Transcript Document

Warm-up
 Find all the solutions over the complex numbers
for this polynomial:
f(x) = x4 – 2x3 + 5x2 – 8x + 4
{1, 1, 2i,  2i}
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.
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.
The polynomial has three variations in sign.
+ to –
+ to –
f(x) = 2x4 – 17x3 + 35x2 + 9x – 45
– to +
f(x) has either three positive real zeros or one positive real zero.
f(– x) = 2(– x)4 – 17(– x)3 + 35(– x)2 + 9(– x) – 45
=2x4 + 17x3 + 35x2 – 9x – 45
f(x) has one negative real zero.
One change in sign
Find Numbers of Positive and Negative
Zeros
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).
p(x) =
–x6
+
yes
– to +
2 or 0 positive real
zeros
4x3
–
yes
+ to –
2x2
–
no
– to –
x
–
no
– to –
1
Find Numbers of Positive and Negative
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.
p(–x) = –x6
no
– to –
4x3
– 2x2
no
– to –
+
yes
– to +
x
–
yes
+ to –
Since there are two sign changes, there are 2 or 0 negative real
zeros.
1
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).
f(x) =
x3
–
x2
yes
f(–x) = –x3
–
no
+
2x
yes
x2
–
no
+
4
2 or 0 positive real
zeros
no
2x
+
yes
4
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) = x – x + 2x + 4.
3
2
x  1  i 3,  1
Possible
roots are
±1, ±2, ±4
There are 3 zeros, we
know there are 2 or 0
positive and 1 negative
-4, -2, -1, 1, 2, 4
Use
Synthetic
Division to
find the roots
Use Quadratic
Formula to find
remaining
solutions
211111- 1- 1- 1 222 444
We need either 2
or 0 positive, so 4
cannot be a zero
21- 1 202 82- 4
1 111 0- 242 4126 0
-1 works, so it is our 1
negative zero. The
other two have to be
imaginary.
2  4  4(1)( 4) 2   12 2  2i 3


 1 i 3
2(1)
2
2
x  1  2i,  1, 3
Solve 0  x 4  2 x 2  16 x  15
There are 4 zeros, We
know there are 1 positive
and 3 or 1 negative
(-3x)-1,
 1,
(3,x)5, 152( x)
-15,f-5,
4
Possible
 16(are
 x±1,
)  15
±3, ±5, ±15
2 roots
35 11 00 --22 --16
16 --15
15
Now try
negatives
35 25
9
115
21
15
495
11 35 23
7
99
5
480
0
 13 11 33 77 55
- 13 -02 -- 21
5
1 02
57 |-016
3 is
Allapositive
solution.soThis
5 isisanour
1upper
positive
bound
zero
-1 Alternates
is our 1 negative.
between
Use
positive
the quadratic
and negative
to find the
so 3remaining
is a lower2 bound
zeros
 2  4  4(1)(5)  2   16  2  4i


 1  2i
2(1)
2
2
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