Warm-up

Find all the solutions over the complex numbers for this polynomial:f(x) = x4 – 2x3 + 5x2 – 8x + 4

}2,2,1,1{ ii

Descartes Rule

2.5

Objectives

I can use Descartes Rule to find the possible combinations of positive and negative real zeros

I can write a polynomial in factor format

Descartes’s Rule of Signs: If f(x) is a polynomial with real

coefficients and a nonzero constant term,

1. The sign changes for f(x) tells the number of positive real

zeros equal to the number of sign changes or less than that

number by an even integer.

2. The sign changes on f(-x) tells the number of negative real

zeros equal to the number sign changes or less than that

number by an even integer.

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.

f(x) = 2x4 – 17x3 + 35x2 + 9x – 45

+ to –

– to +

+ to –

f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45 =2x4 + 17x3 + 35x2 – 9x – 45

f(x) has either three positive real zeros or one positive real zero.

f(x) has one negative real zero.One change in sign

State the possible number of positive real zeros, negative real zeros, and imaginary zeros of p(x) = –x6 + 4x3 – 2x2 – x – 1.

Since p(x) has degree 6, it has 6 zeros. However, some of them may be imaginary. Use Descartes’ Rule of Signs to determine the number and type of real zeros. Count the number of changes in sign for the coefficients of p(x).

Find Numbers of Positive and Negative Zeros

p(x) = –x6 + 4x3 – 2x2 – x –1

yes– to +

yes+ to –

no– to –

no– to –2 or 0 positive real

zeros

Since there are two sign changes, there are 2 or 0 positive real zeros. Find p(–x) and count the number of sign changes for its coefficients.

Since there are two sign changes, there are 2 or 0 negative real zeros.

Find Numbers of Positive and Negative Zeros

p(–x) = –x6 - 4x3 – 2x2 + x –1

no– to –

no– to –

yes– to +

yes+ to –

Find all of the zeros of f(x) = x3 – x2 + 2x + 4.

Since f(x) has degree of 3, the function has at lost three zeros. To determine the possible number and type of real zeros, examine the number of sign changes in f(x) and f(–x).

yes yes no

no no yes

f(x) = x3 – x2 + 2x + 4

f(–x) = –x3 – x2 – 2x + 4

2 or 0 positive real zeros

1 negative real zero

Descarte’s Rule of Signs

Find how many positive and negative roots there are in f(x).

f(x) = 3x4 + 2x3 + 1

0 positive roots

2 or 0 negative roots

Descarte’s Rule of Signs

Find how many positive and negative roots there are in f(x).f(x) = 3x3 + 2x2 - x + 3

2 or 0 positive roots

1 negative root

Descarte’s Rule of Signs

How can this rule help us with the rational root theorem?

It helps us guess which possible root to try.

f(x) = x3 – x2 + 2x + 4.

There are 3 zeros, we know there are 2 or 0

positive and 1 negative

Possible roots are ±1, ±2, ±4

Use Synthetic Division to

find the roots

-4, -2, -1, 1, 2, 4

6 2 0 1

2 0 1

4 2 1- 1 1

12 4 1 1

8 2 2

4 2 1- 1 2

We need either 2 or 0 positive, so 4 cannot be a zero

0 4 2- 1

4- 2 1-

4 2 1- 1 1

-1 works, so it is our 1 negative zero. The

other two have to be imaginary.

Use Quadratic Formula to find

remaining solutions 31

2

322

2

122

)1(2

)4)(1(442i

i

1,31 ix

151620 Solve 24 xxx

There are 4 zeros, We know there are 1 positive

and 3 or 1 negative

15)(16)(2)()( 24 xxxxfPossible

roots are ±1, ±3, ±5, ±15

-15, -5, -3 -1, 1, 3, 5, 15

480 99 23 5 1

495 115 25 5

15- 16- 2- 0 1 5

All positive so 5 is an upper bound

0 5 7 3 1

15 21 9 3

15- 16- 2- 0 1 3

3 is a solution. This is our 1 positive zero

Now try negatives

16- 7 0 1

21- 0 3-

5 7 3 1 3Alternates between

positive and negative so -3 is a lower bound0| 5 2 1

5- 2- 1-

5 7 3 1 1-1 is our 1 negative. Use the quadratic to find the

remaining 2 zeros

ii

212

42

2

162

)1(2

)5)(1(442

3,1,21 ix

Linear Factorization

If we know the solutions, we can work backwards and find the Linear Factorization

Example:Given the solutions to a polynomial are:{-2, 3, 6, 2+3i, 2-3i} write the polynomial as a product of its factors

f(x) = (x+2)(x-3)(x-6)(x-(2+3i))(x-(2-3i))

Also Note: If we know all the factors and multiply them together, we get the polynomial function.

Homework

WS 4-3