# Section 2.5 Part Two - PowerPoint PPT Presentation

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Section 2.5 Part Two. Other Tests for Zeros Descartes’s Rule of Signs Upper and Lower Bounds. Descartes’s Rule of Signs. Let f(x) be a polynomial with real coefficients and a o ≠ 0.

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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

### Descartes’s Rule of Signs

• 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.

### Apply Descarte’s Rule of Signs

• 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}

### Variation in Sign

• 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

### Find the Zeros of T(x)

• 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

### Find the Zeros of T(x)

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.

### Find the Zeros of T(x)

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

### Find the Zeros of T(x)

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

### Find the Zeros of T(x)

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}

### Homework 2.5

• Zeros of Polynomial Functions

• page 160

• 1 - 31 odd,

• 37 - 85 odd,

• 91 - 94 all

### #5 Find all of the zeros of the function.

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

### #15 Find all of the real zeros of the function.

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

With a cubic we expect another turn down to the left

### #25 f(x) = x3 + x2 – 4x – 4

• 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)

### #42. Find the polynomial function with integer coefficients that has the given zeros.

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