2. Chapter 3: Polynomial Functions
3.1 Complex Numbers
3.2 Quadratic Functions and Graphs
3.3 Quadratic Equations and Inequalities
3.4 Further Applications of Quadratic Functions and Models
3.5 Higher-Degree Polynomial Functions and Graphs
3.6 Topics in the Theory of Polynomial Functions (I)
3.7 Topics in the Theory of Polynomial Functions (II)
3.8 Polynomial Equations and Inequalities; Further Applications and Models

3. 3.7 Topics in the Theory of Polynomial Functions (II)
Complex Zeros and the Fundamental Theorem of Algebra
It can be shown that if a + bi is a zero of a polynomial function with real coefficients, then so is its complex conjugate, a – bi.
Conjugate Zeros Theorem
If P(x) defines a polynomial function having only real coefficients, and if a + bi is a zero of P(x), then the conjugate a – bi is also a zero of P(x).

4. 3.7 Topics in the Theory of Polynomial Functions (II)
Example Find a polynomial having zeros 3 and 2 + i that
satisfies the requirement P(–2) = 4.
Solution Since 2 + i is a zero, so is 2 – i. A general solution is
Since P(–2) = 4, we have

5. 3.7 Fundamental Theorem of Algebra
If P(x) is a polynomial of degree 1 or more, then there is some number k such that P(k) = 0. In other words,
where Q(x) can also be factored resulting in
Fundamental Theorem of Algebra
Every function defined by a polynomial of degree 1 or more has at least one complex (real or imaginary) zero.

6. 3.7 Zeros of a Polynomial Function
Example
Find all complex zeros of
given that 1 – i is a zero.
Solution
Number of Zeros Theorem
A function defined by a polynomial of degree n has at most n distinct (unique) complex zeros.

7. 3.7 Zeros of a Polynomial Function
Using the Conjugate Zeros Theorem, 1 + i is also a zero.
The zeros of x2 – 5x + 6 are 2 and 3. Thus,
and has four zeros: 1 – i, 1 + i, 2, and 3.

8. 3.7 Multiplicity of a Zero
The multiplicity of the zero refers to the number of times a zero appears.
e.g.
– x = 0 leads to a single zero
– (x + 2)2 leads to a zero of –2 with multiplicity two
– (x – 1)3 leads to a zero of 1 with multiplicity three

9. 3.7 Polynomial Function Satisfying Given Conditions
Example Find a polynomial function with real coefficients
of lowest possible degree having a zero 2 of multiplicity 3, a
zero 0 of multiplicity 2, and a zero i of single multiplicity.
Solution By the conjugate zeros theorem, –i is also a zero.
This is one of many such functions. Multiplying P(x) by any
nonzero number will yield another function satisfying these
conditions.

10. 3.7 Multiplicity of Zeros
Observe the behavior around the zeros of the polynomials
The following figure illustrates some conclusions.
By observing the dominating term and noting the parity of
multiplicities of zeros of a polynomial in factored form, we
can draw a rough graph of a polynomial by hand.

11. 3.7 Sketching a Graph of a Polynomial Function by Hand
Example
Solution The dominating term is –2x5, so the end behavior
will rise on the left and fall on the right. Because –4 and 1 are
x-intercepts determined by zeros of even multiplicity, the
graph will be tangent to the x-axis at these x-intercepts. The
y-intercept is –96.

12. 3.7 The Rational Zeros Theorem
Example
List all possible rational zeros.
Use a graph to eliminate some of the possible zeros listed in part (a).
Find all rational zeros and factor P(x).
The Rational Zeros Theorem

13. 3.7 The Rational Zeros Theorem
Solution
(a)
From the graph, the zeros
are no less than –2 and no
greater than 1. Also, –1 is
clearly not a zero since the
graph does not intersect the
x-axis at the point (-1,0).

14. 3.7 The Rational Zeros Theorem
(c) Show that 1 and –2 are zeros.
Solving the equation 6x2 + x – 1 = 0, we get x = –1/2, 1/3.

15. 3.7 Descartes’ Rule of Sign
Descartes’ Rule of Signs
Let P(x) define a polynomial function with real coefficients
and a nonzero constant term, with terms in descending
powers of x.
The number of positive real zeros either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer.
(b) The number of negative real zeros either equals the number of variations in sign occurring in the coefficients of P(x) or is less than the number of variations by a positive even integer.

16. 3.7 Applying Descartes’ Rule of Signs
Example Determine the possible number of positive real zeros and negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1.
We first consider the possible number of positive zeros by observing that P(x) has three variations in signs.
+ x4 – 6x3 + 8x2 + 2x – 1
Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros.
For negative zeros, consider the variations in signs for P(x).
P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x)  1
= x4 + 6x3 + 8x2 – 2x – 1
Since there is only one variation in sign, P(x) has only one negative real root. 1 2 3

17. 3.7 Boundedness Theorem Boundedness Theorem
Let P(x) define a polynomial function of degree n  1 with
real coefficients and with a positive leading coefficient. If
P(x) is divided synthetically by x – c, and
if c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then P(x) has no zero greater than c;
(b) if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then P(x) has no zero less than c.

18. 3.7 Using the Boundedness Theorem
Example Show that the real zeros of P(x) = 2x4 – 5x3 + 3x + 1 satisfy the following conditions.
(a) No real zero is greater than 3.
(b) No real zero is less than –1.
Solution
a) c > 0
b) c < 0
All are nonegative.
No real zero greater than 3.
The numbers alternate in sign.
No zero less than 1.