2
9=9, so x=1 is a solution.
Now have to solve the cubic equation 2x3-x2-7x+6=0. We know x=1 is a solution, so (x-1) is a factor. Using synthetic division,
So the quadratic factor is 2x2+x-6. This can be factored to
(2x-3)(x+2). Thus the remaining solutions are 3/2 and -2.

48. Review-The Remainder Theorem
The Remainder Theorem If a polynomial f(x) is divided by x – k, the remainder is equal to f(k).

49. The Factor Theorem

50. k is Zero of a Polynomial Function if P(k) = 0
Example Decide whether the given number is a zero of P.
Analytic Solution
(a)
(b)
The remainder is zero, so
x = 2 is a zero of P.
The remainder is not zero, so
x = –2 is not a zero of P.

51. The Factor Theorem From the previous example, part (a), we have
indicating that x – 2 is a factor of P(x).
Factor Theorem:
(x – a) is a factor of f(x) if and only if the
remainder (or f(a)) is equal to zero.
This is a good way to find the first factor of
a polynomial.

52. The Factor Theorem
When a polynomial division results in a zero remainder
The divisor is a factor
f(x) = (x – k) q(x) + 0
This would mean that f(k) = 0
That is … k is a zero of the function
It also means the quotient, q(x), is a factor of f(x).

53. Example using the Factor Theorem
Determine whether the second polynomial is a
factor of the first.
Solution Use synthetic division with k = –2.
Since the remainder is 0, x + 2 is a factor of P(x),
where

54. Example using the Factor Theorem
Use Remainder Theorem (evaluate f(1)) to determine whether x – 1 is a factor of x3 – x2 – 5x – 3.
Let x3 – x2 – 5x – 3 = 0
f(1) = (1)3 – (1)2 – 5(1) - 3
f(1) = 1 – 1 – 5 - 3
f(1) = -8
Since f(1) does not equal zero, x – 1 is not a factor.

55. Factoring a polynomial using the Factor Theorem
Factor f(x) = 2x3 + 11x2 + 18x + 9
Given f(-3)=0
Since f(-3)=0
x-(-3) or x+3 is a factor
So use synthetic division to find the others!!

56. Factoring a polynomial using the Factor Theorem cont.
So…. 2x3 + 11x2 + 18x + 9 factors to:
(x + 3)(2x2 + 5x + 3)
Now keep factoring gives you:
(x+3)(2x+3)(x+1)

57. Your Turn:
Show that (x+3) is a factor of f(x)=x3-19x-30. Then find the remaining factors f(x).
Solution: f(-3) = 0, since the remainder is 0, (x+3) is a factor.
Remaining factors are (x-5) and (x+2).

58. Example using the Factor Theorem
The function f(x)=x3+ax2-2x+b has (x+1) as a factor, and leaves a remainder of -3 when divided by x-2. Find the values of a and b.
a b a b
a 1+a a 4+4a
1 a-1 -1-a 1+a+b a 2+2a 4+4a+b
1+a+b= a+b=-3
Solve the system: a+b=1
4a+b=-7 ∴ a=-2, b=1

59. Review-The Factor Theorem
The Factor Theorem A polynomial f(x) has a factor (x – k), if and only if f(k)=0.

60. The Rational Zero Theorem

61. The Rational Zeros Theorem

62. The Rational Zero Theorem
What it Means:
The Rational Zero Theorem allows us to identify possible rational zeros of the polynomial. We then check the possible zeros, using the Remainder and Factor Theorems, to determine if they are actual zeros.
Does not agree the polynomial has any rational zeros.

63. Example Rational Zeros Theorem

64. Your Turn: Find all possible rational zeros of f(x) = 6x3-x2+9x+4.
Solution: ±1, ±2, ±4, ±1/6, ±1/3, ±1/2, ±2/3, ±4/3

65. Finding Rational Zeros
Example: Find the rational zeros for f(x) = x3 + x2 – 10x + 8
p/q represents all possible zeros where p are all factors of the constant, and q are all factors of the leading coefficient.
p : all the factors of q : all factors of 1
+ 1, + 2, + 4,
Here are all of the possible zeros for the function:
p , + 2, + 4, + 8 q
Possible zeros are , + 2, + 4, + 8
So which one do you pick?
Pick any. Find one that is a zero using synthetic division...

66. f(x) = x3 + x2 – 10x + 8 possible zeros + 1, + 2, + 4, + 8
Let’s try 1: –
– –
1 is a zero of the function
The depressed polynomial is x2 + 2x – 8
Find the zeros of x2 + 2x – 8 by factoring or (by using the quadratic formula)…
(x + 4)(x – 2) = 0
x = –4, x = 2
The zeros of f(x) are 1, –4, and 2

67. Finding Zeros of a Polynomial Function
Use Rational Zeros Theorem to find all possible rational zeros.
Use Synthetic Division and Remainder Theorem to try to find one rational zero (remainder will be zero).
If “n” is a rational zero, factor original polynomial as
(x – n)q(x).
Test remaining possible rational zeros in q(x). If one is found, factor again as in previous step.
Continue in this way until all rational zeros have been found.
See if additional irrational or non-real complex zeros can be found by solving a quadratic equation.

68. Example Find all zeros of: Find all solutions to:
Rational Zeros Theorem says the only possible rational zeros are:
See if -1 is a zero:
Conclusion:

69. Example Continued
This new factor has the same possible rational zeros:
Check to see if -1 is also a zero of this:
Conclusion:

70. Example Continued This new factor has as possible rational zeros:
Check to see if -1 is also a zero of this:
Conclusion:

71. Example Continued
Check to see if 1 is a zero:
Conclusion:

72. Example Continued
Check to see if 2 is a zero:
Conclusion:

73. Example Continued
Summary of work done:

74. Your Turn: Find rational zeros of f(x)=x3+2x2-11x-12 List possible
q= p=-12
p/q= ±1,± 2, ± 3, ± 4, ± 6, ±12
Test:
p/q= p/q=
Since -1 is a zero: (x+1)(x2+x-12)=f(x)
Factor: (x+1)(x-3)(x+4)=0
x= x=3 x=-4

75. Your Turn: Find the zeros of g(x) = 6x3 + 4x2 – 14x + 4
Factors of p: + 1, + 2, Factors of q: + 1, + 2, + 3, + 6
Possible zeros are + 1, + 2, + 3, + 4, + 4/3, + 1/2, + 1/3, + 1/6, + 2/3 –
– –
1 is a zero, and 6x2 + 10x – 4 is the depressed polynomial.
Factor 6x2 + 10x – 4 … 2(3x2 + 5x – 2)
We’ll try 1
2(3x + 1)(x – 2)
Find the zeros… 2(3x + 1)(x – 2) = 0
3x + 1 = 0 x – 2 = 0
x = –1/ x = 2
The zeros are 1, –1/3, and 2

76. Example : f(x)=10x4-3x3-29x2+5x+12 List: q=10 p=12
p/q= ± 1/1, ± 2/1, ± 3/1, ± 4/1, ± 6/1,±12/1, ± 3/2, ± 1/5, ± 2/5, ± 3/5, ± 6/5, ± 12/5, ± 1/10, ± 3/10, ± 12/10
Check possible zeros using synthetic division:
Check: /
p/q= -3/
Yes it works
* (x+3/2)(10x3-18x2-2x+8)*
(x+3/2)(2)(5x3-9x2-x+4) -factor out GCF
(2x+3)(5x3-9x2-x+4) multiply 1st factor by 2

77. Repeat finding zeros for:
g(x)=5x3-9x2-x+4
q= p=4
p/q:±1, ±2, ±4, ±1/5, ±2/5, ±4/5
Check possible zeros of depressed equation: 4/
p/q=4/
(2x+3)(x-4/5)(5x2-5x-5)=
(2x+3)(x-4/5)(5)(x2-x-1)= mult.2nd factor by 5
(2x+3)(5x-4)(x2-x-1)=
-now use quad for last-
*-3/2, 4/5, 1± , 2

78. Your Turn: Find all zeros of f(x) = x4-6x3-7x2+6x-8
Solution: 1, -1, 2, 4

79. Review-Theorems The Remainder Theorem
If a polynomial f(x) is divided by x – k, the remainder is equal to f(k).
The Factor Theorem
The polynomial x – k is a factor of the polynomial P(x) if and only if P(k) = 0.
The Rational Zeros Theorem

80. Other Tests For Zeros of Polynomials
Descartes’s Rule of Signs
Used to help identify the number of real zeros of a polynomial.
Upper and Lower Bound Rules
Used to give an upper or lower bound of real zeros of a polynomial, which can help eliminate possible real zeros.

81. Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and a nonzero constant term. The number of positive real zeros of P(x) is either:
1. The same as the number of variations of sign in P(x), or
2. Less than the number of variations of sign in P(x) by a positive even integer.
The number of negative real zeros of P(x) is either:
3. The same as the number of variations of sign in P(x), or
4. Less than the number of variations of sign in P(x) by a positive even integer.
A zero of multiplicity m must be counted m times.

82. Descartes’ Rule of Signs
A variation in sign means that two consecutive (nonzero) coefficients have opposite signs.
The polynomial x3-3x+2 has two variations in sign.

83. The polynomial has three variations in sign.
Example: Use Descartes’s Rule of Signs to determine the possible number of positive and negative real zeros of f(x) = 2x4 – 17x3 + 35x2 + 9x – 45.
The polynomial has three variations in sign.
+ to –
+ to –
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45
– to +
f(x) has either three positive real zeros or one positive real zero.
f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45
=2x4 + 17x3 + 35x2 – 9x – 45
One change in sign
f(x) has one negative real zero.
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45 = (x + 1)(2x – 3)(x – 3)(x – 5).

84. Example
What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros?
There are two variations of sign, so there are either two or zero positive real zeros to the equation.

85. Example continued
The number of negative real zeros is either two or zero.
What this tells us;
Total Number of Zeros 4
Positive
Negative
Nonreal

86. Your Turn:
Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1. We first consider the possible number of positive zeros by observing that P(x) has three variations in signs. + x4 – 6x3 + 8x2 + 2x – 1 Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros. For negative zeros, consider the variations in signs for P(x). P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x)  1 = x4 + 6x3 + 8x2 – 2x – 1 Since there is only one variation in sign, P(x) has only one negative real root. 1 2 3
Total number of zeros 4
Positive:
Negative:
Nonreal:

87. Upper and Lower Bound Rules
Let P(x) define a polynomial function of degree n  1 with real coefficients and with a positive leading coefficient. If P(x) is divided synthetically by x – c, and
(a) if c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then P(x) has no zero greater than c;
(b) if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then P(x) has no zero less than c.

88. Example:
Show that the real zeros of P(x) = 2x4 – 5x3 + 3x + 1 satisfy the following conditions.
(a) No real zero is greater than 3.
(b) No real zero is less than –1.
Solution
a) c > 0
b) c < 0
All are nonegative.
No real zero greater than 3.
The numbers alternate in sign.
No zero less than 1.

89. How to Use the Upper and Lower Bound Rules
Finding the real zeros of
we check the possible zero 2 by synthetic division:
Notice bottom row. What does this tell us about zeros?
Now we don’t need to check any possible zeros greater than 2.

90. How to Use the Upper and Lower Bound Rules
Using the same function:
check the possible zero -3
Notice bottom row. What does this tell us about zeros?
So we don’t need to check any possible zeros less than -3.