January 19, 2012At the end of today, you will be able to find ALL zeros of a polynomial.

Warm-up: Check HW 2.4 Pg. 167 #20-21, 27-29, 46-49, 55, 65, 6720. -3 – 11i 47. 8/41 + (10/41)i21. 3 -3i√2 48. 5/2 + (5/2)i27. 5 + i 49. 4/5 + (3/5)i28. 6 – 22i 55. -1/2 – (5/2)i29. 12 + 30i 65. 1 ± i46. 7i 67. -2 ± (1/2)i

Lesson 2.5 Zeros of Polynomial Functions

The Fundamental Theorem of Algebra tells you how many zeros or factors of a polynomial exist.

If f(x) is a polynomial of degree n, where n > 0, then f has at least one zero in the complex number system.

For example:a) f(x) = x – 2, b) f(x) = x2 – 6x + 9 = (x – 3)(x – 3),

c) f(x) = x3 + 4x = x(x2 + 4) = x(x -2i)(x +2i),

has 2 zeros: x = 3 (multiplicity of 2)

has exactly 1 zero: x = 2

has 3 zerosx = 0, x = 2i, x = -2i

Using the Rational Zero Test to find real zeros. The Rational Zero Test helps you find the rational zeros

of polynomials. Rational zero = , where p and q have

no common factors other than 1, andp = a factor of the constant term a0

q = a factor of the leading coefficent an

for f(x) = anxn + an-1xn-1 +…+ a2x2 + a1x + a0€

pq

Example 1: Use the Rational Zero Test to find possible zeros for f(x) = 2x3 + 3x2 – 8x + 3

€

pq

€

=factors of 3factors of 2

€

=±1, ±3±1, ±2

=

€

±1,

€

±3,

€

±12,

€

±32

Finding the zeros

• After you use the Rational Zero Test, use synthetic division to find zeros.

• Pick one of the zeros from the test and hope you get a remainder of 0! (1 is the easiest)

Example 1: Use the Rational Zero Test to find possible zeros for f(x) = 2x3 + 3x2 – 8x + 3

€

pq

€

=factors of 3factors of 2

€

=±1, ±3±1, ±2

=

€

±1,

€

±3,

€

±12,

€

±32

1 2 3 -8 3

225

5-3

-30

= 2x2 + 5x – 3

So your factors are:(x – 1)( 2x2 + 5x – 3)

(x – 1)( 2x – 1)(x + 3) Zeros = 1, 1/2, -3

Use the Rational Zero Test and Synthetic/Long Division to find the factors and zeros.

Example 2: Find all the zeros of the function and write the polynomial as a product of linear factors.

f(x) = x5 + x3 + 2x2 – 12x + 8

Try 1

€

pq=±1,±2,±,4,±8

±1

1 1 0 1 2 -12 8

111

12

24

4-8

-80

x4 + x3 + 2x2 + 4x – 8

Try another zero: -2

-2 1 1 2 4 -8

1-2-1

24

-8-4

80

x3 – x2 + 4x – 4

So far we have (x – 1)(x + 2)(x3 – x2 + 4x – 4)

NOW FACTOR

x2(x – 1) + 4(x – 1) (x – 1)(x2 + 4) x = 1 and x = ±2iFinally!Factors: (x – 1)(x + 2)(x + 1)( x + 2i)(x – 2i)Zeros: x = 1, -2, -1, 2i, -2i

(x – 1)(x + 2)(x3 – x2 + 4x – 4)

*Complex Zeros occur in Conjugate Pairs! (a + bi)(a – bi)