1. 3.3 Zeros of Polynomials

2. Complete factorization theorem for Polynomials
If f(x) is a polynomial of degree n > 0, then there exist n complex numbers c1, c2, …., cn such that
f (x) = a (x – c1) (x – c2)…(x - cn)

3. Ex. Find a polynomial of degree 3 that has the indicated zeros and satisfies the given condition.
Zeros at x = -5, 2, 4
f(3) = -24

4. Solution f (x) = a (x + 5) (x-2) (x-4)
Solve for a by plugging in f(3) = -24
-24 = (3+5)(3-2)(3-4)
-24 = a8(1)(-1)
3 = a
f(x) = 3 (x+5)(x-2)(x-4) *multiply out
f (x) = 3x3 – 3x2 – 66x + 120

5. 3.3 The Fundamental Theorem of Algebra
Every function defined by a polynomial of degree 1 or more has at least one complex (real or imaginary) zero.
Conjugate Zeros Theorem
If P(x) is a polynomial having only real coefficients, and if a + bi is a zero of P, then the conjugate a – bi is also a zero of P.

6. 3.3 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

7. 3.3 Polynomial Function Satisfying Given Conditions
Example Find a polynomial function with real coefficients
of lowest possible degree having a leading coefficient 1, 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.

8. 3.3 Observation: Parity of Multiplicities 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.

9. 3.3 Sketching 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.

10. Descartes’ Rule of Sign
Let P(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x.
(a) The number of positive real zeros of P 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 of P is either 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.

11. 3.3 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

12. 3.3 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
(a) 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.

13. 3.3 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.