3.3 (2) Zeros of polynomials

Descarte’s Rule of Signs

Two theorems on bounds for zeros

First, a little PODUse CAS to factor one of the following:

What do you notice about one of the factors in each case?

99

77

55

33

cx

cx

cx

cx

First, a little POD

Use the Factor Theorem to show that (x + c) is a factor of xn + cn for every positive odd integer n.

The Factor Theorem: f(x) has a factor (x- c) if and only if f(c) = 0.

First, a little PODUse the Factor Theorem to show that (x + y) is a factor of xn + yn for

every positive odd integer n.

The Factor Theorem: f(x) has a factor (x- c) if and only if f(c) = 0.

f(-y) = (-y)n + yn

= -yn + yn (Since n is odd)

= 0

TodayWe’ll look at three tools to determine the nature of roots

for a polynomial—not what the roots actually are.

Mathematicians look for these patterns all the time.

One tool (Descartes Rule of Signs) will give us an idea of how many roots can be positive, negative, zero, or imaginary.

Two tools will give us an idea of where roots are located by helping us find possible bounds for the roots.

Descarte’s Rule of Signs

This rule helps us determine the nature of roots: positive, negative, zero, or imaginary.

Consider f(x) = 2x5 - 7x4 + 3x2 + 6x - 5

How many changes of sign (variations of sign) are there in f(x)?

How many are in f(-x)?

Descarte’s Rule of SignsConsider f(x) = 2x5 - 7x4 + 3x2 + 6x - 5

How many changes of sign (variations of sign) are there in f(x)?Three: between the first two terms, the second two terms, and the last two terms.

How many are in f(-x)?Two: between the first two terms, and between the third and fourth

Descarte’s Rule of SignsConsider f(x) = 2x5 - 7x4 + 3x2 + 6x - 5

According to Descarte’s Rule, this means there are:

3 or 1 positive real solutions2 or 0 negative real solutions.

How many solutions must there be? Then how many imaginary solutions might there be?

Descarte’s Rule of Signs

Consider f(x) = 2x5 - 7x4 + 3x2 + 6x – 5

3 or 1 positive real solutions

2 or 0 negative real solutions

A chart helps:

Real + Real - Imaginary Total

3 2 0 5

1 2 2 5

3 0 2 5

1 0 4 5

Descarte’s Rule of SignsLet f(x) be a polynomial with real coefficients and a non-

zero constant term,

1. The number of positive real roots is either equal to the number of variations in sign in f(x), or is less than that number by an even integer.

2. The number of negative real roots is either equal to the number of variations in sign in f(-x) or is less than that number by an even integer.

Caution!If f(x) has a final term that is not a constant, then factor

out the lowest power of x.

x4 - 3x3 + 2x2 - 5x = 0

x(x3 - 3x2 + 2x - 5) = 0

So, one solution is 0, and we can use Descarte’s Rule to determine the nature of the rest.

Descarte’s Rule of Signs

x(x3 - 3x2 + 2x - 5) = 0

So, one solution is 0, and we can use Descarte’s Rule to determine the nature of the rest.

Note: no negative real roots.

Zero Real + Real - Imaginary Total

1 3 0 0 4

1 1 0 2 4

Bound Definition

By definition, a real number b is an upper bound for zeros, if no zero is greater than b.

A real number a is a lower bound for zeros if no zero is less than a.

So that, if r is a real zero of f(x), then a ≤ r ≤ b.

Bound #1 Using synthetic divisionUse it:

Find upper and lower bounds for real solutions for the equation f(x) = 2x3 + 5x2 - 8x - 7 = 0.

Let’s try some trial and error and pick numbers: x = 1, 2, and -4. Do a little synthetic division.

What are the numbers on the third (bottom) line? What does that mean?

What three points do we know lie on the curve?

Bound #1 Using synthetic divisionUse it:

Find upper and lower bounds for real solutions for the equation f(x) = 2x3 + 5x2 - 8x - 7 = 0.

Let’s try some trial and error and pick numbers: x = 1, 2, and -4. Do a little synthetic division with them.

The bottom line of synthetic division is all positive with x = 2, a positive number, so that number is an upper bound for zeros.

The bottom line of synthetic division alternates signs with x = -4, a negative number, so that number is a lower bound for zeros.

We could find a tighter interval for zeros with other tools, but this is a start.

Bound #1 Using synthetic divisionIf r is a real zero of f(x), then a ≤ r ≤ b.

Suppose f(x) is a polynomial with real coefficients and a positive leading coefficient, and that f(x) is divided by (x - c):

1. If c > 0, and if all the numbers in the third row of the synthetic division are either positive or zero, the c is an upper bound for the real zeros of f(x).

2. If c < 0, and if the numbers in the third row of the synthetic division are alternately positive and negative (0 is considered to be either positive or negative), then c is a lower bound for the real zeros of f(x).

Bound #2 Using coefficientsSuppose

f(x) = anxn + an-1xn-1 + an-2xn-2 + ……. + a1x + a0

is a polynomial with real coefficients.

All of the zeros of f(x) are in the interval (-M, M)

where M = (the absolute value of the largest coefficient)/( the absolute value of the leading coefficient) + 1

Try it with f(x) = 2x3 + 5x2 - 8x - 7

Bound #2 Using coefficientsAll of the zeros of f(x) are in the interval (-M, M)

where M = (the largest coefficient (in magnitude))/( the absolute value of the leading coefficient) + 1

Try it with f(x) = 2x3 + 5x2 - 8x - 7

M = 8/2 + 1 = 5 So, the zeros fall in the interval (-5, 5).

Again, we could eventually find a tighter fit for the interval holding the zeros of the polynomial, but this is a start.

LastlyLet’s use some of all of what

we’ve seen in this section.

1. This is a graph of f(x). Find a factored form for f that has minimal degree. If the leading coefficient is 1, what is the y-intercept? (The x-scale is 1 and the y-scale is 5.)

LastlyLet’s use some of all of what we’ve seen in this

section.

1. This is a graph of f(x). Find a factored form for f that has minimal degree. If the leading coefficient is 1, what is the y-intercept? (The x-scale is 1 and the y-scale is 5.)

f(x) = 1(x + 2)2(x – 1)3(x – 3)

The y-intercept will be 1(x + 2)2(x – 1)3(x – 3) =

(4)(-1)(-3) = 12

Lastly2. Find the zeros of f(x) = x3 - 1000x2 - x + 1000. (One method

here.) Don’t factor– try another set of tools.

How many zeros will there be? Use Descartes’ Rule to determine possible categories.

Graph to find a first zero, then use it in synthetic division.

Perhaps we can factor the depressed equation to find another zero. Or we can use synthetic division again to test a zero and find the final zero (and factor).

Lastly2. Find the zeros of f(x) = x3 -

1000x2 - x + 1000.

How many zeros will there be? 3

Graph to find a first zero, then use it in synthetic division. The graphed zero could be 1 or -1.

Use synthetic division.

Lastly

2. Find the zeros of f(x) = x3 - 1000x2 - x + 1000.

Final factorization:

f(x) = (x + 1)(x – 1)(x – 1000)