Download PCH (3.3)(2) Zeros of Polynomial 10

3.3 (2) Zeros of polynomials
Descarte’s Rule of Signs
Two theorems on bounds for zeros
First, a little POD
Use CAS to factor one of the following:
x3  c3
x5  c5
x7  c7
x9  c9
What do you notice about one of the factors in each case?
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 POD
Use 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
=0
(Since n is odd)
Today
We’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 Signs
Consider
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 Signs
Consider
f(x) = 2x5 - 7x4 + 3x2 + 6x - 5
According to Descarte’s Rule, this means there are:
3 or 1 positive real solutions
2 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 +
3
1
3
1
Real 2
2
0
0
Imaginary
0
2
2
4
Total
5
5
5
5
Descarte’s Rule of Signs
Let f(x) be a polynomial with real coefficients and a nonzero 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
1
1
Real +
3
1
Real -
0
0
Imaginary
0
2
Total
4
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 division
Use 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 division
Use 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 division
If 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.
2.
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).
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 coefficients
Suppose
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 coefficients
All 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.
Lastly
Let’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.)
Lastly
Let’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
Lastly
2. 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).
Lastly
2. 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)