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Section 2.5 Part Two

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Section 2.5 Part Two

Other Tests for Zeros

Descartes’s Rule of Signs

Upper and Lower Bounds

- Let f(x) be a polynomial with real coefficients and ao≠ 0.
- The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer.
- The number of negative real zeros of f is either equal to the number of variations in the sign of f(-x) of less than that number by an even integer.

- Consider f(x) = x2 + 8x + 15
- Since there are no variations in sign of f(x) there are no positive roots.
- Since f(-x) = (-x)2 + 8(-x) + 15
= x2 – 8x + 15

has two variations, f(x) may have two or zero negative roots.

x = {-3. -5}

- A variation in sign means that when the polynomial is written in standard form that one term has a different sign than the next.
- T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
- For Variations of T(x) just look at the signs
T(x) has one variation in signs

- For Variations of T(-x) you must change the sign of terms with odd numbered exponents
- T(-x) = -x5 + 9x4 – 19x3 – 21x2 + 92x – 60
T(-x) has four variation in signs

- T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60
- 60 + 1, + 2 , + 3, + 4, + 5 , + 6 , + 10 , + 12 , + 15 , + 20, + 30, + 60
- Since there is only one sign variation for T(x) there is at most, one positive root
- SYN Program

T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60

T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)

1

9

19

-21

-92

-60

+2

2

22

122

60

82

61

30

0

1

11

41

Now that we have found the positive root we know that any other real roots must be negative.

T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60

T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)

- 30 -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30
T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)

1

11

41

61

30

-1

-1

-10

-30

-31

30

0

1

10

31

T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60

T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)

T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)

- 30 -1, -2 , -3, -4, -5 , -6 , -10 ,-15, - 30
T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15)

1

10

31

30

-2

-2

-16

-30

0

1

8

15

T(x) = x5 + 9x4 + 19x3 – 21x2 – 92x – 60

T(x) = (x – 2)(x4 + 11x3 + 41x2 + 61x + 30)

T(x) = (x – 2)(x + 1)(x3 + 10x2 + 31x + 30)

T(x) = (x – 2)(x + 1)(x + 2)(x2 + 8x + 15)

T(x) = (x – 2)(x + 1)(x + 2)(x + 3)(x + 5)

x = {-5, -3, -2, -1, 2}

- Zeros of Polynomial Functions
- page 160
- 1 - 31 odd,
- 37 - 85 odd,
- 91 - 94 all

F(x) = (x + 6)(x + i)(x – i)

x = {-6, -i, i}

#10 Use the rational zero test to list all of the possible rational zeros of f. Verify that the zeros of f shown are contained in the list.

f(x) = 4x5 – 8x4 – 5x3 + 10x2 + x – 2

h(t) = t3 + 12t2 + 21t + 10

Caution graph may be misleading.

With a cubic we expect another turn down to the left

- List the possible rational zeros of f,
- sketch the graph of f so that some possibilities can be eliminated
- determine all the real zeros

Graph eliminates -4, 1, 4

f(x) = x3 + x2 – 4x – 4

f(x) = x2 (x +1)– 4(x + 1)

f(x) = (x + 1)(x2– 4)

f(x) = (x + 1)(x +2)(x – 2)

Since imaginary solutions always appear in conjugate pairs we know to include