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
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.