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

2. Descartes Rule
2.5

3. I can write a polynomial in factor format
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

4. Descartes’s Rule of Signs
Descartes’s Rule of Signs: If f(x) is a polynomial with real coefficients and a nonzero constant term,
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.
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.
Descartes’s Rule of Signs

5. Example: Descartes’s Rule of Signs
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
One change in sign
f(x) has one negative real zero.
Example: Descartes’s Rule of Signs

6. 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 + 4x3 – 2x2 – x – 1
yes
– to +
yes
+ to – no
– to – no
– to –
2 or 0 positive real zeros

7. 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) = –x x3 – 2x x – 1 no
– to – no
– to –
yes
– to +
yes
+ to –
Since there are two sign changes, there are 2 or 0 negative real zeros.

8. 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 + 2x + 4
2 or 0 positive real zeros
yes
yes no
f(–x) = –x3 – x2 – 2x + 4
1 negative real zero no no
yes

9. 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

10. 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

11. 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.

12. 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
-4, -2, -1, 1, 2, 4
We need either 2 or 0 positive, so 4 cannot be a zero
Use Synthetic Division to find the roots
-1 works, so it is our 1 negative zero. The other two have to be imaginary.
Use Quadratic Formula to find remaining solutions

13. Possible roots are ±1, ±3, ±5, ±15
There are 4 zeros, We know there are 1 positive and 3 or 1 negative
-15, -5, -3 -1, 1, 3, 5, 15
3 is a solution. This is our 1 positive zero
All positive so 5 is an upper bound
Now try negatives
-1 is our 1 negative. Use the quadratic to find the remaining 2 zeros
Alternates between positive and negative so -3 is a lower bound

14. 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.

15. Homework
WS 4-3