1. Zeros of Polynomials
2.5

2. Properties of Polynomial Equations
A polynomial of degree n, has n roots (counting multiple roots separately)
If a + bi is a root, then a – bi is also a root. Complex roots occur in conjugate pairs!

3. Descartes’ Rule of Signs

4. 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

5. 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)6 + 4(–x)3 – 2(–x)2 – (–x) – 1 no
– to – no
– to –
yes
– to +
yes
+ to –
Since there are two sign changes, there are 2 or 0 negative real zeros. Make a chart of possible combinations.

6. Find Numbers of Positive and Negative Zeros
Answer:

7. Find all of the zeros of f(x) = x3 – x2 + 2x + 4.
Since f(x) has degree of 3, the function has 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

8. Rational Zero Theorem

9. Identify Possible Zeros
A. List all of the possible rational zeros of f(x) = 3x4 – x3 + 4.
Answer:

10. B. List all of the possible rational zeros of f(x) = x3 + 3x + 24
B. List all of the possible rational zeros of f(x) = x3 + 3x <YU TRY> A. B. C. D.

11. Use the Factor Theorem
Show that x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial.
The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division.
1 4 –15 –18

12. x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Use the Factor Theorem
Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored.
x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial.
Answer: So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1).

13. Example
Repeated zero at x = 2

14. Upper and Lower Bounds
Suppose f(x) is divided by x – c using synthetic division
If c > 0 and each number in the last row is either positive or zero, then c is an upper bound for real zeros
If c < 0 and each number in the last row are alternatively positive or negative (zero counts as both), then c is a lower bound

15. The degree is 4, so there are 4 roots!
Possible roots are ±1, ±13
The degree is 4, so there are 4 roots!
Use Synthetic Division to find the roots
Use Quadratic Formula to find remaining solutions
Multiplicity of 2

16. 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

17. There are 4 zeros, We know there are 1 positive and 3 or 1 negative
Possible roots are ±1, ±3, ±5, ±15
-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

18. Linear Factorization Theorem
Find a fourth degree polynomial with real coefficients that has 2, -2 and i as zeros and such that f(3) = -150
Foil
Foil
Substitute f(3) = -150
Solve for a
Substitute back in equation

19. Linear Factorization Theorem
An nth – degree polynomial can be expressed as the product of a nonzero constant and n linear factors